Sample Chem(5)

46

Objective MHT-CET Chemistry

2

Solutions and Colligative Properties

2.1

Introduction

2.7

Lowering of Vapour Pressure

2.2

Types of Solutions

2.8

Boiling Point Elevation

2.3

Concentration of Solutions of Solids in Liquids

2.9

Freezing Point Depression

2.4

Solubility of Gases in Liquids

2.10 Osmosis and Osmotic Pressure

2.5

Solid Solutions

2.11 Abnormal Molecular Masses

2.6

Colligative Properties

2.12 van't Hoff Factor

2.1

Introduction

•

wo-third wo-third of the earth is covered with water, the most essential, vital component for life. Water in the sea is not in pure form. It contains sodium chloride, magnesium chloride, calcium salts and gases like oxygen, carbon dioxide in dissolved state. Tus, sea water is a ‘solution’. Solution consists of at least two components, a medium and dissolved phase. Te physical properties of solution soluti on are mainly originated by intermolecular forces of attraction between the solvent molecules. Tese intermolecular forces of attraction are changed by the presence of dissolved substances. Te physical properties include vapour pressure, boiling point, freezing point and osmotic pressure.

Solutions Solutions are mixtures of two or more components. Depending on sizes of the components, the mixtures are classified into three types : A coarse mixture : It is formed when the sizes of the constituent components are relatively bigger, e.g .;.; mixture of salt and sugar. A colloidal dispersion : It is formed when the sizes of the particles dispersed in solvent are in the range of 10–7 cm to 10–4 cm. Colloidal particles carry positive or negative charge which stabilizes colloidal dispersion e.g .; .; ferric hydroxide sol, arsenic sulphide sol, etc. Colloidal solutions are heterogeneous and can be easily separated. •

•

•

•

•

A true solution : A true solution is formed when soluble substances are dissolved in the solvent. Te sizes of the particles dissolved are very small of the order of 10–8 cm. rue solutions are homogeneous and cannot be separated into components by simple mechanical methods. f A true solution is defined as a homogeneous mixture of two or more substances, subst ances, the composition of which is not fixed and may be varied within certain limits. f A solution soluti on is formed by two components, solvent and one or more solutes. Te component of the solution which constitutes larger part of the solution solut ion is called solvent and the other component that constitutes smaller part is called solute . Homogeneous solution : Solution is homogeneous if its composition is uniform throughout the body of the solution. Heterogeneous solution : Solution is heterogeneous when two or more phases are present in it. Solvation : Te process of interaction of solvent molecules with solute particles to form aggregates. When water is a solvent, the process of solvation is called hydration or aquation. f Te extent to which solute dissolves d issolves in solvent sol vent to form homogeneous solution depends on nature of solute and solvent. Te general rule is, like dissolves like i.e .,., polar f

46


2


2.1

Introduction

2.7


2.2

Types of Solutions

2.8


2.3

Concentration of Solutions of Solids in Liquids

2.9


2.4


2.10 Osmosis and Osmotic Pressure

2.5

Solid Solutions

2.11 Abnormal Molecular Masses

2.6


2.12 van't Hoff Factor

2.1

Introduction

•

wo-third wo-third of the earth is covered with water, the most essential, vital component for life. Water in the sea is not in pure form. It contains sodium chloride, magnesium chloride, calcium salts and gases like oxygen, carbon dioxide in dissolved state. Tus, sea water is a ‘solution’. Solution consists of at least two components, a medium and dissolved phase. Te physical properties of solution soluti on are mainly originated by intermolecular forces of attraction between the solvent molecules. Tese intermolecular forces of attraction are changed by the presence of dissolved substances. Te physical properties include vapour pressure, boiling point, freezing point and osmotic pressure.

Solutions Solutions are mixtures of two or more components. Depending on sizes of the components, the mixtures are classified into three types : A coarse mixture : It is formed when the sizes of the constituent components are relatively bigger, e.g .;.; mixture of salt and sugar. A colloidal dispersion : It is formed when the sizes of the particles dispersed in solvent are in the range of 10–7 cm to 10–4 cm. Colloidal particles carry positive or negative charge which stabilizes colloidal dispersion e.g .; .; ferric hydroxide sol, arsenic sulphide sol, etc. Colloidal solutions are heterogeneous and can be easily separated. •

•

•

•

•

A true solution : A true solution is formed when soluble substances are dissolved in the solvent. Te sizes of the particles dissolved are very small of the order of 10–8 cm. rue solutions are homogeneous and cannot be separated into components by simple mechanical methods. f A true solution is defined as a homogeneous mixture of two or more substances, subst ances, the composition of which is not fixed and may be varied within certain limits. f A solution soluti on is formed by two components, solvent and one or more solutes. Te component of the solution which constitutes larger part of the solution solut ion is called solvent and the other component that constitutes smaller part is called solute . Homogeneous solution : Solution is homogeneous if its composition is uniform throughout the body of the solution. Heterogeneous solution : Solution is heterogeneous when two or more phases are present in it. Solvation : Te process of interaction of solvent molecules with solute particles to form aggregates. When water is a solvent, the process of solvation is called hydration or aquation. f Te extent to which solute dissolves d issolves in solvent sol vent to form homogeneous solution depends on nature of solute and solvent. Te general rule is, like dissolves like i.e .,., polar f

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•

•

•

•

solutes are soluble in polar solvents, (e.g.; NaCl in water) while non-polar solutes are soluble in nonpolar solvents (e.g.; iodine in CCl4). Water Water is called universal solvent , as it is polar and has very high dielectric constant, hence, dissolves dissol ves most of the polar solutes. Solutions containing two, three or four components are called binary , ternary and quaternary solutions respectively. Solutions prepared in water are called aqueous solutions and solutions in other solvents are called non-aqueous solutions . As the solute particles are very ver y small, smal l, the components of true solutions cannot be separated by simple physical methods like centrifugation, filtration, etc.

2.2

Types of Solutions

Liquid Liqui quid Liquid Gas Gas Gas 2.3

•

•

Solvent Solute Examples Solid Solid Allo lloys like bras rass, bronze, ze, copper in in gold etc. Soli Solid d Liqui iquid d Amal Amalga gams ms of mer mercu cury ry with with me meta tals ls

Name

Solid

Symbol

•

Formula

Gas

Hydrogen gas in palladium metal, pumice stone Solid Iodine in CCl4, benzoic acid in C6H6, sugar in water Liquid uid Ethanol in water Gas Oxy Oxygen, carbon dioxide in water Solid Iodine in air Liquid Chloroform in nitrogen Gas Air, mixtures of of non-reacting gases

Concentration Concentration of Solutions of Solids in Liquids

Te concentration concentration of a solution is defined as the amount of solute dissolved in a specific amount of solvent. Solutions containing relatively less amount of solute are called dilute solutions and and if it contains relatively more amount of solute then the solution is called concentrated solution. Concentration of solutions may be expressed in different ways as discussed below :

Definition

Percentage %(w/W ) by mass

Mass of solute × 100 Total mass of solution

Volume percentage

%(v /V )

Mass by volume percentage Strength

% (w/V )

Volume of solute in mL dissolved in 100 mL of the solution. Amount of solute in grams Mass of solute × 100 dissolved in 100 mL of the Total volume of solution in mL solution. Amount of solute in grams Mass of solute in grams present in one litre (or dm3) of 3 Volum Volumee of solut solutio ion n in L (or (or dm ) solution. Number of parts of solute Mass or volume of solute × 1006 present in million (106) parts Total mass or volume of solution of solution.

g/L (o (or g/dm3)

Parts per million

ppm

Molarity

M

Volume of solute × 100 Total volume of solution

Moles of solute Volume Volume of solution in L (or dm 3 )

Molality

m

Mole fraction

x

Amount of solute in grams present in 100 g of solution.

Moles of solute Mass of solvent in kg x A

=

n A

n A + nB

Effect of temperature No effect Changes with change of temperature. Changes with change of temperature. Changes with change of temperature. No effect

Number of moles of solute dissolved in one litre (or one dm3) of solution.

Changes with change of temperature.

Number of moles of solute dissolved in 1 kg of the solvent. Ratio of number of moles of one component to the total number of moles of all the components.

No effect

No effect

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Illustration

: 1.23 g of sodium hydroxide (molar mass = 40) are dissolved in water and the solution is made to 100 cm3. Calculate the molarity of the solution.

Soln.: Amount of NaOH = 1.23 g Volume of solution = 100 cm3 Moles of NaOH =

Mass of NaOH 1.23 = = 0.0307 Molar mass of NaOH 40

Moles of NaOH × 1000 Volume of solution 0.0307 0307 = × 1000 = 0.31 31 M 100

•

Molarity =

•

Illustration

: 1.8 g of glucose (molar mass = 180) are dissolved in 60 g of water. Calculate (a) the molality (b) mole fraction of glucose and water.

Soln.: (a) Molality of solution =

Moles of solute × 1000 Mass of solvent in g

Moles of glucose × 1000 Mass of water 1.8 × 1000 = = 0.167 m 180 × 60 (b) Mole fraction

\ Molality =

Moles of solute = Moles of solute + Moles of solvent 1.8 = 0.01 ; 180 60 Moles of water = = 3.33 18 0.01 Mole fraction of glucose = = 0.003 ; 0.01 01 + 3.33 33

Moles of glucose =

Mole fraction of water =

3.33 = 0.997 0.01 01 + 3.33 33

Illustration

: 4 g of sodium chloride was dissolved in 300 g of water. Calculate percentage by mass of sodium chloride in solution. Soln.: Percentage by mass of sodium chloride (w/W )

= = •

mass of sodium chloride × 100 mass of sodium chloride + mass of w water 4g 4 g + 300 g

× 100 =

400 = 1.316% by mass 304

Depending upon the quantity of solute dissolved in a liquid solvent, solutions can be of three types : Saturated solution : A solution which cannot f dissolve any further amount of solute at a given temperature is called saturated solution.

Unsaturated solution : A solution in which more amount of the solute can be dissolved at a given temperature is called unsaturated solution. Supersaturated solution : A solution in which f amount of solute is more than it can dissolve at a particular temperature is called supersaturated solution. Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. Factors affecting solubility of a solid in a liquid : f Effect of temperature : − If the dissolution process is endothermic (DsolH > 0), the solubility increases solubility increases with rise in temperature. − If dissolution process is exothermic (DsolH < < 0) the solubility decreases solubility decreases with rise in temperature. Effect of pressure : Pressure does not have any f significant effect on solubility of solids in liquids as these are highly incompressible. f

2.4


Gases are soluble in liquids including water. Te solubility of gases like O2, N2, etc. are much low. O2 molecules being non-polar, have less solubility in polar solvent, water. CO2 and NH3 gases are more soluble in water as CO2 reacts with water to form carbonic acid and NH3 reacts with water to form ammonium hydroxide. HCl gas is polar, its solubility is very high in water, forming hydrochloric acid. Factors affecting solubility of a gas in a liquid : f Effect of pressure : Te solubility of gas increases with the increase in external pressure. Henry’s law - It states that the solubility of a gas in a liquid at constant temperature is proportional to the pressure of the gas above the solution. = K ⋅P S ∝ P i.e. P i.e., S = where S is the solubility of the gas in mol dm–3, is the pressure of the gas in atm, K is is constant of P is proportionality and has the unit of mol dm–3 atm–1. If P = = 1 atm, then S = = K . Hence, Henry’s constant K is defined as solubility of gas in mol dm–3 at 1 atmospheric pressure at reference temperature. If several gases are present then the solubility of any gas may be evaluated by using P as as partial pressure of that gas in the mixture. Effect of temperature : According to Charles’ f law, volume of a given mass of a gas increases with increase of temperature. Terefore, volume •

49


of a given mass of dissolved gas in solution also increases with increase of temperature, so that it becomes impossible for the solvent in solution to accommodate gaseous solute in it and gas bubbles out. Hence, solubility of gas in liquid decreases with increase of temperature. Effect of addition of soluble salt : Solubility of dissolved gas is suppressed when a soluble salt is added to the solution of gas.

f

Spiegeleisen (5-20% Mn in Iron ) and Ferromanganeous (70 - 80% Mn + 30 - 20% Fe) : −

Manganin (84% Cu + 12% Mn + 4% Ni) : − −

–4

–3

Soln.: (i) S = 6.8 × 10 mol dm , P N2 = 1 atm S = K ⋅P N2

6.8 × 10–4 mol dm–3 = K × 1 atm K = 6.8 × 10–4 mol dm–3 atm–1

(ii) P = 0.78 atm, S = ? S = 6.8 × 10–4 mol dm–3 atm–1 × 0.78 atm

= 5.304 × 10–4 mol dm–3 i.e. Solubility of N2 is reduced to

−

2.6

•

•

• • •

2.7 •

A solid solution of two or more metals or of a metal or metals with one or more non-metals is called an alloy or solid solution. All the properties of the pure metals are improved when they form solid solutions, i.e ., alloys. Duralumin (Al + Cu + Mg + Mn) : Light and strong as steel. Used in the construction of aircrafts.

Aluminium bronze (Al + Cu + Mn) Lead alloy (Pb + 10 - 20% Sb) : − Acid resistant. − Used for bearings, bullets, shrapnel and for manufacturing lead storage battery plates.

Babbitt metal (Sb + Sn + Cu) : − Antifriction alloy. − Used in machine bearings.

Stainless steel (Steel + Cr + Ni) : − −

Resistant to corrosion. Used in cutlery.


•

Solid Solutions

− −

Used for extracting metals from ores.

Tese are the properties that depend on the number of solute particles in solution and not on the nature of the solute particles. Tese are : Lowering of vapour pressure, Elevation of boiling point of solvent in solution, Depression of freezing point of solvent in solution and Osmotic pressure.

= 5.304 × 10–4 mol dm–3 2.5

Has almost zero temperature coefficient of electrical resistance. Used for making electrical measurements.

Amalgams (Hg + metals) :

Illustration

: Te solubility of nitrogen gas at 1 atm pressure at 25°C is 6.8 × 10–4 mol dm–3. Calculate the solubility of N2 gas from atmosphere at 25°C if atmospheric pressure is 1 atmosphere and partial pressure of N2 gas at this temperature and pressure is 0.78 atm.

Used for making very hard steels and to manufacture rails, safes and heavy machinery.

•


Vapour pressure of liquids : f Te vapour pressure of a substance is defined as the pressure exerted by the gaseous state of that substance when it is in equilibrium with the solid or liquid phase. Vapour pressure of a liquid, increases with the f increase of temperature. Te boiling point of a liquid is a temperature at f which vapour pressure of liquid becomes equal to external pressure. If the boiling is carried out in an open atmosphere then external pressure is the atmospheric pressure. Vapour pressure lowering : f Te vapour pressure of a liquid solvent is lowered when a non-volatile solute is dissolved in it to form a solution. Tis is due to the fact that in case of pure solvent, f its surface area is completely occupied by volatile solvent molecules. While in case of solution of non volatile solute, its surface area is not completely available for volatile solvent; partly it is occupied by non-volatile solute. f Hence, rate of evaporation of the solution will be less as compared to that of pure solvent and vapour pressure of solution is lower than that of the pure solvent. If p 1° is the vapour pressure of pure solvent and p is f the vapour pressure of the solution of non-volatile

50


f

f

solute in the same solvent, then p < p 1° and the lowering of vapour pressure is, D p = p 1° – p Te difference between vapour pressure of pure solvent and the vapour pressure of solvent from solution is called vapour pressure lowering . Te ratio of vapour pressure lowering of solvent from solution to the vapour pressure of pure solvent is called the relative lowering of vapour pressure .

solution. Tus the lowering of vapour pressure depends on nature of pure solvent and concentration of solute in mole fraction. Now the relative lowering of vapour pressure is given by, D p p1° − p p1° x 2 = = = x 2 p1° p1° p1° Hence, relative lowering of vapour pressure = x 2 Equation (2) proves that the lowering of vapour pressure is a colligative property because it depends on the concentration of non-volatile solute.

D p p1° − p = p1° p1° Raoult’s law Te law states that, the partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution. p 1 = p 1° x 1 and p 2 = p° 2x 2 where p 1° and p 2° are vapour pressures of pure components 1 and 2 respectively, at the same temperature. otal vapour pressure, p total of solutions of two volatile components is the sum of partial vapour pressures of the two components, p total = p 1 + p 2 = x 1 p 1° + x 2 p 2° = (x 1) p 1° + (1 – x 1) p 2° = p 2° + ( p 1° – p 2°)x 1 Te solution which obeys Raoult’s law over the entire range of concentration is called an ideal solution. If a solution does not obey Raoult’s law, the solution is nonideal . If y 1 and y 2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, p 1 = y 1 p total and p 2 = y 2 p total •

Molar mass of solute and relative lowering of vapour pressure •

For dilute solutions n1 >> n2

\

•

•

Raoult’s law for a solution of non-volatile solute Non-volatile solute does not evaporate and does not contribute to the total vapour pressure of solution. Te vapour pressure of pure component A 1 is p 1° and that of component A 2 is p 2° = 0. p = p 2° + ( p 1° – p 2°)x 1 (as p 2° = 0) p = 0 + ( p 1° – 0)x 1 …(1) i.e., p = p 1°x 1 Te lowering of vapour pressure D p is given by, D p = p 1° – p = p 1° – p 1°x 1 = p 1°(1 – x 1) But 1 – x 1 = x 2 Hence, D p = p 1°x 2 …(2) Lowering of vapour pressure is the product of vapour pressure of pure solvent and mole fraction of non volatile solute dissolved in volatile solvent to form a •

•

•

n W2 / M 2 D p p1° − p = = x 2 = 2 = n1 + n2 W1 / M 1 + W2 / M2 p1° p1°

D p n2 W2 / M 2 W2 M 1 = = =

p1°

n1

W1 / M 1

W1M 2

Knowing the masses of non-volatile solute and the solvent in dilute solutions and by determining experimentally vapour pressure of pure solvent and the solution it is possible to determine molar mass of a non-volatile solute. Illustration

: Calculate the weight of a non-volatile solute (mol. wt. 40), which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. Soln.: Given: p = (80/100) p °, M 2 = 40 W 1 = 114 g, M 1 = 114

We know,

/ ) p° p° − (80100

\

p°

\

=

W 2 40

×

114 114

W 2 = 18 g (weight of solute required)

2.8 •

p° − p W 2 M 1 = × p° M 2 W 1


Boiling point is defined as the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure. It is a characteristic property of liquids and is a f criterion to check the purity of liquid. f f

It increases with increase in external pressure. Liquids having greater intermolecular forces have high boiling points.

51

Solutions and Colligative Properties •

Elevation of boiling point : Solution has lower vapour pressure and hence higher boiling point than pure solvent. Te increase in the boiling point, DT b = T b – T b° is known as elevation of boiling point .

f

For dilute solutions, DT b ∝ m

f

 W × 1000  or DT b = K b m = K b  B  M B × W A   1000 × WB × K b   DTb × W A 

or M B = 

where m is molality of solution and K b is called boiling point elevation constant or molal elevation constant or ebullioscopic constant , having unit K kg mol–1. Illustration : Te

molal elevation constant for water is –1 0.51 K kg mol . Calculate the b. pt. of solution made by dissolving 6 g urea in 200 g water.

•

Depression in freezing point and molar mass of the solute : K f × W B × 1000  W × 1000  or DT f = K f  B M = B DT f × W A  M B × W A 

Illustration : Te freezing point of a solution containing

50 cm3 of ethylene glycol in 50 g water is found to be –34°C. Assuming ideal behaviour, calculate the density of ethylene glycol. (K f for H2O = 1.86 K kg mol–1) Soln.: Given: V ethylene glycol = 50 cm3, W 1 = 50 g K f = 1.86 K kg mol–1, M 2 = 62 (glycol) (H2O) DT f = 34°C, W 2 = 50 × d (glycol) 1000 × K f × W 2 We know, DT f = M2 × W 1 Substituting values, 34 =

\

d = 1.133 g/cm 3

Soln.: Given: W 2 = 6 g, M 2 = 60 (urea), W 1 = 200 g,

2.10 Osmosis

K b = 0.51 K kg mol –1

•

We know,

DT b =

1000 × K b × W 2 M2 × W 1

1000 × 0.51 × 6 Substituting values, DT b = = 0.255°C 60 × 200

•

As elevation in b. pt. = 0.255°C

\

B. pt. of solution = 100 + 0.255 = 100.255°C

2.9 •

•


Freezing point of a liquid is the temperature at which the vapour pressure of solid is equal to the vapour pressure of liquid.

Depression of freezing point : Te lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent. Te decrease in freezing point, DT f = T f ° – T f is known as depression in freezing point .

Freezing point depression and vapour pressure lowering : For dilute solutions, DT f ∝ p 1° – p DT f ∝ m or DT f = K f m where K f is known as freezing point depression constant or molal depression constant or cryoscopic constant , having unit K kg mol–1. •

1000 × 1.86 × (50 × d ) 62 × 50

and Osmotic Pressure

Osmosis : Te net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane. Osmotic pressure : Te minimum excess pressure that has to be applied on the solution to prevent the entr y of the solvent into the solution through the semipermeable membrane. wo or more solutions having same osmotic f pressure at a given temperature are called isotonic solutions . If one solution is of lower osmotic pressure, f it is called hypotonic with respect to the more concentrated solution. Te more concentrated solution is said to be hypertonic with respect to the dilute solution. If a pressure higher than the osmotic pressure f is applied on the solution, the solvent will flow from the solution into the pure solvent through the semipermeable membrane and the process is called reverse osmosis . It is used in desalination of sea water .

Laws of osmotic pressure van’t Hoff–Boyle’s law : It states that, at constant temperature the osmotic pressure (p) of a dilute solution is directly proportional to its molar concentration or inversely proportional to the volume of the solution. At constant T , p ∝ C (in mol L–1) •

52


If n moles of solute is dissolved in V litres then, n C = V 1 n \ p ∝ or p ∝ for constant n V V

\ pV = constant or •

p

C

= constant

van’t Hoff-Charles’ law :It states that, the concentration remaining constant, the osmotic pressure of a dilute solution is directly proportional to the absolute temperature. At constant C , p ∝ T i.e.,

•

•

•

•

p = constant

T van’t Hoff general solution equation : p = CRT R = 8.314 J K –1 mol–1 if p is in N m–2 and V in m3 R = 0.082 L atm K –1 mol–1 if p is in atm and V in dm3

Determination of molar mass from osmotic pressure : n W RT W RT , M 2 = 2 p = 2 RT , p = 2 V M 2 V p V van’t Hoff–Avogadro’s law : It states that, two solutions of equal concentrations of different solutes exert same osmotic pressure at the same temperature. It can also be, stated as, equal volumes of isotonic solutions contain an equal number of solute particles at the given temperature. For a given solution pV = nRT …(i) p1V 1 = n1RT 1 for solution 1 …(ii) p2V 2 = n2RT 2 for solution 2 If p1 = p2, T 1 = T 2 and V 1 = V 2 then from equations (i) and (ii), n1 = n2 Since, number of moles are equal, number of molecules are also equal. Hence, osmotic pressure and temperature remaining the same, equal volumes of solutions would contain equal number of moles of the solute. For isotonic solutions of equal volume, n1 = n2

2.11 Abnormal •

multiple number of ions/particles. Tis results in increase of number of solute particles in solution which in turn results in increase of colligative properties and increase of DT f /m value and the value of DT f /m is approximately equal to integral multiple of K f value. Te value of integral is equal to total number of ions produced on dissociation as shown : (i) HCl → H+ + Cl–; 2 particles; DT f /m = 2 × 1.86 K mol–1 kg (ii) NH4Cl → NH4+ + Cl–; 2 particles; DT f /m = 2 × 1.86 K mol–1 kg (iii) CoCl2 → Co2+ + 2Cl–; 3 particles; DT f /m = 3 × 1.86 K mol–1 kg (iv) K 2SO4 → 2K + + SO42–; 3 particles; DT f /m = 3 × 1.86 K mol–1 kg (v) AlCl3 → Al3+ + 3Cl–; 4 particles; DT f /m = 4 × 1.86 K mol–1 kg DT f /m = K f value observed in case of solutions of electrolytes may not be exactly two fold, three fold etc of theoretical K f value observed in case of solutions of non-electrolyte solute. Te value fluctuates with degree of dissociation of solute in solution.

Molecular Masses

For substances undergoing association or dissociation in solution, the molecular mass determined by studying any of the colligative properties is different than the theoretically expected value, and the substance is said to show abnormal molecular mass . Dissociation of electrolyte solutes : Electrolytic solutes when dissolved in solvent dissociate to produce

•

Association of solutes : In some non-polar solvents, two or more molecules of solute associate to form bigger molecules. For example, in benzene solutes like acetic acid, benzoic acid, etc. associate to form dimers. Tis association is due to the hydrogen bonding between these molecules. 2CH3COOH = (CH3COOH)2 2C6H5COOH = (C6H5COOH)2 Hence, numbers of solute particles are reduced to almost half. Observed molecular masses of these species are almost twice the expected values in dilute solutions. Due to association the total number of molecules in solution will be almost half of the number of molecules of the substance dissolved. Hence such solutions show abnormally low colligative properties. Te observed molecular masses are almost double.

2.12 van’t Hoff Factor •

Te colligative properties of electrolytes may be expressed in relation to colligative properties of nonelectrolytes by using van’t Hoff factor i . It is defined as the ratio of observed colligative property produced by a given concentration of electrolyte solution to the property observed for the same concentration of nonelectrolyte solution.

53

Solutions and Colligative Properties i =

Observed value of the colligative property Theoretical vallue of the colligative property

or i = Theoretical molecular mass = M th Observed molecular mass

M o

Total number of moles of particles i =

after association / dissociation

,

Number of moles of particles before association / dissociation

for association, i < 1; for dissociation, i > 1 •

van’t Hoff factor (i ) and degree of dissociation ( a) : i −1 ; a= n −1 M (theoretical) = [1 + (n − 1) a] i= M (observed)

a= •

•

M (theoretical) − M (observed) M (observed) (n − 1)

Illustration

: A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate the osmotic pressure of the solution ( R = 8.314 JK –1 mol–1). Soln.: Initial number of moles of K 4[Fe(CN)6] = 0.1, (as the solution is decimolar). a = degree of dissociation = 50% = 1/2 Tus, we may represent the facts as below : K 4[Fe(CN)6] → 4K + + [Fe(CN)6]4– total Initial no. 0.1 of moles After 0.1(1 – a) dissociation

0

0.1 (4a)

0

0.1

0.1 (a) 0.1(1 + 4a) = 0.1 [1 + (4 × 1/2)] = 0.3

van’t Hoff factor, Number of particles after dissociation Actual number of partticles before dissociation 0.3 = 0.1 pV = inRT

i =

=3

van’t Hoff factor (i ) and degree of association (a) : 1−i a= 1 −1 / n

p = inRT = iCRT (C = n/V = molarity = 0.1 mol L–1)

Modified equations for colligative properties : p1° − p1 n = i ⋅ 2 , DT b = iK b m, DT f = iK f m, p = i n2RT /V p1° n1

or,

V 0.1 mol or = 100 mol m–3 3 3 − 10 m

p = 3 × 100 × 8.314 × 300 = 74.826 × 10 4 = 7.4826 × 105 Nm–2 = 7.4826 atm

54


, t e e h d f d s l e e a t h g l e o n n l e b m u l a u n a a . e v o a r o l c b r t e e r c d e f o h i o r s l s s m t u t r u o m s s f n n i e e s o m e e i n o o p h i e e t m l o e t v t i r w n n o a t p n r a o m a e e s t n a l c e r t t r o f v l s i f n t n l b t b n e o c e a o o e e h o s e w n c m r n g m p f e u s o m l n o e u s o c o s r o m h s f t o d c e e w e r r p o r e r e e r h h l o i l p s e l p t l T f m n w h b o a f t o l g a l e o i s n c i e o t s t : a s h o i n l u f m s a i t e o a o f s h o o r i s u e t e t n . i s n p s a n l s c o i i c h e n r x e s o o o s o i r g h b i i m o e i t t t t u e d m m u u u o e o o s l l m s i e n r t s h l t n o n m t i s o s a o T o s i O u h

x

K K

K K x

S E I T R E P O R P E V I T A G I L L O C D N A S N O I T U L O S

n i h t i w

i

s g s n a n n a n n i o o h h i t i t v t a u e t u e h l r l s o u o r u s n S s S s s o i e t : r : e r u l . s p s p o e c n i c i o t S r n o t i : u i o o t s s s t u m u m n e l l s o r o s o o o S i t p S r c r u i e e c h l o t i c w o g n i S i o r . o h r . n l m c t s o i t e r g e h n o o g h e n t n t o e p i o p i o t e e o m y v y v a a h s a h I s H h t H h t

y y t r t : r s e e s s n s p p a o o a i r o t m r p a p e r m i a r e i v l a c v l u t i a c u o s t s a i e c g l e A g l l i l o o d l l o m n o m a c c f f o o n n f f o i / o o e e e i u 1 t e l 1  1 u a : l u u l a l   1  i r a a v a c i v n v 1 o v l l t a d o s c d a   c s e c i v i a e i ) ) t f v t n n D i e f e o o r r i e f r f r e t t s o i a o s o o a i e b e h c c e H b h o o e O s s t O T T r i s s ’ a g d (

n a v

e t e a o h h l t t t a d t i n e a u i o h q t v o t i l r o b a P s  e a p o t r s a a K t p n s i i s g = e S t s I a e r h . , t u e g : t i . r f a a o P e w f p a e  r L o m u S e t s y . s n s t i t ’ n e l o y a r i r i b t s p t n u u n e l o e l s c h t o s H o

A B x  x  ° A ° B p p = = B A p p

 i



e D

(





e v i t a g i l l o C f o s n o i t a u 2 1 q n n E i :  s V d i m / e 1 f T e t b m p i K K f r R i i i e 2 1  = n d p  p = b f i 1  o o r T T =    M P p

, : e n m o i u t l a o u v t q l n : e : a a t w u e t n n t s : n q r a o a a n l : t o w t i e u t s s s f s a n u ’ s w n c l o o e l a o = ’ o l r s r c ) s c o s d n s e = l P ’ = 1 l o a i a e c l  t r g i r L a  T e u t y V o l o o  v o n , l o h , e s B m g A - T - C - i c t  m C t f s f f f n n f f f n o a f n O o a i o T o t t t o . f ( o H s H s H R H o 2 C n n s ’ t o t t t s n ’ o C ’ i w n c c ’ n = n r = t a a t r n a a o 1 a L v A o v A v  v F n

0 0 1 0 0 1



n

= i o n t o e l u i t t o u l u s : l e o s o s f g f a o f o t o s n s s % e a ) s c a m r W e / M l P ( a s w t

s a M

o T

0 0 1



n o i t = u e l n t : i o l u o e t s g l u o s f a o t o s f e n f o e o e m c r m l u e % u o ) l P V e / o v v V l m ( a u t l o o T V



L m : n i e g n a = t o i e n t n t e o u u c i l l t r o u o e l s s p o f f s o o e f m o s s e u a m l o % ) u v V M l y / o b W v ( s l s a a t o M T

s e l o M

s e l o M

) 3 m s d r m ( o a r g L n i n i e n t o u i l t o u s l f o s o f s s o a e M m u l o V

= h t g n e r t S

55


Multiple Choice Questions LEVEL - 1 2.1 1.

2.

3.

Introduction Te mixture o salt and sugar is called a (a) coarse mixture (b) homogeneous mixture (c) racemic mixture (d) solution mixture. In a solution, the larger proportion o the component is known as (a) solution (b) solute (c) solvent (d) mixed solution. A solution is defined as a (a) homogeneous mixture o two or more substances (b) heterogeneous mixture o two or more substances (c) homogeneous mixture o liquid and solid components only (d) homogeneous mixture consisting water as one o the components.

4.

Te substances involved in solution are called (a) mole (b) atoms (c) molecules (d) components.

5.

Te scientist who won the first Noble prize in chemistry in the year 1901, or his work in solutions was (a) van’t Hoff (b) Beckmann (c) Raoult (d) Jean Nollet.

6.

7.

8.

In solution, the molecular size o particles is o the order o (a) 10–8 cm (b) 10–9 m (c) > 10–9 m (d) 10–9 cm Te new method to determine alcohol content o a wine was discovered in 1870 by (a) Newton (b) Joule (c) Raoult (d) C.A. Wurtz. A solution having three components is called a (a) quaternary solution (b) binary solution (c) single solution (d) ternary solution.

2.2

Types of Solutions

9.

An example o a solution having liquid in gas is (a) moist air (b) dry air (c) Au-Hg (d) C2H5OH + H2O

10.

Which o the ollowing is not a solution? (a) Air (b) A gold ring (c) Smoke (d) Salt solution

11.

Solutions are o ____ types. (a) three (b) six (c) nine (d) eleven

12.

Which o the ollowing possesses physical states o solute and solvent as liquid and solid respectively? (a) Solution o sugar in water (b) Zinc amalgam (c) Solution o naphthalene in benzene (d) Brass

13.

Which o the ollowing pairs o solutions having different physical states o solute? (a) Homogeneous mixture o chloroorm in N2 gas, solution o CO2 in water (b) Brass, homogeneous mixture o camphor in N2 gas (c) Sodium amalgam, moist air (d) Solution o H2 gas in Pd metal, mixture o N2 and O2.

14.

In vaporization o liquids into air, liquid dissolves into air hence liquid is (a) solution (b) solute (c) solvent (d) mixture.

15.

Sugar dissolved in water is a type o (a) solid in solid solution (b) solid in gas solution (c) solid in liquid solution (d) gas in solid solution.

2.3

16.

Concentration of Solutions of Solids in Liquids Dissolution o calcium chloride in water is exothermic process while that o ammonium nitrate is endothermic process, the observation is

56

Objective MHT-CET Chemistry (a)

solubility o calcium chloride decreases with increase in temperature and that o ammonium nitrate increases with increase in temperature (b) solubilities o both decrease with increase in temperature (c) solubilities o both increase with increase in temperature (d) solubility o calcium chloride increases with increase in temperature and that o ammonium nitrate decreases with increase in temperature. 17.

22 cm3 o C2H5OH on dissolving in 200 cm3 o water orms 220 cm3 solution o C2H5OH. What would be the percentage by volume o C 2H5OH in the solution? (a) 22% (b) 11% (c) 01% (d) 10%

18.

Te volume o 2 M NaOH solution is 0.1 dm3. What will be the volume o decimolar NaOH in dm3? (a) 0.1 (b) 0.2 (c) 1.5 (d) 2

19.

A molal solution is one that contains one mole o a solute in (a) 1000 g o the solvent (b) one litre o the solvent (c) one litre o the solution (d) 1000 g o the solution.

(a) 0.196 (c) 0.392 26.

When 0.5 g o solute is present in 10 7 g o a solution, the ppm o solute will be (a) 0.5 (b) 0.05 6 (c) 10 (d) 107

27.

6 g o urea was dissolved in 500 g o water. Te percentage (by mass) o urea in the solution is (a) 0.86% (b) 1.186% (c) 11.86% (d) 0.08%

28.

58 cm3 o ethyl alcohol was dissolved in 400 cm3 o water to orm 454 cm3 o solution o ethyl alcohol. Te percentage by volume o ethyl alcohol in water is (a) 0.12.% (b) 1.2% (c) 12.78% (d) 0.01%

29.

23 g o ethyl alcohol (molar mass 46 g mol–1) is dissolved in 54 g o water (molar mass 18 g mol–1). Te mole raction o ethyl alcohol and water in solution are respectively (a) 0.7 and 0.3 (b) 0.8571 and 0.1429 (c) 0.846 and 0.154 (d) 0.1429 and 0.8571

30.

35% (w/W ) solution o ethylene glycol in water, an antireezer is used in cars as coolant. It lowers reezing point o water to –17.6 °C. Te mole raction o ethylene glycol is (a) 0.1529 (b) 0.1431 (c) 0.1343 (d) 0.1634

31.

A solution o NaOH (molar mass 40 g mol–1) was prepared by dissolving 1.6 g o NaOH in 500 cm 3 o water. Te molarity o solution is (a) 0.2 mol dm–3 (b) 0.02 mol dm–3 (c) 0.08 mol dm–3 (d) 0.8 mol dm–3

32.

11.11 g o urea was dissolved in 100 g o water. Te molality o solution is (N = 14, H = 1, C = 12, O = 16) (a) 18.52 mol kg–1 (b) 1.852 mol kg–1 (c) 18.52 mol g–1 (d) 1.852 mol g–1

33.

34.2 g o sugar was dissolved in water to produce 214.2 g o sugar syrup. Te molality and mole raction o sugar in the syrup are respectively (C = 12, H = 1, O = 16) (a) 0.224 m, 0.002 (b) 0.05 m, 0.02 (c) 0.556 m, 0.0099 (d) 0.05 m, 0.99

34.

Te molarity and molality o sulphuric acid solution o density 1.198 g cm–3 containing 27% by mass o sulphuric acid (molar mass o H2SO4 = 98 g mol–1) are respectively

20. When 10 g o caustic soda is dissolved in

250 cm3 o water, molarity o the solution is (a) 1 M (b) 0.5 M (c) 0.25 M (d) 0.1 M

21.

A solution contains 0.5 mole o a solute in 400 g o water, its molality would be (a) 1.25 m (b) 0.125 m (c) 2.51 m (d) 0.025 m

22. One part o solute in one million parts o solvent is

expressed as (a) ppm (c) grams/litre

(b) milligrams/100 cc (d) grams/100 cc.

23. 36 g o glucose (molar mass = 180 g/mol) is present in

500 g o water, the molarity o the solution is (a) 0.2 M (b) 0.4 M (c) 0.8 M (d) 1.0 M 24.

25.

Te molality o 648 g o pure water is (a) 36 m (b) 55.5 m (c) 3.6 m (d) 5.55 m An aqueous solution contains 25% ethanol and 50% acetic acid by mass. Calculate the mole raction o acetic acid in this solution.

(b) 0.301 (d) 0.503

57


(a) 3.301 M, 3.77 m (c) 0.33 M, 7.3 m 35.

(b) 0.03 M, 0.3 m (d) 33.1 M, 0.7 m

Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. I its density is 1.1 g cm–3, the molarity and mole raction o HCl solution are respectively (a) 1.14 M and 0.768 (b) 0.114 M and 0.232 (c) 11.45 M and 0.232 (d) 11.45 M and 0.768

36. A sample o drinking water was ound to be severely

contaminated with chloroorm which is supposed to be carcinogen. I the level o contamination was 15 ppm (by mass) then the molality o chloroorm in the water sample is (a) 1.26 × 10–2 m (b) 1.26 × 10–4 m (c) 12.6 × 10–4 m (d) 0.12 × 10–4 m 37.

(a) (c) 44.

Sulphuric acid is 95.8% by mass. What is the mole raction and molarity o H2SO4 solution having density 1.91 g cm–3? (H = 1, S = 32, O = 16) (a) 0.0807 and 1.867 M (b) 0.8073 and 18.67 M (c) 0.07 and 7.86 M (d) 18.67 and 0.8073 M

45.

Aqueous solution o NaOH is marked 10% (w/W ). Te density o the solution is 1.070 g cm–3. Te molarity and molality o solution are respectively (Na = 23, H = 1, O = 16) (a) 2.675 M and 2.77 m (b) 0.27 M and 0.26 m (c) 0.26 M and 0.02 m (d) 20.7 M and 20.6 m

46.

Battery acid is 4.22 M aqueous H2SO4 solution, and has density o 1.21 g cm–3. What is the molality o H2SO4? (H = 1, S = 32, O = 16) (a) 5.298 mol/g (b) 52.98 mol/kg (c) 5.298 mol/kg (d) 59.98 mol/kg

47.

A molal solution is one that contains one mole o solute in (a) one litre o the solvent (b) 1000 g o the solvent (c) one litre o the solution (d) 22.4 litres o solution.

48.

Which o the ollowing is independent temperature? (a) Molarity (b) Molality (c) Both (a) and (b) (d) None o these

49.

20 g o NaOH (molar mass = 40 g mol–1) is dissolved in 500 cm3 o water. Molality o resulting solution is (a) 0.1 m (b) 0.5 m (c) 1.5 m (d) 1.0 m

50.

Molarity o solution depends on (a) temperature (b) nature o solute dissolved (c) mass o solvent (d) pressure.

51.

Density o water is 1 g/mL. Te concentration o water in mol/litre is (a) 1000 (b) 18 (c) 0.018 (d) 55.5

52.

Te solutions A and B are 0.1 M and 0.2 M in a substance. I 100 mL o A is mixed with 25 mL o B and there is no change in volume, the final molarity o the solution is (a) 0.15 M (b) 0.18 M (c) 0.30 M (d) 0.12 M

34.2 g o glucose is dissolved in 400 g o water. What is the percentage by mass o glucose solution? (a) 78.7% (b) 7.87% (c) 87.7% (d) 77%

38. A solution is prepared by dissolving certain amount

o solute in 500 g o water. I the percentage by mass o a solute in solution is 2.38 then the mass o solute will be (a) 1.121 g (b) 121.9 g (c) 12.19 g (d) 1219 g 39. 4.6 cm3 o methyl alcohol is dissolved in 25.2 g o

water. Te % by mass o methyl alcohol and mole raction o methyl alcohol are respectively (Given density o methyl alcohol = 0.7952 g cm–3 and C = 12, H = 1, O = 16) (a) 1.26% and 0.755 (b) 12.68% and 0.0755 (c) 12% and 0.0075 (d) 1.26% and 0.0755 40.

12.8 cm3 o benzene is dissolved in 16.8 cm3 o xylene. What is the % by volume o benzene? (a) 24.43% (b) 4.324% (c) 43.24% (d) 0.43%

41. What is the mole raction o HCl in aqueous

solution o HCl containing 24.8% o HCl by mass? (H = 1, Cl = 35.5) (a) 1.399 (b) 0.1399 (c) 0.0139 (d) 0.0013 42. What is the mole raction o solute in its 2 molal

aqueous solution? (a) 0.0347 (c) 0.043 43.

(b) 3.47 (d) 0.074

Te molality and molarity o HNO3 in a solution containing 12.2% HNO3 are respectively (Given density o HNO3 = 1.038 g cm–3, H = 1, N = 14, O = 16)

2.01 m and 2.206 M (b) 1.2 m and 2.6 M 2.206 m and 2.01 M (d) 20.6 m and 20 M

o

58 53.

Objective MHT-CET Chemistry Te molarity o a solution that contains 49 g H3PO4 in 2.0 L o a solution is (a) 0.25 M (b) 0.50 M (c) 0.75 M (d) 1.0 M

63.

A solution o CaCl2 is 0.5 mol/litre, then the moles o chloride ions in 500 mL will be (a) 0.25 (b) 0.50 (c) 0.75 (d) 1.00

54. Te molarity o the solution containing 7.1 g o

64.

Te sum o mole ractions o A, B and C in an aqueous solution containing 0.2 moles o each A, B and C is (a) 0.6 (b) 0.2 (c) 1.0 (d) 3

Na2SO4 in 100 mL o aqueous solution is (a) 1 M (b) 2 M (c) 0.05 M (d) 0.5 M 55.

56.

I 60 cm3 o ethyl alcohol is dissolved in 300 cm3 o water then the percentage by volume o ethyl alcohol is (a) 16.66% (b) 15.76% (c) 17.86% (d) 18.96% What volume o 0.8 M solution contains 0.1 mol o the solute? (a) 62.5 mL (b) 100 mL (c) 500 mL (d) 125 mL

2.4 65.

58.

Te amount o anhydrous Na2CO3 present in 250 mL o 0.25 M solution is (a) 6.225 g (b) 66.25 g (c) 6.0 g (d) 6.625 g

59.

20 mL o HCl solution requires 19.85 mL o 0.01 M NaOH solution or complete neutralization. Te molarity o HCl solution is (a) 0.0099 M (b) 0.099 M (c) 0.99 M (d) 9.9 M

61.

I 1 M and 2.5 litre NaOH solution is mixed with another 0.5 M and 3 litre NaOH solution, then molarity o the resultant solution will be (a) 1.0 M (b) 0.73 M (c) 0.80 M (d) 0.50 M

62. o prepare a solution o concentration o 0.03 g/mL

o AgNO3, what amount o AgNO3 should be added in 60 mL o solution? (a) 1.8 g (b) 0.8 g (c) 0.18 g (d) 0.108 g

S = P /K

(d) S =

KP

66.

I Henry's law constant or oxygen is 1.1 × 10–3 mol dm–3 atm–1 and its partial pressure is 0.20 atm, the concentration o dissolved oxygen at NP will be (a) 2.2 × 10–3 mol dm–3 (b) 2.2 × 10–4 mol dm–3 (c) 1.1 × 10–4 mol dm–3 (d) 0.22 × 10–4 mol dm–3

67.

Te unit o Henry's constant is (a) mol dm–3 atm–1 (b) mol dm–3 atm (c) mol–1 dm–3 atm (d) mol–1 dm–3 atm–1

68.

What is the concentration o dissolved oxygen at 25°C at 1 atmospheric pressure i partial pressure o oxygen is 0.22 atm? Te Henry’s law constant or oxygen is 1.3 × 10–3 mol dm–3 atm–1. (a) 1.3 × 10–3 mol dm–3 (b) 0.13 × 10–3 mol dm–3 (c) 0.28 × 10–4 mol dm–3 (d) 2.86 × 10–4 mol dm–3

69.

What is the Henry’s law constant o dissolved O2 at 10°C at 1 atmospheric pressure, i partial pressure o oxygen is 0.24 atm? Te concentration o dissolved oxygen is 3.12 × 10–4 mol dm–3. (a) 2.5 × 10–3 mol dm–3 atm–1 (b) 1 × 10–4 mol dm–3 atm–1 (c) 1.3 × 10–3 mol dm–3 atm–1 (d) data insufficient.

70.

Solubility o a gas in a liquid increases with (a) increase o pressure and increase o temperature (b) decrease o pressure and increase o temperature (c) increase o pressure and decrease o temperature (d) decrease o pressure and decrease o temperature.

71.

Which law states that the amount o gas dissolved in a given mass o solvent at any temperature is directly proportional to pressure o the gas above t he solution? (a) Raoult’s law (b) Boyle’s law (c) Charles’ law (d) Henry’s law

60. I 5.85 g o NaCl (molecular weight 58.5 g mol–1) is

dissolved in water and the solution is made up to 0.5 litre, the molarity o the solution will be (a) 0.2 M (b) 0.4 M (c) 1.0 M (d) 0.1 M

According to Henry's law (a) S = KP (b) S = K /P (c)

57. 25 mL o 3.0 M HNO3 are mixed with 75 mL o 4.0 M

HNO3. I the volumes are additive, the molarity o the final mixture would be (a) 3.25 M (b) 4.0 M (c) 3.75 M (d) 3.50 M


59


2.5 72.

Solid Solutions

(a) (b) (c) (d) (a) (b) (c) (d)

77.

78.

79.

2.6 80.

83.

Te colligative properties o a solutions depend on (a) nature o solute particles present in it (b) nature o solvent used (c) number o solute particles present in it (d) number o moles o solvent only.

84.

Which o the ollowing is not the colligative property? (a) DT f (b) DT b (c) K b (d) Osmotic pressure

85.

Colligative properties are used or the determination o (a) molar mass (b) equivalent weight (c) arrangement o molecules (d) melting point and boiling point.

86.

Which o the ollowing is a colligative property? (a) Conductance o a solution (b) Surace tension o a solution (c) Osmotic pressure o a solution (d) Radioactivity o a solution

87.

Which is not a colligative property? (a) Reractive index (b) Lowering o vapour pressure (c) Depression o reezing point (d) Elevation o boiling point

bismuth with tin and copper antimony with tin and copper lead with arsenic aluminium with copper and manganese.

74. Amalgam is

76.

Colligative properties are applicable to (a) ideal dilute solutions (b) non-ideal concentrated solutions (c) non-ideal solutions (d) ideal concentrated solutions.

Lead is hardened by the addition o (a) 10-20% aluminium (b) 5-20% manganese (c) 10-20% antimony (d) 20-30% iron

73. Babbitt metal is an alloy o

75.

82.

a solution o gas in gas a solution o liquid in gas a solution o metals in liquid metal a solution o solid in gas.

Stainless steel contains (a) chromium (c) both (a) and (b)

(b) nickel (d) none o these.

An alloys containing zero temperature coefficient o electrical resistance is (a) manganin (b) duralumin (c) bronze (d) spiegeleisen. An alloy is (a) a solution o solid in liquid (b) a solution o solid in gas (c) a solution o solid in plasma (d) a solution o solid in solid. 5 to 20% manganese in iron is (a) erromanganeous (b) manganin (c) duralumin (d) spiegeleisen. An alloy manganin contains (a) 84% Cu, 12% Mn and 4% Ni (b) 70-80% Mn, 30-20% Fe (c) 10-20% Sb, 90-80% Pb (d) Al, Cu, Mg, and Mn.

2.7 88.

Which o the ollowing is incorrect? (a) Relative lowering o vapour pressure is independent o the nature o the solute and the solvent (b) Relative lowering in vapour pressure is a colligative property (c) Vapour pressure o a solution is lower than the vapour pressure o the solvent (d) Relative lowering o vapour pressure is directly proportional to the original pressure.

89.

Which one o the ollowing is not an expression or Raoult’s law or a solution containing two volatile components A and B? p A and pB are the partial ° and p° vapour pressures o A and B, p A B are the vapour pressures o pure A and B, x A and x B are mole ractions o A and B in solution.

Colligative Properties Which o the ollowing is not a colligative property? (a) Osmotic pressure (b) Elevation in boiling point (c) Vapour pressure (d) Depression in reezing point

81. Which o the ollowing is a colligative property?

(a) Osmotic pressure (c) Vapour pressure

(b) Boiling point (d) Freezing point


60

Objective MHT-CET Chemistry (a) p A = p° A x A (c) D p = p°Bx B

90.

(b) P total = p A° x A + p°Bx B (d) D p = p A° x A

97.

Te vapour pressure o a solution containing 13 × 10–3 kg o solute in 0.1 kg o water at 298 K is 27.371 mm Hg. Given that the vapour pressure o water at 298 K is 28.065 mm Hg. Te molar mass o the solute will be (a) 69 g mol–1 (b) 95 g mol–1 (c) 60 g mol–1 (d) 90 g mol–1

98.

A solution is prepared rom 26.2 × 10–3 kg o an unknown substance and 112.0 × 10–3 kg acetone at 313 K. Te vapour pressure o pure acetone at this temperature is 0.526 atm. Te vapour pressure o solution is 273.52 × 10–3 kg mol–1 i the molar mass o substance is (a) 0.400 atm (b) 0.600 atm (c) 0.500 atm (d) 0.300 atm

99.

Te vapour pressure o 2.1% solution o a nonelectrolyte in water at 100°C is 755 mm Hg. What is the molar mass o the solute? (a) 58.69 kg/mol (b) 58.69 g/mol (c) 5.86 kg/mol (d) 586.9 g/mol

Lowering o vapour pressure is a colligative property because, it depends on the concentration o (a) volatile solute (b) non-volatile solute (c) volatile solvent (d) non-volatile solvent.

91. At 25°C, the total pressure o an ideal solution

obtained by mixing 3 moles o A and 2 moles o B, is 184 torr. What is the vapour pressure (in torr) o pure B at the same temperature? (vapour pressure o pure A, at 25°C, is 200 torr.) (a) 180 (b) 160 (c)

16

(d) 100

92. Te relative lowering o vapour pressure is equal to

93.

(a)

p°1 D p

(b)

(c)

D p p°1

(d)

p − p1° p°1 p°1

−p

D p

Te aqueous solution that has the highest value o relative lowering o vapour pressure at a given temperature is (a) 0.1 molal sodium phosphate (b) 0.1 molal barium chloride (c) 0.1 molal sodium chloride (d) 0.1 molal glucose.

94. For dilute solutions, Raoult’s law states that

(a)

lowering o vapour pressure is equal to the mole raction o the solute (b) relative lowering o vapour pressure is equal to the mole raction o the solvent (c) relative lowering o vapour pressure o the solvent is equal to the mole raction o the solute (d) vapour pressure o the solution is equal to the vapour pressure o the solvent. 95. Vapour pressure o dilute aqueous solution o glucose

is 750 mm o mercury at 373 K. Te mole raction o solute is (a) 1/76 (b) 1/7.6 (c) 1/38 (d) 1/10 96.

Vapour pressure o a solution o 5 g o non-electrolyte in 100 g o water at a particular temperature is 2985 N/m2. Te vapour pressure o pure water is 3000 N/m2, the molecular weight o the solute is (a) 90 g mol–1 (b) 200 g mol–1 (c) 180 g mol–1 (d) 380 g mol–1

100. Te vapour pressure o water at 20°C is 17 mm Hg.

What is the vapour pressure o a solution containing 2.8 g o urea (NH2CONH2) in 50 g o water? (N = 14, C = 12, H = 1) (a) 16.71 mm o Hg (b) 1.671 mm o Hg (c) 0.1671 mm o Hg (d) 0.01671 mm o Hg 101. In an experiment, 18.04 g o mannitol was dissolved

in 100 g o water. Te vapour pressure o water was lowered by 0.309 mm Hg rom 17.535 mm Hg. Te molar mass o mannitol is (a) 18.42 g/mol (b) 18.42 kg/mol (c) 1.842 g/mol (d) 184.27 g/mol 102. According to the Raoult’s law, the relative lowering o

vapour pressure is equal to the (a) mole raction o solvent (b) mole raction o solute (c) independent o mole raction o solute (d) molality o solution. 103. Partial pressure o solvent in solution o non-volatile

solute is given by equation, (a) p = x 2 p1° (b) p1° = xp (c) p = x 1 p1° (d) p1° = x 1 p 104. When partial pressure o solvent in solution o non-

volatile solute is plotted against its mole raction, nature o graph is (a) a straight line passing through origin (b) a straight line parallel to mole raction o solvent

61


(c)

a straight line parallel to vapour pressure o solvent (d) a straight line intersecting vapour pressure axis. 105. Relative lowering o vapour pressure o solution is a

(a) (b) (c) (d)

property o solute property o solute as well as solvent property o solvent colligative property.

106. Vapour pressure o solution o a non-volatile solute is

always (a) equal to the vapour pressure o pure solvent (b) higher than vapour pressure o pure solvent (c) lower than vapour pressure o pure solvent (d) constant. 107. Te vapour pressure at equilibrium o a liquid in a

closed vessel depends on (a) pressure (b) concentration (c) temperature (d) volume. 108. Te vapour pressure o water at 300 K in a closed

container is 0.4 atm. I the volume o the container is doubled, its vapour pressure at 300 K will be (a) 0.8 atm (b) 0.2 atm (c) 0.4 atm (d) 0.6 atm 109. At 300 K when a solute is added to a solvent its vapour

pressure over the mercury reduces rom 50 mm to 45 mm. Te value o mole raction o solute will be (a) 0.005 (b) 0.010 (c) 0.100 (d) 0.900 110. I p1° and p are the vapour pressures o a solvent and

its solution respectively, and i x 1 and x 2 are the mole ractions o the solvent and solute respectively then (a) p = p1°x 1 (b) p = p1°x 2 (c) p1° = p x 2 (d) p = p1° (x 1/x 2) 111. Te vapour pressure o pure liquid A is 0.80 atm.

On mixing a non-volatile solute B to A, its vapour pressure becomes 0.6 atm. Te mole raction o B in the solution is (a) 0.150 (b) 0.25 (c) 0.50 (d) 0.75 112. Te vapour pressure o two liquids P and Q are 80

and 60 torr respectively. Te total vapour pressure o solution obtained by mixing 3 moles o P and 2 moles o Q would be (a) 140 torr (b) 20 torr (c) 68 torr (d) 72 torr

113. 60 g o urea (mol. wt. 60) was dissolved in 9.9 moles

o water. I the vapour pressure o pure water is p1° , the vapour pressure o solution is (a) 0.10 p1° (b) 1.10 p1° (c) 0.90 p1° (d) 0.99 p1° 114. Te vapour pressure o a solvent is decreased by

10 mm o mercury when a non-volatile solute was added to the solvent. Te mole raction o the s olute in the solution is 0.2. What should be the mole raction o the solvent, i the decrease in the vapour pressure is 20 mm o mercury? (a) 0.8 (b) 0.6 (c) 0.4 (d) 0.2 115. Te vapour pressure o a pure liquid A is 70 torr at

27°C. It orms an ideal solution with another liquid B. Te mole raction o B is 0.2 and the total vapour pressure o the solution is 84 torr at 27°C. Te vapour pressure o pure liquid B at 27°C is (a) 14 torr (b) 56 torr (c) 70 torr (d) 140 torr 116. At 40°C the vapour pressure o pure liquids, benzene

and toluene, are 160 mm Hg and 60 mm Hg respectively. At the same temperature, the vapour pressure o an equimolar solution o the two liquids, assuming the ideal solution should be (a) 140 mm Hg (b) 110 mm Hg (c) 220 mm Hg (d) 100 mm Hg 117. One mole o sugar is dissolved in two moles o water.

Te vapour pressure o the solution relative to that o pure water is (a) 2/3 (b) 1/3 (c) 3/2 (d) 1/2 118. Te vapour pressure o benzene at 80°C is lowered by

10 mm by dissolving 2 g o a non-volatile substance in 78 g o benzene. Te vapour pressure o pure benzene at 80°C is 750 mm. Te molecular mass o the substance will be (a) 15 g mol–1 (b) 150 g mol–1 (c) 1500 g mol–1 (d) 148 g mol–1 119. Vapour pressure o CCl4 at 25°C is 143 mm o Hg.

0.5 g o a non-volatile solute (mol. wt. = 65) is dissolved in 100 mL CCl4. Find the vapour pressure o the solution. (Density o CCl4 = 1.58 g/cm3) (a) 141.43 mm (b) 94.39 mm (c) 199.34 mm (d) 143.99 mm 120. Te vapour pressure o a solution increases when

(a) (b) (c) (d)

the temperature is raised the volume is increased the number o moles o the solute is increased the temperature is lowered.

62


121. Pressure cooker reduces cooking time or ood

128. A solution is prepared by dissolving 1.9 × 10–2 kg

because (a) heat is more evenly distributed in the cooking space (b) boiling point o water involved in cooking is increased (c) the higher pressure inside the cooker crushed the ood material (d) cooking involves chemical changes helped by a rise in temperature.

ammonia in 400 g o water. Te elevation in boiling point i K b or water is 0.52 K kg mol –1 (N = 14, H = 1, O = 16) is (a) 145°C (b) 1.45 K (c) 14.5 K (d) 45.1 K

122. Te relative lowering o the vapour pressure is equal

to the ratio between the number o (a) solute molecules and solvent molecules (b) solute molecules and the total molecules in the solution (c) solvent molecules and the total molecules in the solution (d) solvent molecules and the total number o ions o the solute. 123. Te vapour pressure lowering caused by the addition

o 100 g o sucrose (molecular mass = 342) to 1000 g o water i the vapour pressure o pure water at 25°C is 23.8 mm Hg is (a) 1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm Hg (d) 00.12 mm Hg

2.8


129. A solution containing 1.21 g o camphor (molar mass

152 g mol–1) in 26.68 g o acetone boils at 329.95 K. I the boiling point o pure acetone is 329.45 K, the molal elevation constant or acetone will be (a) 167 K kg mol–1 (b) 1.67 K mol–1 (c) 16.76 K kg mol–1 (d) 1.676 K kg mol–1 130. 3.795 g o sulphur is dissolved in 100 g o CS2. Tis

solution boils at 319.81 K. What is the molecular ormula o sulphur in solution? Te boiling point o CS2 is 319.45 K. (Given : K b or CS2 = 2.42 K kg mol–1 and atomic mass o S = 32.) (a) S2 (b) S4 (c) S8 (d) None o these. 131. A solution prepared by dissolving certain amount o

compound in 31.8 g o CCl 4 has a boiling point o 0.392 K higher than that o pure CCl4. I the molar mass o compound is 128 g mol –1, the mass o the solute dissolved will be (Given K b or CCl4 = 5.02 K kg mol–1) (a) 317 g (b) 0.317 kg (c) 0.317 g (d) 31.7 g

124. Boiling point o water is 373.11 K. What isK b o water,

132. Boiling point o water at 750 mm Hg is 96.63 °C. How

i 0.15 molal aqueous solution o a substance boils at 373.20 K? (a) 0.09 K kg mol–1 (b) 0.6 K kg mol–1 (c) 6.0 K kg mol–1 (d) 60 K kg mol–1

much sucrose is to be added to 500 g o water such that it boils at 100°C? Molal elevation constant or water is 0.52 K kg mol–1. (a) 11.08 kg (b) 1108.2 g (c) 1108.2 kg (d) 11.08 g

125. K b is the elevation in boiling point produced when

1 mol o solute is dissolved in (a) one litre solvent (b) 1000 g solvent (c) 500 mL solvent (d) 0.5 kg solvent. 126. I 0.15 molal solution o a substance boils at 373.23 K

the molal elevation constant o water is (Given boiling point o water = 373.15 K) (a) 0.35 K kg mol–1 (b) 5.3 K kg mol–1 (c) 0.53 K kg mol–1 (d) 0.053 K kg mol–1 127. A solution o a substance on dissolving in benzene

boils at 353.71 K. I K b or benzene is 2.53 K kg mol –1 and boiling point o pure benzene is 353.35 K, then the molality o the solution is (a) 0.1423 m (b) 0.4123 m (c) 0.04 m (d) 0.004 m

133. A solution containing 0.5126 g o naphthalene

(molar mass = 128.17 g mol–1) in 50.00 g o CCl4 gives a boiling point elevation o 0.402 K. While a solution o 0.6216 g o unknown solute in the same mass o the solvent gives a boiling point elevation o 0.647 K. Te molar mass o the unknown solute is (K b or CCl4 = 5.03 K kg mol–1) (a) 0.096306 kg/mol (b) 96.31 kg/mol (c) 0.009630 kg/mol (d) 0.9631 kg/mol 134. Te mass in grams o an impurity o molar mass

100 g mol–1 which would be required to raise the boiling point o 50 g o chloroorm by 0.30 K (K b or chloroorm = 3.63 K kg mol –1) is (a) 0.4132 g (b) 41.32 g (c) 43.1 g (d) 42.3 g

63

Solutions and Colligative Properties 135. Boiling point o a solvent is 80.2°C. When 0.419 g o –1

the solute o molar mass 252.4 g mol was dissolved in 75 g o the solvent, the boiling point o the solution was ound to be 80.256°C. What is the molal elevation constant ? (a) 25.4 K kg/mol (b) 2.54 K g/mol (c) 25.4 K g/mol (d) 2.54 K kg/mol 136. When NaCl is added to water

(a) (b) (c) (d)

reezing point is raised boiling point is depressed reezing point does not change boiling point is raised.

137. Molal elevation constant is elevation in boiling point

produced by (a) 1 g o solute in 100 g o solvent (b) 100 g o solute in 1000 g o solvent (c) 1 mole o solute in one litre o solvent (d) 1 mole o solute in one kg o solvent. 138. Te determination o molar mass rom elevation in

boiling point is called as (a) cryoscopy (c) ebullioscopy

(b) osmometry (d) spectroscopy.

139. I mass is expressed in gram then K b is given by

(a)

M2 DTb × W 2 1000 × W 1

(b)

W 2 × 1000 DTb × W1 × M 2

(c)

M2 DTb × W 1 1000 × W 2

(d)

W 1 × 1000 DTb × W2 × M 2

100.25°C. Te aqueous solution containing 3 g o glucose in the same volume will boil at (molecular weights o urea and glucose are 60 and 180 respectively) (a) 100.75 °C (b) 100.5 °C (c) 100.25 °C (d) 100 °C 144. I 0.15 g o a solute dissolved in 15 g o solvent is

boiled at a temperature higher by 0.216 °C than that o the pure solvent, the molecular weight o the substance (molal elevation o boiling point constant or the solvent is 2.16 °C) is (a) 1.01 g mol–1 (b) 10 g mol–1 (c) 1000 g mol–1 (d) 100 g mol–1 145. A liquid is in equilibrium with its vapours at its

boiling point. On average, the molecules in the two phases have equal (a) intermolecular orces (b) potential energy (c) total energy (d) kinetic energy. 146. Which o the ollowing aqueous solutions containing

10 g o solute in each case has the highest boiling point? (a) NaCl solution (b) KCl solution (c) Sugar solution (d) Glucose solution 147. Elevation in boiling point was 0.52°C when

140. Unit o boiling point elevation constant (K b) is

(a) kg mol–1 (c) g mol–1

143. An aqueous solution containing 1 g o urea boils at

(b) K mol–1 (d) K kg mol–1

141. Te rise in the boiling point o a solution containing

1.8 g o glucose in 100 g o a solvent is 0.1°C. Te molal elevation constant o the liquid is (a) 0.01 K/m (b) 0.1 K/m (c) 1 K/m (d) 10 K/m 142. Te molal elevation o boiling point constant or

water is 0.513°C kg mol–1. When 0.1 mole o sugar is dissolved in 200 mL o water, the solution boils under a pressure o one atmosphere at (a) 100.513 °C (b) 100.0513 °C (c) 100.256 °C (d) 101.025 °C

6 g o a compound X was dissolved in 100 g o water. Molecular weight o X is (K b or water is 0.52 K kg mol–1). (a) 120 g mol–1 (b) 60 g mol–1 (c) 180 g mol–1 (d) 600 g mol–1 148. At higher altitudes the boiling point o water is lower

because (a) atmospheric pressure is low (b) temperature is low (c) atmospheric pressure is high (d) temperature is high. 149. Te boiling point o C6H6, CH3OH, C6H5NH2

and C6H5NO2 are 80°C, 65°C, 184°C and 212°C respectively. Which will show highest vapour pressure at room temperature? (a) C6H6 (b) CH3OH (c) C6H5NH2 (d) C6H5NO2 150. Te boiling point o 0.1 molal aqueous solution o

urea is 100.18°C at 1 atm. Te molal elevation constant o water (in °C kg mol–1) is (a) 1.8 (b) 0.18 (c) 18 (d) 18.6

64


151. I the solution boils at a temperature T 1 and the

solvent at a temperature T 2, the elevation o boiling point is given by (a) T 1 + T 2 (b) T 1 – T 2 (c) T 2 – T 1 (d) T 1 ÷ T 2 152. Te molal elevation constant is the ratio o the

elevation in boiling point to (a) molarity (b) molality (c) mole raction o solute (d) mole raction o solvent. 153. Elevation o boiling point was 0.52°C, when 6 g o

compound ‘ X ’ was dissolved in 100 g o H2O. Molar mass o ‘ X ’ is (K b o water is 5.2°C per 100 g water) (a) 120 g mol–1 (b) 60 g mol–1 (c) 600 g mol–1 (d) 180 g mol–1 154. Te elevation o boiling point method is used or the

determination o molecular weight o (a) non-volatile and soluble solute (b) non-volatile and insoluble solute (c) volatile and soluble solute (d) volatile and insoluble solute. 155. When a substance is dissolved in a solvent, the vapour

pressure o the solvent is decreased. Tis results in (a) an increase in the boiling point o the solution (b) a decrease in the boiling point o the solvent (c) the solution having a higher reezing point than the solvent (d) the solution having a lower osmotic pressure than the solvent. 156. When 10 g o a non-volatile solute is dissolved in

100 g o benzene, it raises boiling point by 1°C, then molecular mass o the solute is (K b or benzene = 2.53 K kg mol–1) (a) 223 g (b) 233 g (c) 243 g (d) 253 g 157. Te value o K b or a solvent is X K kg mol –1. A 0.2 m

solution o a non-electrolyte in this solvent will boil at (Given : boiling point o solvent = A °C). X   (a) ( A + X ) °C (b)  A +  °C  10   X   X  (c)  A +  °C (d)  A +  K   5  5  158. At certain hill-station pure water boils at 99.725°C.

I K b or water is 0.513 °C kg mol–1, the boiling point o 0.69 m solution o urea will be (a) 100.079°C (b) 103°C (c) 100.359°C (d) Unpredictable

2.9


159. Which o the ollowing solutions shows maximum

depression in reezing point? (a) 0.5 M Li2SO4 (b) 1 M NaCl (c) 0.5 M Al2(SO4)3 (d) 0.5 M BaCl2 160. Te unit o reezing point depression constant is

(a) (c)

K mol–1 K kg mol–1

(b) K kg–1 mol–1 (d) K kg–1

161. Te reezing point o equimolal aqueous solution will

be highest or + (a) C6H5NH3 Cl– (c) glucose

(b) La(NO3)3 (d) Ca(NO3)2

162. A 0.5 molal solution o ethylene glycol in water is used

as coolant in a car. I K f or water is 1.86 K kg mol–1 the mixture will reeze at (a) 0.93°C (b) –0.93°C (c)

1.86°C

(d) –1.86°C

163. 5.08 g o a substance dissolved in 50 g o water lowered

the reezing point by 1.2 K. I the molal depression constant or water is 1.86 K kg mol–1, the molar mass o the substance, will be (a) 157.48 × 10–3 g mol–1 (b) 157.48 g mol–1 (c) 15.7 × 10–3 g mol–1 (d) 15.7 kg mol–1 164. 0.440 × 10–3 kg o a substance (molar mass =

178.9 × 10–3 kg mol–1) dissolved in 22.2 × 10–3 kg o benzene lowered the reezing point o benzene by 0.567 K. Te molal depression constant or benzene is (a) (c)

512 K kg mol–1 51.2 K kg mol–1

(b) 5.12 K kg mol–1 (d) none o these.

165. Te observed depression in the reezing point o water

or a particular solution is 0.087 K. Te molality o the solution i molal depression constant or water is 1.86 K kg mol–1 will be (a) (c)

0.4 m 0.0467 m

(b) 4.67 m (d) 4 m

166. 1.02 g o urea when dissolved in 98.5 g o certain

solvent decreases its reezing point by 0.211 K. 1.60 g o unknown compound when dissolved in 86.0 g o the same solvent depresses the reezing point by 0.34 K. Te molar mass o the unknown compound is (Urea NH2CONH2, N = 14, C = 12, O = 16, H = 1) (a) (c)

60 g mol–1 78 g mol–1

(b) 68 g mol–1 (d) 40 g mol–1

65

Solutions and Colligative Properties 167. A solution o glucose (C6H12O6) was prepared

by dissolving certain amount o glucose in 100 g o water. Te depression in reezing point was 0.0410 K. I molal depression constant or water is 1.86 K kg mol–1, the mass o glucose dissolved will be (C = 12, H = 1, O = 16) (a) 0.396 g (b) 0.396 kg (c) 39.6 g (d) 39.6 kg 168. An aqueous solution containing 12.5 × 10–3 kg o

non-volatile compound in 0.1 kg o water reezes at 272.49 K. Te molar mass o the compound is (K f or water = 1.86 K kg mol–1. (a) 35.27 g mol–1 (b) 352.27 g mol–1 (c) 452.27 g mol–1 (d) 45.27 g mol–1 169. Te reezing point o solution prepared by dissolving

(a) (b) (c) (d)

reducing viscosity reducing specific heat reducing reezing point reducing boiling point.

176. Te molar mass o the solute using depression o

reezing point may be calculated using 10K f W 2 100K f W 2 (a) M 2 = (b) M 2 = DT f m DT f W 1 (c) M 2

=

K f W 1

DT f W 2

(d) M 2 =

1000K f W 2

DT f W 1

177. Te reezing point o 0.05 molal solution o non-

electrolyte in water is (K b = 1.86 °C kg mol–1) (a) –1.86 °C (b) –0.93 °C (c) –0.093 °C (d) 0.93 °C

4.5 g o glucose (Molar mass = 180 g mol–1) in 250 g o bromoorm is (Given : Freezing point o bromoorm = 7.8°C and K f or bromoorm = 14.4 K kg mol–1) (a) 6.36°C (b) 4.23°C (c) 3.66°C (d) 0.06°C

178. What is the molality o solution o a certain solute

170. A temperature at which the vapour pressure o a solid

179. I all the ollowing our compounds were sold at the

is equal to the vapour pressure o liquid is called (a) elevation o boiling point (b) reezing point (c) melting point (d) depression o reezing point 171. Te reezing point o 1 percent solution o lead nitrate

in water will be (a) below 0 °C (c) 1 °C

(b) 0 °C (d) 2 °C

172. How much polystyrene o molecular weight 9000

in a solvent, i there is a reezing point depression o 0.184 °C, and the reezing point constant is 18.4? (a) 0.01 m (b) 1 m (c) 0.001 m (d) 100 m same price, which would be cheapest or preparing an antireeze solution or a car radiator? (a) Methanol (b) Ethanol (c) Ethylene glycol (d) Glycerol 180. Equimolal solutions o A and B show depression

o reezing point in the ratio o 2 : 1. A remains in normal state. Te state o B in the solution is (a) normal (b) dissociated (c) associated (d) unpredictable. 181. A solution containing 6.8 g o a non-ionic solute

would have to be dissolved in 100 grams o C 6H6 to lower its reezing point by 1.05°C? (K f C6H6 = 4.9) (a) 19.3 g (b) 193 g (c) 38.6 g (d) 77.2 g

in 100 g water was ound to reeze at –0.93 °C. Te reezing point depression constant o water is 1.86. Te molar mass o the solute is (a) 13.6 g mol–1 (b) 34 g mol–1 (c) 68 g mol–1 (d) 136 g mol–1

DT f can be written

182. Which o the ollowing aqueous solutions has

173. Relationship between K f , m and

as (a) (c)

DT f = K f /m DT f = K f

(b) (d)

DT f = K f m DT f = m

174. Which will show maximum depression in reezing

point when concentration is 0.1 M? (a) NaCl (b) Urea (c) Glucose (d) K2SO4 175. In cold countries, ethylene glycol is added to water in

the radiators o cars during winter. It results in

minimum reezing point? (a) 0.01 m NaCl (b) 0.005 m C2H5OH (c) 0.005 m MgI2 (d) 0.001 m MgSO4 183. When mercuric iodide is added to the aqueous

solution o potassium iodide, the (a) reezing point is lowered (b) reezing point is raised (c) reezing point does not change (d) boiling point does not change.

66


184. An aqueous solution reezes at 0.186°C. Te boiling

point o the same solution is (K f = 1.86 K m–1, K b = 0.512 K m–1) (a) 100.186 °C (b) 100.512 °C 100.512 (c) (d) 100.0512 °C 0.186 185. An aqueous solution o a non-electrolyte boils at 100.52°C. Te reezing point o the solution will be (K f = 1.86 K m–1, K b = 0.512 K m–1) (a) 0°C (b) –1.86°C (c) 1.86°C (d) –0.52°C 186. A solution o urea (mol. mass 56 g mol –1) boils at

100.18° C at the atmospheric pressure. I K f and K b or water are 1.86 and 0.512 K kg mol –1 respectively the above solution will reeze at (a) –6.54°C (b) 6.54°C (c) 0.654°C (d) –0.654°C 187. Afer adding a solute reezing point o solution

decreases to –0.186. Calculate DT b i K f = 1.86 and K b = 0.521. (a) 0.521 (b) 0.0521 (c) 1.86 (d) 0.0186

188. Which o the ollowing will have the highest reezing

point at one atmosphere? (a) 0.1 M NaCl solution (b) 0.1 M sugar solution (c) 0.1 M BaCl2 solution (d) 0.1 M FeCl3 solution 189. 6 g o substance x dissolved in 100 g o water reezes

at –0.93 °C. Te molar mass o x is (K f = 1.86 K m–1) (a) 60 g mol–1 (b) 120 g mol–1 (c) 180 g mol–1 (d) 140 g mol–1

190. What is the molality o ethyl alcohol (mol. wt. = 46)

in aqueous solution which reezes at –10°C? (K f or water = 1.86 K molality –1) (a) 3.540 m (b) 4.567 m (c) 5.376 m (d) 6.315 m 191. I K f value o H2O is 1.86, the value o DT f or 0.1 m

solution o non-volatile solute is (a) 18.6 (b) 0.186 (c) 1.86 (d) 0.0186 192. 1% solution o Ca(NO3)2 has the reezing point

(a) 0°C (c) greater than 0°C

(b) less than 0°C (d) 273 K

193. I a substance exists as trimer in the solution, then

which o the ollowing alternative is possible or depression in reezing point o m molal solution? mK f mK f (a) (b) 3 4 mK f mK f (c) (d) 5 8

2.10 Osmosis

and Osmotic Pressure

194. 0.6 g o a solute is dissolved in 0.1 L o a s olvent which

develops an osmotic pressure o 1.23 atm at 27°C. Te molecular mass o the substance is (a) 149.5 g mol–1 (b) 120 g mol–1 (c) 430 g mol–1 (d) 102 g mol–1 195. Solutions having same osmotic pressure are known

as (a) (c)

hypotonic isotonic

(b) hypertonic (d) isomeric.

196. Te osmotic pressure o a dilute solution is directly

proportional to the (a) diffusion rate o the solute (b) concentration (c) boiling point (d) flow o solvent rom a concentrated to a dilute solution. 197. At a given temperature isotonic solutions have the

same (a) density (b) volume (c) normality (d) molar concentration. 198. Te osmotic pressure is expressed in units o

(a) (c)

MeV cm

(b) calories (d) atmosphere.

199. Te scientist, who discovered the phenomenon o

osmosis in natural membranes is (a) Raoult F. Marie (b) E. Otto (c) van't Hoff (d) Abbe Nollet. 200. What happens when blood cells are placed in pure

water? (a) Te fluid in blood cells rapidly moves into water (b) Te water molecules rapidly move into blood cells (c) Te blood cells dissolve in water (d) No change takes place. 201. Calculate the osmotic pressure o a solution

containing 0.153 g glucose in 0.1 litre o the solution at 298 K? (a) 0.027 atm (b) 0.107 atm (c) 0.172 atm (d) 0.207 atm 202. Te osmotic pressure o 4.5 g o glucose (molar mass

= 180 g mol–1) dissolved in 100 mL o water at 298 K (R = 0.0821 L atm mol–1 K–1) is (a) 5.116 atm (b) 61.6 atm (c) 6.116 atm (d) 611.6 atm

67

Solutions and Colligative Properties 203. A solution containing 17.8 g L–1 o cane sugar (molar –1

mass 342 g mol ) has an osmotic pressure 1.2 atm. What is the temperature o the solution? (R = 0.082 L atm mol–1 K–1). (a) 281.4°C (b) 428.1 K (c) 281.4 K (d) 428.1°C 204. 30 g o glucose dissolved in one litre o water has an

osmotic pressure 4.91 atm at 303 K. I the osmotic pressure o the glucose solution is 1.5 atm at the same temperature, what would be its concentration? (Molar mass o glucose is 180 g mol–1) (a) 0.5 M (b) 0.0509 M (c) 5 M (d) None o these 205. A solution has an osmotic pressure o 3.90 × 105 Nm–2

at 300 K. Te volume containing 1 mole o solute i solution o same solute has an osmotic pressure o 2.82 × 105 Nm–2 and contains 1 mole o solute in 10.5 m3 is (a) 7.59 m3 (b) 7.59 L 3 (c) 7.59 cm (d) 75.9 m3 206. What is the mass o sucrose in its 1 L solution

(molar mass = 342 g mol–1) which is isotonic with 6.6 × 10–3 kg L–1 o urea (NH2CONH2)? (Given : atomic masses H = 1, C = 12, N = 14, O = 16 in g mol–1) (a) 37.62 kg (b) 3.762 g (c) 37.62 g (d) 376.2 g

207. Osmotic

pressure o a solution containing –3 6.8 × 10 kg o protein per 1 × 10–4 m3 o solution is 3.02 × 103 Pa at 37°C. What is the molar mass o protein? (R = 8.314 J K–1 mol–1) (a) 55.06 kg mol–1 (b) 58.06 kg mol–1 (c) 68.05 kg mol–1 (d) 65.08 kg mol–1

208. At 298 K, 1000 cm3 o a solution containing 4.34 g o

solute shows osmotic pressure o 2.55 atm. What is the molar mass o solute? (R = 0.0821 L atm K–1 mol–1) (a) 44.16 g mol–1 (b) 41.46 g mol–1 (c) 41.64 g mol–1 (d) 46.14 g mol–1 209. What is the volume o a solution containing 34.2 g o

cane sugar (molar mass = 342 g mol –1) which has an osmotic pressure 2.42 atm at 20°C? (a) 0.9945 L (b) 0.9945 × 10–3 L (c) 0.09945 L (d) 0.09945 × 10–3 L

210. A solution o particular amount o organic substance

o molar mass 196 g mol–1 dissolved in 2 litres o water gave an osmotic pressure o 0.54 atm at 12°C. Te mass o the solute dissolved is [R = 0.0821 L atm K–1 mol–1]

(a) (c)

10.5 g 9.04 g

(b) 9.04 kg (d) 0.9 kg

211. Which o the ollowing 0.1 M aqueous solutions will

exert highest osmotic pressure? (a) NaCl (b) BaCl2 (c) MgSO4 (d) Al2(SO4)3 212. In osmosis

(a)

solvent molecules pass rom high concentration o solute to low concentration. (b) solvent molecules pass rom a solution o low concentration o solute to a solution o high concentration o solute. (c) solute molecules pass rom low concentration to high concentration (d) solute molecules pass rom high concentration to low concentration. 213. A solution contains non-volatile solute o molecular

mass M B. Which o the ollowing can be used to calculate molecular mass o the solute in terms o osmotic pressure? (W B = Mass o solute, V = Volume o solution and p = Osmotic pressure) W W RT (a) M B = B VRT (b) M B = B p pV W V W (c) M B = B (d) M B = B pRT p RT V 214. A membrane which allows solvent molecules but not

the solute molecules to pass through it is called as (a) semipermeable membrane (b) permeable membrane (c) filter membrane (d) porous membrane. 215. Which inorganic precipitate acts as semipermeable

membrane? (a) Calcium sulphate (c) Nickel phosphate

(b) Barium oxalate (d) Copper errocyanide

216. Te osmotic pressure o a solution increases i

(a) (b) (c) (d)

temperature is decreased solution constant is increased number o solute molecule is increased volume is increased.

217. Te solution in which the blood cells retain their

normal orm are _______ to the blood. (a) isotonic (b) isomotic (c) hypertonic (d) equinormal 218. In equimolar solutions o glucose, NaCl and BaCl2,

the order o osmotic pressure is (a) Glucose > NaCl > BaCl2 (b) NaCl > BaCl2 > Glucose

68

Objective MHT-CET Chemistry (c) BaCl2 > NaCl > Glucose (d) Glucose > BaCl2 > NaCl

219. A solution having a higher osmotic pressure than

another solution, is called a (a) hypotonic solution (b) isotopic solution (c) isotonic solution (d) hypertonic solution. 220. Water transportation in plants takes place by the

phenomenon o (a) diffusion (c) reverse osmosis

(b) osmosis (d) reverse diffusion.

221. Te solution containing 4.0 g o a polyvinyl chloride

polymer in 1 litre o dioxane was ound to have an osmotic pressure 6.0 × 10–4 atmosphere at 300 K, the value o R used is 0.082 L atm mol–1 K–1. Te molecular mass o the polymer was ound to be (a) 3.0 × 102 (b) 1.6 × 105 (c) 5.6 × 104 (d) 6.4 × 102 222. I mole raction o the solvent in a solution decreases

then (a) (b) (c) (d)

228. A 5% solution o cane sugar (mol. wt. = 342) is isotonic

with 1% solution o a substance X . Te molecular weight o X is (a) 34.2 (b) 171.2 (c) 68.4 (d) 136.8 229. Te hard shell o an egg is dissolved in acetic acid,

and then egg is placed in saturated solution o NaCl. Choose the correct statement. (a) Te egg will shrink. (b) Te egg will become harder. (c) Te egg will swell. (d) Tere will be no change in the size o egg. 230. wo solutions o KNO3 and CH3COOH are prepared

separately. Molarity o both is 0.1 M and osmotic pressures are P 1 and P 2 respectively. Te correct relationship between the osmotic pressures is (a) P 2 > P 1 (b) P 2 = P 1 P 1 P = 2 (c) P 1 > P 2 (d) P1 − P 2 P1 + P 2 231. A plant cell shrinks when it is kept in

vapour pressure o solution increases boiling point decreases osmotic pressure increases none o these.

223. Te concentration in g/L o a solution o cane sugar –1

(a) (b) (c) (d)

a hypotonic solution a hypertonic solution a solution isotonic with the cell sap water.

232. 20 g o a solute was dissolved in 500 mL o water

(molar mass = 342 g mol ) which is isotonic with a solution containing 6 g o urea (molar mass = 60 g mol–1) per litre is (a) 3.42 (b) 34.2 (c) 5.7 (d) 19

and osmotic pressure o the solution was ound to be 600 mm o Hg at 15 °C. Molecular weight o the solute is (a) 1000 g mol–1 (b) 1200 g mol–1 (c) 1400 g mol–1 (d) 1800 g mol–1

224. Te osmotic pressure o 10% solution o cane sugar at

233. Which statement is incorrect or osmotic pressure

69°C in atmospheres is (a) 724 (c) 8.21

(b) 824 (d) 7.21

225. Te relationship between osmotic pressure at 273 K

when 10 g glucose (P 1), 10 g urea (P 2) and 10 g sucrose (P 3) are dissolved in 250 mL o water is (a) P 1 > P 2 > P 3 (b) P 3 > P 1 > P 2 (c) P 2 > P 1 > P 3 (d) P 2 > P 3 > P 1

(p), volume (V ) and temperature (T )? 1 (a) p ∝ i T is constant V (b) p ∝ V i T is constant (c) p ∝ T i V is constant (d) pV is constant i T is constant

234. Te osmotic pressure o a decinormal solution o

(b) atmospheric pressure (d) vapour pressure.

BaCl2 in water is (a) inversely proportional to its Celsius temperature (b) inversely proportional to its absolute temperature (c) directly proportional to its Celsius temperature (d) directly proportional to its absolute temperature.

227. Osmotic pressure o a solution containing 3 g o

235. Afer swimming or a long time in salt water the skin o

glucose in 60 g o water at 15°C is (density = 1 g/mL, molecular weight o glucose = 180 g/mol) (a) 0.34 atm (b) 0.65 atm (c) 6.56 atm (d) 5.57 atm

one’s finger tips wrinkles. Which one o the ollowing property is responsible or this observation? (a) Osmosis (b) Dialysis (c) Electrodialysis (d) Coagulation

226. Pressure at which reverse osmosis starts is

(a) (c)

very high pressure osmotic pressure

69

Solutions and Colligative Properties 236. Te osmotic pressure o which solution is maximum?

(consider that deci-molar solution o each is 90% dissociated) (a) Aluminium sulphate (b) Barium chloride (c) Sodium sulphate (d) A mixture o equal volumes o (b) and (c)

243. Te molar mass ( M 2) o W 2 g solute and the osmotic

pressure (p) o the solution by the solute at temperature relationship W RT (a) M 2 = 2 (b) pV mRT (c) M 2 = (d)

p

237. Te osmotic pressure o 0.4% urea solution is

prepared in V litres T has the ollowing M 2 =

W2 R pT

M 2RT = p

1.66 atm and that o 3.42% solution o sugar is 2.46 atm. When both the solution are mixed, then the osmotic pressure o the resultant solution will be (a) 1.64 atm (b) 2.46 atm (c) 2.06 atm (d) 0.82 atm

244. I osmotic pressure o 1 M o the ollowing in water

238. I a 0.1 M solution o glucose (mol. wt. 180 g mol–1) and

245. Which o the ollowing has the highest osmotic

–1

0.1 M solution o urea (mol. wt. 60 g mol ) are placed on the two sides o a semipermeable membrane to equal heights, then it will be correct to say that (a) there will be no net movement across the membrane (b) glucose will flow across the membrane into urea solution (c) urea will flow across the membrane into glucose solution (d) water will flow rom urea solution into glucose solution.

can be measured, which one will show the maximum osmotic pressure? (a) AgNO3 (b) MgCl2 (c) (NH4)3PO4 (d) Na2SO4 pressure? (a) M /10 HCl (c) M /10 BaCl2

(b) M /10 urea (d) M /10 glucose

246. According to van’t Hoff — Avogadro’s law, volume

occupied by a solution is (a) directly proportional to molar mass o solute (b) inversely proportional to mass o solute (c) directly proportional to number o moles o solute (d) inversely proportional to number o molecules o solute.

239. Solutions containing 1.63 g o boric acid in 450 mL

and 20 g o sucrose (molecular mass = 342) per litre are isotonic. Te molar mass o boric acid is (a) (c)

342 × 1.63 20 1.63 × 342 × 450 1000 × 20

(b) (d)

1.63 × 1000 × 342 20 × 450 20 × 342 × 450 1000 × 1.63

240. Te osmotic pressure o a solution at 276 K is 2.5 atm.

Its osmotic pressure at 546 K under similar conditions will be (a) 0.5 atm (b) 1.0 atm (c) 2.5 atm (d) 5.0 atm 241. 0.5 M solution o urea is isotonic with

(a) (b) (c) (d)

0.5 M NaCl solution 0.5 M sugar solution 0.5 M BaCl2 solution 0.5 M solution o benzoic acid in benzene.

242. Te osmotic pressure o a solution at 0°C is

4 atmospheres. What will be its osmotic pressure at 273°C under similar conditions? (a) 4 atm (b) 2 atm (c) 8 atm (d) 1 atm

2.11 Abnormal

Molecular Masses

247. Abnormal colligative properties are observed only

when the dissolved non-volatile solute in a given dilute solution (a) is a non-electrolyte (b) offers an intense colour (c) associates or dissociates (d) offers no colour. 248. Te expression to compute molar mass o a s olute rom

the elevation o boiling point o a solvent is (where the various symbols have their usual meanings) D T b W 2 K W (a) M 2 = b 1 (b) M 2 = K b W 1 D T b W 2 (c) M 2

=

K b W 2 DT b W 1

(d) M 2

=

DT b W 1 K b W 2

249. Abnormal molar mass is produced by

(a) (b) (c) (d)

association o solute dissociation o solute both association and dissociation o solute separation by semipermeable membrane.

70


250. How many grams o KCl should be added to 1000 g

o water, so that the reezing point reduces to –10°C? (K f or water = 1.86 °C kg mol–1 molar mass o KCl= 74.5 g/mole). (a) 74.5 g (b) 745 g (c) 268 g (d) 199.66 g 251. Acetic acid dissolved in benzene shows a molecular

mass o (a) 30 (c) 120

(b) 60 (d) 180

2CH3COOH → (CH3COOH)2 represents (a) association (b) polymerisation (c) condensation (d) evaporation. 253. Te molecular mass o acetic acid dissolved in water

is 60 and when dissolved in benzene it is 120. Tis difference in behaviour o CH3COOH is because (a) water prevents association o acetic acid (b) acetic acid does not dissolve ully in water (c) acetic acid ully dissolves in benzene (d) acetic acid does not ionize in benzene.

2.12 van’t Hoff Factor 254. Te van’t Hoff actor i or a dilute aqueous solution o

(b) 1.0 (d) 2.0

pcal = theoretical colligative property assuming normal behaviour o solute then van’t Hoff actor (i) is given by (a) i = pobs × pcal (b) i = pobs + pcal i = pobs – pcal

(d) i =

(b) 78% (d) 98%

259. 8 g o benzoic acid when dissolved in 100 g o

benzene lowers its reezing point by 1.62 K. What is the degree o association o benzoic acid i it, orms dimers in benzene, K f or Benzene is 4.9 K kg mol –1 (C = 12, H = 1, O = 16)? (a) 99.16% (b) 9.9% (c) 91.6% (d) 9.16% is dissolved in 1 kg water and the solution roze at –0.0205°C. What is the van’t Hoff actor i K f or water is 1.86 K kg mol–1? (a) 10.41 (b) 2.41 (c) 1.041 (d) 2 261. Te van’t Hoff actor i or a 0.2 molal aqueous solution

o urea is (a) 0.2 (c) 1.2

(b) 0.1 (d) 1.0

262. van’t Hoff actor is

(a) (b) (c) (d)

less than one in case o dissociation more than one in case o association always less than one less than one in case o association.

263. Which o the ollowing compounds has van’t Hoff

actor ‘i’ equal to 2 or dilute solution? (a) K2SO4 (b) NaHSO4 (c) Sugar (d) MgSO4

255. I pobs = observed colligative property and

(c)

87% 89%

260. 0.6 mL o glacial acetic acid with density 1.06 g mL–1

252. 2C6H5COOH → (C6H5COOH)2 and

sucrose is (a) zero (c) 1.5

(a) (c)

pobs pcal

256. I a is the degree o dissociation o Na 2SO4 the van’t

Hoff actor (i) used or calculating molecular mass is (a) 1 + a (b) 1 – a (c) 1 + 2a (d) 1 – 2a 257. 0.2 m aqueous solution o KCl reezes at –0.680°C.

Te van’t Hoff actor and observed osmotic pressure o solution at 0°C (K f = 1.86 K kg mol–1) are respectively (a) 1.83, 8.19 atm (b) 8.13, 9.18 atm (c) 3.18, 19.8 atm (d) 3.81, 89.1 atm 258. 0.01 molal aqueous solution o K3[Fe(CN)6] reezes

at –0.062°C. What is the percentage dissociation o solute, K f or water is 1.86 K kg mol–1?

264. van’t Hoff actor (i) is the ratio o

(a) observed molar mass to theoretical molar mass (b) observed value o colligative property to theoretical value (c) theoretical value o colligative property to observed value (d) none o these. 265. Te van’t Hoff actor will be highest or

(a) (c)

sodium chloride sodium phosphate

(b) magnesium chloride (d) urea.

266. Te depression in reezing point or 1 M urea, 1 M

glucose and 1 M NaCl are in the ratio (a) 1 : 1 : 2 (b) 3 : 2 : 2 (c) 2 : 1 : 1 (d) 1 : 1 : 1 267. Te degree o dissociation (a) o a weak electrolyte

Ax B y is related to van’t Hoff actor (i) by the expression i −1 i −1 (a) a = (b) a = (x + y − 1) x + y + 1 (c)

a=

x + y − 1 i −1

(d)

a=

x + y + 1 i −1

71

Solutions and Colligative Properties 268. Te van’t Hoff actor i or 0.2 m aqueous glucose

solution is (a) 0.2 (c) 0.6

272. Which o the ollowing compounds corresponds

to van’t Hoff actor (i) to be equal to 2 or dilute solution? (a) Na2SO4 (b) AlCl3 (c) Glucose (d) BaSO4

(b) 0.4 (d) 1.0

269. Te van’t Hoff actor calculated rom association

data is always ____ than calculated rom dissociation data. (a) less (b) more (c) same (d) more or less

273. Which one o the ollowing salt, will have the same

270. I van’t Hoff actor or dissolution o Ca(NO 3)2 in

274. Te van’t Hoff actor 0.1 M CaCl2 solution is 2.74. Te

water is 2.5, then degree o dissociation is (a) 25% (b) 50% (c) 75% (d) 84%

value o van’t Hoff actor (i) as that o K4[Fe(CN)6]? (a) Fe2(SO4)3 (b) NaNO3 (c) Ca(NO3)2 (d) K2SO4 degree o dissociation is (a) 61% (c) 100%

271. Te substance A when dissolved in solvent B gave

molar mass corresponding to A3. Te van’t Hoff actor will be 1 (a) (b) 1 2 1 (c) 2 (d) 3

(b) 87% (d) 54%

275. What will be the ratio o any o the colligative

properties o 1.0 m aqueous solutions o NaCl, Na2SO4 and K4[Fe(CN)6] [Assume that solute completely (100%) dissociates in the solution] (a) 2 : 3 : 4 (b) 1 : 2 : 4 (c) 2 : 3 : 5 (d) 1 : 3 : 5

LEVEL - 2 1.

During osmosis, flow o water through a semipermeable membrane is (a) rom solution having lower concentration only

5.

I the osmotic pressure o a dilute solution is 7 × 105 Pa at 273 K then its osmotic pressure at 283 K will be (a) 7.256 × 105 Nm–2 (b) 7.562 × 104 Nm–2 (c) 7.256 × 103 Nm–2 (d) 7.652 × 105 Nm–2

6.

Te solubility o a gas in liquid increases with (a) increase in temperature (b) reduction o gas pressure (c) decrease in temperature (d) amount o liquid taken.

7.

Arrange the ollowing aqueous solutions in the order o their increasing boiling points (i) 10–4 M NaCl (ii) 10–4 M Urea (iii) 10–3 M MgCl2 (iv) 10–2 M NaCl (a) (i) < (ii) < (iv) < (iii) (b) (ii) < (i) = (iii) < (iv) (c) (ii) < (i) < (iii) < (iv) (d) (iv) < (iii) < (i) = (ii).

8.

A solution containing 0.5126 g naphthalene (mol. mass = 128) in 50 g o carbon tetrachloride yields a boiling point elevation 0.402°C while a solution o 0.6216 g o an unknown solute in the same weight o sample solvent gives a boiling point elevation o 0.647°C. Te molecular mass o unknown solute is (a) 98.48 g mol–1 (b) 46.25 g mol–1 (c) 96.46 g mol–1 (d) 46.52 g mol–1

(b) rom solution having higher concentration only (c)

rom both sides o semipermeable membrane with equal flow rates (d) rom both sides o semipermeable membrane with unequal flow rates. 2.

3.

4.

Find the percentage o aqueous cane sugar solution which will have the same reezing point as that o 3% aqueous solution o urea (molecular weight o urea = 60 g mol–1, and molecular weight o cane sugar = 342 g mol–1). (a)

17.1%

(b) 16.6%

(c)

77%

(d) 20%

Which o the ollowing salts has the same value o vant’ Hoff actor as that o K3[Fe(CN)6]? (a)

Na2SO4

(b) Al(NO3)3

(c)

Al2(SO4)3

(d) Fe3O4

I equimolar solutions o CaCl 2 and AlCl3 in water have boiling point o T 1 and T 2 respectively then (a) T 1 > T 2 (b) T 2 > T 1 (c)

T 1 = T 2

(d) can't say

72 9.

Objective MHT-CET Chemistry I 0.5 m solution o Ca(NO3)2 and 0.75 m solution o KOH is taken, then the depression in reezing point is (a) greater in Ca(NO3)2 because number o ions are greater (b) greater in KOH because concentration is high (c) equal in both and reezing point is less than 0°C because ionic concentration is same (d) equal to 0°C in both because ionic concentration is negligible.

(c)

Raoult’s law states that the vapour pressure o a component over a solution is proportional to its mole raction. (d) wo sucrose solutions o same molality prepared in different solvents will have the same reezing point depression. 15.

When 2 g o a non-volatile solute was dissolved in 90 g o benzene the boiling point o benzene is raised by 0.88 K. Which o the ollowing may be the solute? (K b or benzene = 2.53 K kg mol–1) (a) CO(NH2)2 (b) C6H12O6 (c) NaCl (d) None o these.

16.

A solution containing 25.6 g o sulphur dissolved in 1000 g o naphthalene whose melting point is 80.1°C gave a reezing point lowering o 0.68°C. Calculate ormula o sulphur (K f or naphthalene = 6.8 K m –1). (a) S6 (b) S4 (c) S8 (d) S2

17.

Phenol associates in benzene to a certain extent to orm dimer. A solution containing 2.0 × 10–2 kg o phenol in 1.0 kg o benzene has its reezing point decreased by 0.69 K. Te degree o association o phenol is (K f or benzene = 5.12 K kg mol –1) (a) 73.4 (b) 50.1 (c) 42.3 (d) 25.1

18.

Te order o increasing reezing point o C 2H5OH, Ba3(PO4)2, Na2SO4, KCl and Li3PO4 is (a) Ba3(PO4)2 < Na2SO4 < Li3PO4 < C2H5OH < KCl (b) Ba3(PO4)2 < C2H5OH < Li3PO4 < Na2SO4 < KCl (c) C2H5OH < KCl < Na2SO4 < Ba3(PO4)2 < Li3PO4 (d) Ba3(PO4)2 < Li3PO4 < Na2SO4 < KCl < C2H5OH

19.

Calculate the weight o ethylene glycol (an effective antireeze) that must be added to 25 litre water to protect its reezing at –24°C. (K f = 1.86 °C m–1) (a) 20 kg (b) 322.5 kg (c) 200 kg (d) 32.25 kg

20.

19.5 g o monofluoroacetic acid was dissolved in 0.5 kg o water. Te reezing point o solution was observed to be –1.0°C. Te van’t Hoff actor, degree o dissociation and dissociation constant o acid are respectively (Atomic masses C = 12, H = 1, F = 19, O = 16 and K f = 1.86 K kg mol–1) (a) 1.076, 0.076 and 0.003126 (b) 0.076, 1.076 and 0.003126 (c) 0.003126, 0.076 and 1.076 (d) None o these.

10. CrCl3·6NH3 can exist as a complex. 0.1 molal aqueous

solution o this complex shows a depression in reezing point o 0.558°C. Assuming 100% ionisation o the complex and co-ordination number o Cr as six, the complex will be (K f or water = 1.86 K kg mol–1) (a) [Cr(NH3)6]Cl3 (b) [Cr(NH3)5Cl]Cl2 (c) [Cr(NH3)4Cl2]Cl (d) [Cr(NH3)3Cl3]Cl 11.

12.

A 0.0020 m aqueous solution o an ionic compound [Co(NH3)5(NO2)]Cl reezes at –0.00732°C. Number o moles o ions which 1 mol o ionic compound produces on being dissolved in water will be (K f = 1.86°C/m) (a) 3 (b) 4 (c) 1 (d) 2 Pure benzene reezes at 5.3°C. A solution o 0.223 g o phenylacetic acid (C6H5CH2COOH) in 4.4 g o benzene (K f = 5.12 K kg mol–1) reezes at 4.47°C. From this observation, one can conclude that (a) phenylacetic acid exists as such in benzene (b) phenylacetic acid undergoes partial ionisation in benzene (c) phenylacetic acid undergoes complete ionisation in benzene (d) phenylacetic acid dimerises in benzene.

13. What is the molarity o H2SO4 solution, that has a

density 1.84 g/cc at 35°C and contains 98% H2SO4 by weight? (a) 18.4 M (b) 18 M (c) 4.18 M (d) 8.14 M 14.

Which one o the ollowing statements is alse? (a) Te correct order o osmotic pressure o 0.01 M aqueous solution o each compound is BaCl2 > KCl > CH3COOH > sucrose. (b) Te osmotic pressure (p) o a solution is given by the equation, p = MRT , where M is the molarity o the solution.

73

Solutions and Colligative Properties 21.

10 g o monochlorobutyric acid was dissolved in 250 g o water. I dissociation constant o the acid is 1.45 × 10–3 and K f = 1.86 K kg mol–1, the depression o the reezing point will be (C = 12, Cl = 35, H = 1, O = 16) (a) 0.56 K (b) 0.65 K (c) 0.065 K (d) 0.056 K

(a) (c)

23.

A solution contains 25% H2O, 25% C2H5OH and 50% CH3COOH by mass. Te mole raction o H2O would be (a) 0.25 (b) 2.5 (c) 0.502 (d) 5.03

How many grams o sulphuric acid is to be dissolved to prepare 200 mL aqueous solution having concentration o [H3O+] ions 1 M at 25°C temperature? [H = 1, O = 16, S = 32 g mol –1] (a) 4.9 g (b) 19.6 g (c) 9.8 g (d) 0.98 g

29.

Henry’s law constant or CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity o CO2 in 500 mL o soda water when packed under 2.5 atm CO 2 pressure at 298 K. (a) 1.854 kg (b) 1.854 g (c) 18.54 kg (d) 18.54 g

30.

How much mL o a 0.1 M HCl is required to react completely with 1 g mixture o Na2CO3 and NaHCO3 containing equimolar amounts o the two? (a) 15.78 mL (b) 157.8 L (c) 157.8 mL (d) 15.78 L

31.

20 g o a binary electrolyte (mol. wt. = 100) are dissolved in 500 g o water. Te depression in reezing point o the solution is 0.74°C (K f = 1.86 K molality –1). Te degree o the ionisation o the electrolyte is (a) 0% (b) 100% (c) 75% (d) 50%

32.

Which o the ollowing aqueous solutions produce the same osmotic pressure? (i) 0.1 M NaCl solution (ii) 0.1 M glucose solution (iii) 0.6 g urea in 100 mL solution (iv) 1.0 g o a non-electrolyte solute ( X ) in 50 mL solution (molar mass o X = 200) (a) (i), (ii), (iii) (b) (ii), (iii), (iv) (c) (i), (ii), (iv) (d) (i), (iii), (iv)

33.

A solution containing 30 g o a non-volatile solute exactly in 90 g water has a vapour pressure o 2.8 kPa at 298 K. Further 18 g o water is then added to solution, the new vapour pressure becomes 2.9 kPa. Te molecular mass o the solute and the vapour pressure o water at 298 K are respectively (a) 25 g and 253 kPa (b) 2.5 g and 2.53 kPa (c) 23 g and 3.53 kPa (d) 2.3 g and 355 kPa

34.

wo elements A and B orm compounds having molecular ormula AB2 and AB4. When dissolved in 20 g o benzene (C6H6), 1 g o AB2 lowers the reezing point by 2.3 K whereas 1.0 g o AB4 lowers it by 1.3 K. Te molar depression constant or benzene is

24. Match the terms given in Column I with the type o

solutions given in Column II. Column I

Column II

A. Soda water

1. A solution o gas in solid

B. Sugar solution 2. A solution o gas in gas C. German silver 3. A solution o solid in liquid D. Air

4. A solution o solid in solid

E. Hydrogen gas 5. A solution o gas in liquid in palladium 6. A solution o liquid in solid (a) (b) (c) (d) 25.

26.

A

B

C

D

E

5 5 1 1

6 3 2 2

4 4 3 3

2 2 4 4

1 1 5 6

Which statement is true or solution o 0.020 M H2SO4? (a) 2 litre o the solution contains 0.20 mole o SO42– (b) 2 litre o the solution contains 0.080 mole o H3O+ (c) 1 litre o the solution contains 0.020 mole o H3O+ (d) 1 litre o the solution contains 0.04 mole o SO42– Te mass o a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80% is (a) 20 g (b) 30 g (c) 10 g (d) 15 g

27. Te partial pressure o ethane over a solution

containing 6.56 × 10–2 g o ethane is 1 bar. I the solution contains 5.00 × 10–2 g o ethane, then what shall be the partial pressure o the gas?

(b) 76.2 bar (d) 7.6 bar

28.

22. Which o the ollowing solutions is a 1 M solution?

(C = 12, H = 1, O = 16, Ca = 39.98, Cl = 35.5,Na = 23) (a) 0.46 g o C2H5OH in 100 mL o solution (b) 110.98 g o CaCl2 in 1000 mL o solution (c) 0.23 g o CH3OH in 100 mL o solution (d) 5.85 g o NaCl in 1000 mL o solution

0.762 bar 0.076 bar

74

Objective MHT-CET Chemistry 5.1 K kg mol–1. Te atomic masses o A and B are respectively (a) 35.29 u and 62.24 u (b) 24.84 u and 43.03 u (c) 21.29 u and 39.49 u (d) 20 u and 40 u

35.

36.

Te depression in the reezing point o water when 10 g o CH3CH2CHClCOOH is added to 250 g o water is (K a = 1.4 × 10–3, K f = 1.86 K kg mol–1) (a) 0.25°C (b) 0.35°C (c) 0.55°C (d) 0.65°C 1 mole each o the ollowing solutes are taken in 5 moles o water. A. NaCl B. K2SO4 C . Na3PO4 D. glucose Assuming 100% ionisation o the electrolyte, relative decrease in vapour pressure will be in the order (a) A < B < C < D (b) D < C < B < A (c) D < A < B < C (d) Equal

41.

Which one o the ollowing pairs o solutions are isotonic? (a) 0.1 M urea and 0.1 M NaCl (b) 0.1 M urea and 0.2 M MgCl2 (c) 0.1 M NaCl and 0.1 M Na2SO4 (d) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4

42.

Which o the ollowing statements is alse? (a) Te value o molal depression constant depends on the nature o solvent. (b) Relative lowering o vapour pressure is a dimensionless quantity. (c) Relative lowering o vapour pressure is equal to mole raction o solvent. (d) Equimolal solutions o different non-electrolyte solutes would elevate the boiling point to the same extent.

43.

Te reezing point o equimolal aqueous solution will be highest or _____ (a) C6H5NH3+Cl– (b) Ca(NO3)2 (c) La(NO3)3 (d) C6H12O6

44.

Repeated measurements o boiling points o separate solutions o 2.36 g o mercurous chloride in 100 g o water produced DT b values in the range o 0.024 to 0.026 K. Te atomic weights o mercury and chlorine are 200 and 35.5 respectively. K b = 0.5 K per mol or water. Tese data suggest that mercurous chloride unctions as a/an (a) covalent compound in aqueous medium (b) ionic compound in aqueous medium (c) reducing agent in aqueous medium (d) oxidising agent in aqueous medium.

45.

Solutions A, B, C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M BaCl2 and 0.1 M AlCl3. Which one o the ollowing pairs is isotonic? (a) A and B (b) B and C (c) A and D (d) A and C

46.

A solution containing 0.52 g o KCl in 100 g o water roze at –0.25°C. Calculate the percentage ionisation o the salt. (K f = 1.86 K m–1, At. masses : K = 39.1, Cl = 35.5) (a) 93 (b) 83 (c) 73 (d) 63

47.

Te van't Hoff actor or 0.1 M Ba(NO 3)2 solution is 2.74. Te degree o dissociation o the salt at this concentration is (a) 91.3% (b) 87% (c) 100% (d) 74%

37. How much C2H5OH must be added to 1.0 L o H2O,

so that solution should not reeze at –4°F? [K f (C2H5OH) = 1.86°C/m] (a) < 10.75 g (b) > 494.5 g (c) < 20 g (d) 494.5 g 38. Assume that 0.1 molal solutions o sodium chloride,

barium chloride, sodium phosphate and aluminium sulphate are all 100% dissociated. Te solution having the highest boiling point is that o (a) NaCl (b) BaCl2 (c) Na3PO4 (d) Al2(SO4)3 39.

In a 0.2 molal aqueous solution o a weak acid H X , the degree o ionisation is 0.3. aking K f or water as 1.85 K kg mol–1, the reezing point o the solution will be nearest to (a) –0.481°C (b) –0.360°C (c) –0.260°C (d) +0.480°C

40. 58.5 g o NaCl and 180 g o glucose were separately

dissolved in 1000 mL o water. Identiy the correct statement regarding the elevation o boiling point o the resulting solutions. (a) NaCl solution will show higher elevation o boiling point. (b) Glucose solution will show higher elevation o boiling point. (c) Both the solutions will show equal elevation o boiling point. (d) Te boiling point elevation will be shown by neither o the solutions.

75


Te vapour pressures o two liquids ‘P ’ and ‘Q’ are 80 and 60 torr respectively. Te total vapour pressure o the solution obtained by mixing 3 moles o P and 2 moles o Q would be (a) 68 torr (b) 140 torr (c) 72 torr (d) 20 torr

55.

When 5% solution o sucrose (molar mass = 342) is isotonic with 1% solution o a compound ‘ A’, the molar mass o A is (a) 180 g (b) 68.4 g (c) 18 g (d) 32 g

56.

A solution is obtained by mixing 300 g o a 25% solution and 400 g o a 40% solution by mass. Calculate the mass percentage o solute in the resulting solution. (a) 33.57 (b) 66.43 (c) 87.23 (d) 19.24

57.

Given K b and K f o water are 0.52 and 1.86 K m–1. An aqueous solution reezes at –0.186°C. What is the boiling point o the solution? (a) 0.52°C (b) 100.52°C (c) 0.052°C (d) 100.052°C

58.

Pure water as well as separate equimolal and dilute aqueous solutions o NaCl, K 3[Fe(CN)6] and K4[Fe(CN)6] represented respectively as I, II, III and IV are available. Which o the ollowing statements is correct, assuming 100% dissociation o all solutes? (a) I < II < III < IV is the increasing order o reezing points (b) IV < III < II < I is the increasing order o reezing points (c) IV < III < II < I is the increasing order o boiling points (d) IV < III < II is the increasing order o van’t Hoff actors.

59.

Solution A contains 1 g o urea in 100 mL o water and solution B contains 2 g o glucose in 100 mL o water. Now, (a) boiling point o solution A will be less than solution B (b) reezing point o solution A will be lower than solution B (c) both will have same reezing point and boiling point (d) osmotic pressure o solution A will be less than solution B.

60.

Which o the ollowing statements is correct? (a) Lowering o vapour pressure takes place only in ideal solutions (b) Lowering o vapour pressure does not depend upon the solvent at a given concentration o the solute. (c) Lowering o vapour pressure depends upon the nature o solute (d) Relative lowering o vapour pressure does not depend upon the solvent at a given concentration o the solute.

–1

49. Ebullioscopic constant o water is 0.52 K kg mol .

Te incorrect statement is (a) 30 g o urea in hal a kg water produces 0.26 K elevation in boiling point (b) 60 g o urea in one kg water produces 0.52 K elevation in boiling point (c) 90 g o glucose in hal a kg water produces 0.52 K elevation in boiling point (d) 120 g o urea in one kg water produces 1.04 K elevation in boiling point. 50. A sugar solution in water is expected to reeze

(a) at 273 K (c) above 273 K

(b) below 273 K (d) at 298 K.

51. An aqueous solution o a weak monobasic acid

containing 0.1 g in 21.7 g o water reezes at 272.813 K. I the value o K f or water is 1.86 K/m, what is the molecular mass o the monobasic acid? (a) 50.0 g/mol (b) 46.2 g/mol (c) 55.5 g/mol (d) 25.4 g/mol 52.

53.

Which one o the ollowing solutions are isotonic? (a) 3.42 g o sugar in 1000 cm3 o water and 0.18 g o glucose in 1000 cm3 o water (b) 3.42 g o sugar in 1000 cm3 o water and 0.18 g o glucose in 100 cm3 o water (c) 3.42 g o sugar in 1000 cm3 o water and 0.585 g o NaCl in 1000 cm3 o water (d) 3.42 g o sugar in 1000 cm3 o water and 5.85 g o NaCl in 100 cm3 o water wo solutions X and Y are separated by a semipermeable membrane. I the solvent flows rom X to Y , then (a) X is more concentrated than Y (b) X is less concentrated than Y (c) both X and Y have the same concentration (d) both X and Y get diluted.

54. Which o the ollowing statements is not correct?

(a)

Osmotic pressure is directly proportional to molar concentration. (b) Hypertonic solutions have lower concentrations with respect to reerence solution. (c) Isotonic solutions have same molar concentration. (d) Osmotic pressure depends upon temperature.

76


Competitive Exams 7.

2015 1.

I M , W and V represent molar mass o solute, mass o solute and volume o solution in litres respectively, which among the ollowing equations is true?

MWR (a) p = TV (c)

p=

TWR VM

TMR (b) p = WV (d)

p=

TRV WM

(MH-CET)

8. ( MH-CET )

2.

Molarity is defined as (a) the number o moles o solute dissolved in one dm3 o the solution (b) the number o moles o solute dissolved in 1 kg o solvent (c) the number o moles o solute dissolved in 1 dm3 o the solvent (d) the number o moles o solute dissolved in 100 mL o the solvent. (MH-CET)

3.

van’t Hoff actor o centimolal solution o K3[Fe(CN)6] is 3.333. Calculate the percent dissociation o K3[Fe(CN)6]. (a) 33.33 (b) 0.78 (c) 78 (d) 23.33 (MH-CET)

4.

What is the mole raction o the solute in a 1.00 m aqueous solution? (a) 1.770 (b) 0.0354 (c) 0.0177 (d) 0.177 (AIPMT)

5.

Te vapour pressure o acetone at 20°C is 185 torr. When 1.2 g o a non-volatile substance was dissolved in 100 g o acetone at 20°C, its vapour pressure was 183 torr. Te molar mass (g mol–1) o the substance is (a) 128 (b) 488 (JEE Main) (c) 32 (d) 64

What is the molality o a solution containing 200 mg o urea (molar mass = 60 g mol–1) dissolved in 40 g o water ? (a) 0.0825 (b) 0.825 (c) 0.498 (d) 0.0013 (MH-CET)

Consider separate solutions o 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25 °C. Which statement is true about these solutions, assuming all salts to be strong electrolytes? (a) 0.500 M C2H5OH(aq) has the highest osmotic pressure. (b) Tey all have the same osmotic pressure. (c) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure. (d) 0.125 M Na3PO4(aq) has the highest osmotic pressure. (JEE Main)

2012 9.

p A and pB are the vapour pressure o pure liquid components, A and B, respectively o an ideal binary solution. I x A represents the mole raction o component A, the total pressure o the solution will be (a) p A + x A( pB – p A) (b) p A + x A( p A – pB) (c) pB + x A( pB – p A) (d) pB + x A( p A – pB) (AIPMT)

10.

K f or water is 1.86 K kg mol–1. I your automobile radiator holds 1.0 kg o water, how many grams o ethylene glycol (C2H6O2) must you add to get the reezing point o the solution lowered to –2.8°C? (a) 93 g (b) 39 g (AIEEE) (c) 27 g (d) 72 g

11.

Te density o a solution prepared by dissolving 120 g o urea (mol. mass = 60 u) in 1000 g o water is 1.15 g/mL. Te molarity o this solution is (a) 1.78 M (b) 1.02 M (c) 2.05 M (d) 0.50 M (AIEEE)

2014 6.

Solubility o which among the ollowing substances in water increases slightly with rise in temperature? (a) Potassium bromide (b) Potassium chloride (c) Potassium nitrate (d) Sodium nitrate

77


45.0 g H2O, the reezing point is changed by 3.82°C. Calculate the van’t Hoff actor or Na2SO4. (a) 2.05 (b) 2.63 (c) 3.11 (d) 0.381 (AIPMT)

2011 12.

13.

An aqueous solution o urea containing 18 g urea in 1500 cm3 o the solution has a density equal to 1.052. I the molecular weight o urea is 60, the molarity o the solution is (a) 0.200 (b) 0.192 (c) 0.100 (d) 1.200 (MH-CET)

18.

34.2 g o cane sugar is dissolved in 1 80 g o water. Te relative lowering o vapour pressure will be (a) 0.0099 (b) 1.1597 (c) 0.840 (d) 0.9901 (MH-CET)

14. Mole raction o the solute in a 1.00 molal aqueous

solution is (a) 0.1770 (c) 0.0344

(b) 0.0177 (d) 1.7700

(AIPMT)

19.

(AIPMT)

15. A 0.1 molal aqueous solution o a weak acid is 30%

ionized. I K f or water is 1.86°C/m, the reezing point o the solution will be (a) – 0.18°C (b) – 0.54°C (c) – 0.36°C (d) – 0.24°C 200 mL o an aqueous solution o a protein contains its 1.26 g. Te osmotic pressure o this solution at 300 K is ound to be 2.57 × 10–3 bar. Te molar mass o protein will be (R = 0.083 L bar mol–1 K–1) (a) 51022 g mol–1 (b) 122044 g mol–1 (c) 31011 g mol–1 (d) 61038 g mol–1 (AIPMT Mains)

17.

Te reezing point depression constant or water is 1.86°C m–1. I 5.00 g Na2SO4 is dissolved in

Te degree o dissociation (a) o a weak electrolyte, Ax B y is related to van’t Hoff actor (i) by the expression i −1 i −1 (a) a = (b) a = (x + y − 1) (x + y + 1) (c)

a=

(x + y − 1) i −1

(d)

a=

(x + y + 1) i −1

(AIEEE)

20.

Ethylene glycol is used as an antireeze in a cold climate. Mass o ethylene glycol which should be added to 4 kg o water to prevent it rom reezing at – 6°C will be : (K f or water = 1.86 K kg mol–1, and molar mass o ethylene glycol = 62 g mol –1) (a) 804.32 g (b) 204.30 g (c) 400.00 g (d) 304.60 g (AIEEE)

21.

A 5.2 molal aqueous solution o methyl alcohol, CH3OH, is supplied. What is the mole raction o methyl alcohol in the solution? (a) 0.100 (b) 0.190 (c) 0.086 (d) 0.050

(AIPMT Mains)

16.

Te van’t Hoff actor i or a compound which undergoes dissociation in one solvent and association in other solvent is respectively (a) less than one and greater than one (b) less than one and less than one (c) greater than one and less than one (d) greater than one and greater than one.

(AIEEE)

78


Hints & Explanations LEVEL - 1

24.

1.

(a)

2.

(c)

3.

(a)

4.

(d)

5.

(a)

6.

(a)

7.

(c)

8.

(d)

9.

(a)

10.

(c) : Smoke is an aerosol, with gas (air) as the

dispersion medium and solid (soot particles mostly) as the dispersed phase. 11.

(c)

13.

(a) :

12.

(b)

Mixture

(a) Chloroorm in N2 gas

Solute

Solvent

liquid

gas

gas

liquid

solid

solid

solid

gas

liquid

solid

liquid

gas

gas

solid

gas

gas

CO2 in water (b) Brass Camphor in N2 gas (c) Sodium amalgam Moist air (d) H2 gas in Pd metal Mixture o N2 and O2 14.

(b)

15.

(c)

16.

17.

(d) : Precentage by volume o C 2H5OH (v /V )

Volume o solute = Volume o solution 22 = × 100 = 10% 220 18.

(a)

1 2 × 0.1 = × V 2 ⇒ V 2 = 2 dm3 10 Moles o solute 19. (a) : Molality = 1000 g o solvent W 2 × 1000

10 × 1000 20. (a) : M = = = 1 M 3 40 250 × M2 × V (incm ) n2 × 1000 0.5 = 21. (a) : m = W 1 in g 400 × 1000 = 1.25 m 22.

(a)

23.

(b) : Molarity =

m = 25.

W 2 × 1000

M2 × V 36 × 1000 = 0.4 M Molarity = 180 × 500 (Qdensity o water = 1 g/cm3)

648 18

×

1000 648

Number of moles of solute Weight of solvent (in g)

×

1000

= 55.5 m

(b) : Moles o C2H5OH =

25 = 0.54 46

50 = 0.83 60 % o H2O = 100 – (50 + 25) = 25 25 \ Moles o H2O = = 1.39 18 otal number o moles = 2.76 0.83 Mole raction o acetic acid = = 0.301 2.76 0.5 × 106 26. (b) : ppm = = 0.05 107 27. (b) : Percentage by mass o urea (w/W ) mass o urea = × 100 mass o urea + mass o water 6g 600 = × 100 = = 1.186% by mass 6 g + 500 g 506 Moles o CH3COOH =

28.

(c) : Percentage by volume o ethyl alcohol

volume of ethyl alcohol × 100 volume of solution 58 cm3 = × 100 = 12.78% by volume 3 454 cm

=

× 100

(d) : M 1V 1 = M 2V 2

(b) : Molality =

29.

(d) : Moles o ethyl alcohol (n2)

mass o ethyl alcohol molar mass o ethyl alcohol 23 g = = 0.5 mol 46 g mol−1

=

mass o water molar mass o water 54 g = 3.0 mol = 18 g mol−1 n2 0.5 = Mole raction o ethanol = x 2 = n1 + n2 3.0 + 0.5 Moles o water (n1)

=

= 0.1429 n1 Mole raction o water = x 1 = n1 + n2 = 0.8571 and x 1 + x 2 = 0.8571+ 0.1429 = 1

=

3 .0 3.0 + 0.5

79


(c) : 35% (w/W ) solution o ethylene glycol in water

means it has 35 g o ethylene glycol and 65 g o water. Number o moles o ethylene glycol 35 Mass o ethylene glycol = = = 0.56 moles Molar mass o ethylene glycol 62 Mass o water Number o moles o water = Molar mass o water 65 = = 3.61 moles 18 \ Mole raction o ethylene glycol 0.56 0.56 = = = 0.1343 0.56 + 3. 61 4.17 31.

(c) : Mass o NaOH = 1.6 × 10–3 kg,

Volume o solution = 500 cm3 = 0.5 dm3, Molar mass o NaOH = 40 × 10–3 kg mol–1 Molarity o NaOH solution

moles o sugar mass o solvent in kg 0.1 mol = = 0.556 mol kg–1 3 − 180 × 10 kg

Molality =

Mole raction o sugar in syrup moles o sugar = moles o sugar + moles o water 0.1 0.1 = = 0.0099 = 0.1 + 10 10.1 34. (a) : H2SO4 solution is 27% by mass Hence, i mass o H2SO4 is 27 g Ten mass o H2O is (100 – 27) = 73 g Molality o H2SO4 mass o H2 SO4 = molar mass o H2 SO4 × mass o solvent in kg 27 × 10−3 kg = = 3.77 mol kg–1 3 1 3 − − − 98 × 10 kg mol × 73 × 10 kg Mass o solute Volume o solution 100 g Tereore, volume o solution = 1.198 g cm−3 = 83.47 cm3 = 83.47 × 10–3 dm3 Molarity o solution Mass o solute = Molar mass o solute × Volume o solution in dm3 27 × 10−3 kg = 98 × 10−3 kg mol −1 × 83. 47 × 10−3 dm3 Density o solution =

mass o NaOH in kg = molar mass o NaOH in kg mol −1 × volume o solution in dm3

= 32.

1.6 × 10−3 kg 40 × 10−3 kg mol −1 × 0.5 dm3

= 0.08 mol dm−3

(b) : Molar mass o urea, NH2CONH2

= 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol–1 Mass o solvent = 100 g = 0.1 kg 11.11 g = 0.1852 mol Moles o urea = 60 g mol−1 moles o urea Molality o solution = mass o solvent in kg 0.1852 mol = = 1.852 mol kg −1 0.1 kg 33.

(c) : Molar mass o sugar, C12H22O11

= (12 × 12) + (1 × 22) + (11 × 16) = 342 g mol–1 = 342 × 10–3 kg mol–1 Mass o water in syrup = mass o syrup – mass o sugar = 214.2 g – 34.2 g = 180 g = 180 × 10–3 kg mass o sugar Number o moles o sugar = molar mass o sugar − 34. 2 × 10 3 kg = = 0.1 mol 342 × 10−3 kg mol −1 mass o water molar mass o water 180 × 10−3 kg = = 10 moles − − 3 1 18 × 10 kg mol

Moles o water =

= 3.301 mol dm–3 35.

(c) : HCl solution is 38% by mass.

Hence, 100 g HCl solution contains 38 g HCl and (100 – 38) = 62 g water. Volume o 100 g HCl solution Mass o solution 100 g = = Density o solution 1.1 g cm−3 = 90.91 cm3 = 90.91 × 10–3 dm3 Molarity o solution =

Mass o HCl

Molar mass o HCl × Volume o solution in dm3 38 × 10−3 kg = 36.5 × 10−3 kg mol −1 × 90. 91 × 10−3 dm3

Moles o HCl , n2

= 11.45 mol dm–3 Mass o HCl = Molar mass o HCl 38 × 10−3 kg = 36.5 × 10−3 kg mol −1 = 1.04 mol

80

Objective MHT-CET Chemistry Mass o H2 O Molar mass o H2O 62 × 10−3 kg = = 3.44 mol 18 × 10−3 kg mol −1 1.04 n2 Mole raction o HCl, x 2 = = n1 + n2 3.44 + 1. 04 = 0.232

3.66 = 0.1144 moles 32 Number o moles o water Mass o water 25.2 = = = 1.4 moles Molar mass o water 18 0.1144 Mole raction o methyl alcohol = 0.1144 + 1.4 = 0.0755

Moles o H2O, n1 =

36.

(b) : 15 ppm (by mass) means 15 g chloroorm in

=

40.

Volume o xylene = 16.8 cm3 Volume o solute × 100 % by volume o solute = Volume o solution  12.8  × 100 % by volume o benzene =   12.8 + 16. 8  = 43.24% by volume

106 g o the solution. Mass o solvent ≈ 106 g Molar mass o CHCl3 = 12 + 1 + 3 × 35.5 = 119.5 g mol–1 Number o moles o solute

Mass o solute 15 = = 0.126 mol Molar mass o solute 119.5 No. o moles o solute Molality = × 1000 Mass o solvent in g 0.126 = × 1000 = 1.26 × 10–4 m 6 10 =

\

37.

41.

(b) : Solution has 24.8% o HCl, that means 24.8 g o

HCl and 75.2 g o H2O are present. Number o moles o HCl Mass o HCl 24.8 = = = 0.6795 moles Molar mass o HCl 36.5

(b) : Mass o glucose = 34.2 g

Number o moles o H2O

Mass o water = 400 g Percentage by mass o glucose in solution (w/W ) Mass o solute = × 100 Mass o solution 34.2 34.2 = × 100 = × 100 34.2 + 400 434.2 = 7.87 % by mass 38.

(c) : Volume o benzene = 12.8 cm3

=

Mass o H2 O 75.2 = = 4.1777 moles Molar mass o H2 O 18

Mole raction o HCl = Mole raction o HCl =

(c) : Mass o water = 500 g

Percentage by mass o a solute in solution = 2.38 Mass o solute Percentage by mass o solute = × 100 Mass o solution Mass o solute × 100 2.38 = Mass o solution Let mass o solute be ‘x ’.  x  \ 2.38 =   × 100 ⇒ 2.38(x + 500) = 100x x + 500  \ 2.38x + 1190 = 100x ⇒1190 = 100x – 2.38 x 1190 \ 1190 = 97.62 x ⇒ x = = 12.19 g 97.62 mass 39. (b) : Density = volume Mass o methyl alcohol = 0.7952 × 4.6 = 3.66 g Percentage by mass o methyl alcohol =

 3.66  × 100 = 12.68% by mass   3.66 + 25. 2 

Number o moles o methyl alcohol =

Mass o methyl alcohol Molar mass o methyl alcohol

42.

no. o moles o HCl total no. o moles 0.6795 0.6795 = 0.6795 + 4. 1777 4.8572 = 0.1399

(a) : 2 molal aqueous solution is given.

It means 2 moles o solute are present in 1000 g o solvent i.e., water. Mass 1000 Number o moles o H2O = = Molar mass 18 = 55.56 moles 2 = 0.0347 2 + 55. 56 43. (c) : 12.2% HNO3 means 12.2 g o HNO3 and 87.8 g o solvent are present. Mass o HNO3 12.2 = No. o moles o HNO3 = Molar mass o HNO3 63 = 0.1937 moles Mole raction o solute =

Molality o HNO3 =

Mass o HNO3

Molar mass o HNO3 × Mass o solvent in kg 12. 2 × 10−3 = = 2.206 mol/kg 63 × 10−3 × 87. 8 × 10−3

81


Mass o solution Volume o solution 100 = 96.34 cm3 \ Volume o solution = 1.038 = 96.34 × 10–3 dm3 Molarity Mass o solute = Molar mass o solute × Volume o solution in dm3 12. 2 × 10−3 = = 2.01 mol/dm3 3 3 − − 63 × 10 × 96. 34 × 10

Molality o solution

Density o solution =

44.

46.

(a) : Density o solution = 1.070 g/cm3

Mass o solution Density o solution = Volume o solution Mass o solution \ Volume o solution = Density o solution 100 = = 93.46 cm3 1.070 \ Volume o solution = 93.46 × 10–3 dm3 Molarity o solution Mass o solute = Molar mass o solute × Volume o solution in dm3 10 = 2.675 M = 40 × 93. 46 × 10−3

(c) : Te given aqueous solution o H 2SO4 is 4.22 M

and has density o 1.21 g/cm3.

1.21 g/cm3 = 1.21 g/mL Now, 1000 mL o solution contains 4.22 moles o H2SO4

(b) : 95.8% by mass sulphuric acid is present.

Mass o H2 SO4 Number o moles o H2SO4 = Molar mass o H2SO4 95.8 = = 0.9776 mol 98 Mass o H2O = 100 – 95.8 = 4.2 Mass o H2O Number o moles o H2O = Molar mass o H2O 4 .2 = 0.2333 = 18 0.9776 \ Mole raction o H2SO4 = 0.9776 + 0. 2333 0.9776 = = 0.8073 1.2109 Mass o solution Density o solution = Volume o solution Mass o solution Volume o solution = Density o solution 100 = = 52.36 cm3 1.91 = 52.36 × 10–3 dm3 Molarity o solution Mass o solute = Molar mass o solute × Volume o solution in dm3 95.8 = = 18.67 M 98 × 52. 36 × 10−3 45.

Mass o NaOH Molar mass o NaOH × Mass o solvent in kg 10 = 2.77 mol kg −1 = 3 − 40 × 90 × 10 =

\ Mass o 1000 mL o H2SO4 solution = 1000 × 1.21 = 1210 g 1 mole o H2SO4 = 98 g o H2SO4

\ 4.22 mole o H2SO4 = 4.22 × 98 = 413.56 g o H2SO4 1210 g o H2SO4 solution = 413.56 g o H2SO4 So, 100 g o H2SO4 solution = 34.18 g o H2SO4

\ 100 g o H2SO4 solution contains 34.18 g o H2SO4 and 65.82 g H2O.

\ Molality o H2SO4 Mass o H2 SO4

=

Molar mass o H2 SO4 × Mass o solvent in kg 34.18 × 10−3 = = 5.298 mol/kg 98 × 10−3 × 65. 82 × 10−3

47.

(b)

48.

49.

(d) : Molality

= =

(b)

Mass o solute Molar mass o solute × mass o solvent in kg g 20 40 × 500 × 10−3

= 1 mol/kg

50.

(a)

51.

(d) : Number o moles o H2O =

1 g/mL 18 g/mol

= 0.0555 mol/mL = 0.0555 × 1000 = 55.5 moles/L 52.

(d) : M 1 = 0.1 M, M 2 = 0.2 M

V 1 = 100 mL, V 2 = 25 mL M V + M2V 2 Resulting Molarity = 1 1 V1 + V 2 =

(0.1 × 100) + (0. 2 × 25) = 0.12 M 100 + 25

nH PO 3 4 V in L

49 1 × = 0.25 M 98 2 7 .1 54. (d) : Number o moles o Na 2SO4 = = 0.05 142 53.

(a) : Molarity =

=

82


Number o moles o solute Volume o solution in L 0.05 = = 0 .5 M 0.1 55. (a) : Percentage by volume Volume o solute 60  × 100 =  =  × 100 Volume o solution 60 + 300  = 16.66%

64.

Molarity =

56.

(d) : 0.8 moles o solute are present in 1000 mL

(c) : x A + x B + x C

=

1 (n A + nB + nC ) = 1.0 n A + nB + nC

65.

(a)

66.

(b) : According to Henry’s law, S = KP

⇒ S = 1.1 × 10–3 × 0.20 = 2.2 × 10–4 mol dm–3 67.

(a)

0.8 1000 1000 = ⇒ x = = 125 mL x 0.1 8 M V + M2V 2 57. (c) : Resulting molarity = 1 1 V1 + V 2 3 × 25 + 4 × 75 375 = = = 3.75 M 25 + 75 100

68.

(d) : According to Henry’s law S = KP

58.

69.

\ 0.1 mole o solute is present in x mL \

(d) : Molecular wt. o Na2CO3 = 106 g/mol

K = 1.3 × 10–3 mol dm–3 atm–1 P = 0.22 atm

Hence, S= 1.3 × 10–3 mol dm–3 atm–1 × 0.22 atm = 2.86 × 10–4 mol dm–3 Given S = 3.12 × 10–4 mol dm–3, P = 0.24 atm S 3. 12 × 10−4 mol dm −3 K = = 0.24 atm P

1 molar means 1000 mL Na2CO3 solution contains 106 g o Na2CO3 0.25 M means 1000 mL Na2CO3 solution contains 26.5 g o Na2CO3 250 mL o 0.25 M Na2CO3 solution contains 6.625 g o Na2CO3 59.

(a) : HCl

NaOH

M 1V 1 M 2V 2 M HCl × 20 = 0.01 × 19.85 M HCl = 0.01 × 19.85 = 0.0099 M 20 60. (a) : 58.5 g o NaCl in 1 litre gives 1 molar solution. 58.5 g o NaCl in 0.5 litre gives 2 molar solution. \ 5.85 g o NaCl in 0.5 litre gives 0.2 molar solution. 61.

62

63.

(b) : M mix =

M1V1 + M2V 2

=

1 × 2.5 + 0. 5 × 3 2.5 + 3.0

V1 + V 2 2.5 + 1.5 4 = = = 0.73 M 5.5 5.5 (a) : Concentration o AgNO3 is 0.03 g/mL. i.e., 1 mL o the solution contains 0.03 g o AgNO3. Tus, to prepare 60 mL o the solution (0.03 × 60) = 1.8 g o AgNO3 is required. (b) : 0.5 moles o CaCl2 are present in

= 13 × 10–4 mol dm–3 atm–1 = 1.3 × 10–3 mol dm–3 atm–1 70.

(c)

73.

(b)

74.

(c) : Alloy o mercury with other metals is known as

1 mole

75.

2 × 0.25 = 0.50 moles o Cl– ions

(d)

72.

(c)

(c) : Stainless steel contains chromium and some

nickel. 76.

(a)

79.

(a)

80.

(c) : Relative lowering o vapour pressure is a

77.

(d)

78.

(d)

colligative property, not the vapour pressure. 81.

(a)

82.

(a)

83.

(c)

84.

(c)

85.

(a)

86.

(c)

87.

(a)

88.

(d) : According to Raoult’s law, the relative lowering

in vapour pressure o a dilute solution is equal to mole raction o the solute present in the solution. 89.

(c)

91.

(b) : P total = p A + pB = x A p° A + x B p°B

90.

184 =

2 moles

\ 500 mL o solution contains

71.

amalgam, where mercury is in liquid state.

1000 mL o solution 500 mL o solution contains 0.5 × 500 moles o CaCl 2 1000 = 0.25 moles o CaCl2 2+ –    CaCl2  Ca + 2Cl

(c) : According to Henry’s law S = KP

⇒

(b)

3 2 2 × 200 + × pBo or, 184 = 120 + pBo 5 5 5

p°B = 160 torr

Sample Chem(5)

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