CLS Aipmt 14 15 XI Che Study Package 1 SET 1 Chapter 1

Chapter 1

Some Basic Concepts of Chemistry Solutions

SECTION - A School/Board Exam. Type Questions Very Short Answer Type Questions:

1.

The SI uni unitt of of vol volum ume e is is m3 whereas litre (L) is the common unit which is not an SI unit.

2.

The thermom thermometers eters with with celsius celsius scale are are callibrat callibrated ed from 0° 0° to 100° where where there there two temper temperature aturess are the freezing and boiling of water.

3.

The Th e SI SI uni unitt of of den densi sity ty is kg kgm m–3 or kg/m3.

4.

A chemical chemical equat equation ion in which which the the number number of atoms atoms of each elemen elementt is equal equal on the the reactant reactant side side and the product side is called a balanced equation.

5.

Molarity Molar ity is is defined defined as the the number number of moles moles of of solute solute per litre litre of the soluti solution. on. Its SI unit unit is mol L–1.

6.

(a) Mo Mola lari rity ty de decr crea ease sess with with in incr crea ease se in in tem tempe pera ratu ture re (b)) Mole fraction (b fraction : It remains remains unchanged unchanged with with change in temperatur temperature. e.

7.

If any reactan reactantt or product product is a liquid liquid,, the volume volume occupie occupied d by a liquid liquid is extrem extremely ely small small and hence, hence, the the law is not obeyed.

8.

1 g molecule of HNO3 = 1 + 14 + (3 × 16) = 63 g 

9.

3 g molecule of HNO3 = 189 g

(a) Tw Two

(b) Fo Four

(c) Tw Two

10. (a) 6.2 × 10–2 km (b) (1 (1.5 .5 × 1 10 01 cm) × (8.0 × 102 cm) × (1 cm) = 12 × 103 cm3 

1.2 × 104 cm3

Short Answer Type Questions:

11. (a) pico = 10–12

(b) nano = 10–9

(c) cen enti ti = 1 10 0–2

(d) deci = 10–1

9

32 12. F  (C)  32 5



9 5

(35)  32 = 63 + 32 = 95°F

K = °C + 273.15 = 35 + 273.15 = 308.15 K Sector- 11, Dwarka, New Delhi-75 Ph.011-47623456 Ph.0 11-47623456 Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11,

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Some Basic Concepts of Chemistry

(School/Board Exams.) Solutions

13. (i) (a) Molar mass of KAl(SO4)2  12H2O = 39 + 27 + (2 × 96) + (12 × 18) = 474 g mol–1 Percent by mass of Al in KAl(SO 4)2  12H2O 

27 474

 100 100  5.69 5.69% %

(b) Mo Mola larr mas masss of of Na Na2SO4  10H2O = 2 × 23 + 96 + 10 × 18 = 322 g mol–1 

Percent by mass of Na 

23 322

100  7.14 7.14% %  100

(ii) (a) 98 g H2SO4 = 1 mol 

24.5 g H2SO4 

24.5

 0.25 mol

98

(b) 32 g of of O2 = 1 mol 

4.00 g of O2 

1 8

 0.125 mol

14.. (i 14 (i)) If the the digit digit drop dropped ped is is greate greaterr than than 5, add add 1 to the the last last remain remaining ing digi digit. t. e.g., 62.138 will become 62.14

(ii) (i i) If the digit digit dropped dropped is less than than 5, the last remaini remaining ng digit is not not changed. changed. e.g., 28.133 will become 28.13

(iii) If the digit dropped is 5, the last remaining remaining digit is left unchanged if itit is even; 1 is added if it is odd. e.g., 1.8245 will become 1.824

1.8235 will become 1.824 15.. Th 15 The e SI SI unit unit for for spe speed ed are are m mss–1 As given, 1 mile = 1.6 km = 1.60 × 103 m 3



Conversion factor  1.60  10

m

1 mile

1 hour = 60 × 60 s = 3.6 × 103 s 3



Conversion factor 

Now speed  

3.6  10

s

1 hour

150 miles hour 150 miles hour



1.60  10

3

1 mile

m



1 hr 3.6  10

3

s

= 66.66 ms–1 16. Law of constant constant proportions proportions was proposed by Joseph Proust. Proust. This law states states that “a chemical chemical compound compound always always contains same elements combined together in same proportion by mass”. For example, pure water obtained from different sources such as, river, well, spring, sea, etc., always contains hydrogen and oxygen combined together in the ratio 1 : 8 by mass. Sector- 11, Dwarka, New Delhi-75 Ph.011-47623456 Ph.0 11-47623456 Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11,



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17. In red oxide (Cu2O) : 16 parts by mass of oxygen combine with 63.5 × 2 parts by mass of copper. In black oxide (CuO) : 16 parts by mass of oxygen combine with 63.5 parts by mass of copper. Ratio of masses of copper that combine with fixed mass of oxygen (16 parts). In these oxides is 63.5 × 2 : 63.5 or 2 : 1, which is a simple whole number ratio. 18. w C H OH  46 g 2

5

Meth = 46 g mol–1 

nC

nH

2O

46



2H5OH

wH

2O



 1 mol

46

MH



2O

90 18

 5 mol

ntotal = 6 mol 

and

xC

2H5OH

xH

2

O





5

neth neth

 nH O



1 6

 0.167

2

 0.833

6

19. 24.5% by mass H2SO4 means 100 g solution contain 24.5 g H2SO4 

wH SO  24.5 g 2

nH

2SO4

4



24.5

 0.25 mol

98

wH O  75.5 g 2



Molality (m) 

nB w A

 1000 

0.25  1000 75.5

m = 3.31 m kg–1

20. Mass in g of NaOH = 0.48 g Number of moles of NaOH 

0.48 40

 1.2  10 2 mol

Volume of solution = 50 cm = 0.05 L 3

Molarity 

1.2  102 mol 5.0  10

2

L

 0.24 M = 0.24 mol L1

21. 46 wt/volume solution means 100 mL solution contains 46 g H2SO4. So 1 L (1000 mL) solution contains 460 g H2SO4. Therefore, number of moles of H2SO4 

460 98

 4.69

and mass of solution = vol. × density = 1400 g Then, mass of the solvent = mass of solution – mass of solute = 1400 – 460 = 940 g We know that Molality (m) 

nsolute w solvent

 1000 

4.69 940

 1000

= 4.98 mol kg–1 Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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22. Molar mass of sugar = 342 g/mol Moles of sugar 

34.2 342

 0.1

Mass of water = mass of solution – mass of sugar = 224.2 – 34.2 = 190 g (i) Molality  

moles of solute  1000 mass of solvent (in grams) 0.1 190

–1  1000  0.52 mol kg

(ii) Mole fraction of sugar 

nsugar nsugar

 nH

2

0.1

 O

0.1 

190



0.1 10.65

3

 9.38 10

18

23. Let the volume of water be 1 L = 1000 cm3 Mass of water = volume × density = 1000 × 1 = 1000 g 

Moles of water 

1000 18

and molarity of water 

 55.55

55.55 1

 55.55 mol/L

24. The reaction given is 2Br –(aq) + Cl2(aq)  2Cl–(aq) + Br 2(aq) Number of moles of Br – = Number of moles of NaBr  0.080 mol L1 

50.0

L

1000

= 4.0 × 10–3 Number of moles of Cl2 required =

1 2

× No. of moles of Br – = 2 × 10–3 mol

The volume of aqueous Cl2 needed 



Number of moles of Cl 2 Molarity 2  10

3

mol 1

0.050 mol L

 0.04 L = 40 mL

25. In 1808, Dalton published “A new system of chemical philosophy” in which he proposed the following : 1. Matter consists of indivisible atoms 2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass. 3. Compounds are formed when atoms of different elements combine in a fixed ratio. 4. Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction. Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



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26. (i) 1 mole of atom of nitrogen = 6.022 × 1023 atoms 4 mole atoms of nitrogen = 4 × 6.022 × 1023 = 24.08 × 1023 atoms



(ii) 1 mole molecule of nitrogen = 6.022 × 1023 molecules 0.4 mole molecules of nitrogen



= 0.4 × 6.022 × 1023 = 2.408 × 1023 molecules = 2 × 2.408 × 1023 atoms = 4.816 × 1023 atoms 27. (i) 1.66 × 105 (ii) 1.660 × 103 (iii) 1.6 × 10–3 (iv) 1.6 × 102 28. (i) 1 mol = 103 millimole 2.35 millimole = 2.35 × 10–3 mole



(ii) 1 day has 24 hr i.e., 24 × 60 min i.e., 24 × 60 × 60 s Therefore, 2 day = 2 × 86400 s = 172800 = 1.72 × 105 s (iii) 1 m3 = 103 L, therefore 1 mL = 10–6 m 3 Hence, 8.45 mL = 8.45 × 10–6 m 3 (iv) 1  = 10–6 g 68 g = 68 × 10–6 g = 6.8 × 10–5 g



(v) 1 inch = 2.54 × 10–2 m 0.0826 inches = 0.0826 × 2.54 × 10–2 m



= 2.09 × 10–3 m 29.

Zn( s)  2HCl(aq)  ZnCl2 (aq)  H2 (g) 0.40 mol

0.62 mol

0.40 mol Zn requires 2 × 0.40 = 0.80 mol HCl but only 0.62 mol of HCl are given. Therefore, HCl is a limiting reactant. Number of mole of H2 produced  

1 2

1 2

 nHCl  0.62  0.31 mol

30. Molar mass of Na2CO3 = 106 g mol–1 Number of moles of Na2CO3 

0.100 g

Volume of solution = 250 cm3  M



n v

106 1 4

4

 9.4  10 mol

L

4



9.4  10 1

 0.00376 M

L

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Long Answer Type Questions

31. (i) 1 mile = 1.6 × 103 m, 1 million = 106 

83 million miles = 83 × 106 × 1.60 × 103 m = 132.8 × 109 = 1.32 × 1011 m

(ii) 1 feet = 12 inch 

7 × 12 + 2 = 86 inch

and 1 metre = 39.37 inch 

86

86 inch =

39.37

 2.18 m

(iii) 1 Å = 10–10 m 

0.64 Å = 0.64 × 10–10 m

or 6.4 × 10–11 m (iv) 1 mile = 1.6 × 103 m and 1 hour = 60 × 60 sec 3



250 miles/hr



250  1.6  10 60  60

3

400  10



3

3.6  10

 111.11 ms

1

(v) 14.7 lb/m2 = 1 atm = 1.013 × 105 pascals 24 pounds per square inch = 24 × 6894.76 Nm–2 = 165474.24 Nm–2 = 1.65 × 105 pascals (vi) 26°C = 26 + 273 = 299 K (vii) 1 lb = 0.45359 kg 

250 pound = 250 × 0.45359 = 113.39 kg

32. Molar mass of washing soda is (2 × 23) + 12 + (3 × 16) + 18 × 10 = 286 g mol–1 or 286 × 10–3 kg mol–1 nsolute 

6.055  103 kg 286  103 kg mol1

 (i) Molarity

nsolute v



 0.211 mol

0.211 1

 0.211 M

Mass of 1 L solution (10–3 m 3) = density × volume = 1077.2 kg m–3 × 10–3 m 3 = 1.0772 kg 

Mass of solvent = (1.0772 – 0.06055) kg = 1.01665 kg

(ii) Molality 

moles of solute mass of solvent in kg



0.211 1.01665

1  0.207 mol kg

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x solute

(iii)





33.

xC

2H5OH

nH O 2

nsolute nso lu te

0.211



nC

H5 OH

2

2.77

18

2H5OH

nC

2H5 OH

1000 18

2H5OH

2H5OH

1016.65

nC

is 1 L H2O 

nC

0.211

 0.00372

0.211  56.48

 0.050 

7

0.211



 ns olv en t

nC




 55.55

 nH2O

 55.55

 0.050

 0.050 nC H OH  2.77 2

5

 0.95 nC H OH 2

nC

2H5OH



2.77 0.95

5

 2.91 mol

Now, 1 L H2O = 1 kg H2O (At 4°C) 

Molality 

nC H OH 2

5

w H O (in kg) 2



2.91 1

 2.91 mol kg1

34. (i) 1 mole of P4 contains 6.022 × 1023 P 4 molecules 

Number of P4 molecules in 0.5 mol 23



6.022  10

molecules

1 mol

 0.5 mol

= 3.011 × 1023 P 4 molecules (ii) Since, 1 P4 molecule contains 4 P atoms Number of P atoms = 4 × 3.011 × 1023 = 12.044 × 1023 P atoms (iii) Moles of P atoms = 0.5 × 4 = 2 mole (iv) Molecular mass of P4 = 4 × 31 = 124 u 1 mole of P4 molecule weighs 124 grams 

mass of 0.5 mol P4 molecules 

124 g 1 mol

 0.5 mol = 62.0 g

35. 1 mole of Zn3(PO4)2 contains 3 mol of zinc 2 mol of phosphorus and 8 mol of oxygen Mass of 3 mol of Zn = 3 × 65.5 = 196.5 g Mass of 2 mol of P = 2 × 31 = 62 g Mass of 8 mol of O = 8 × 16 = 128 g Mass of 1 mol of Zn3(PO4)2 = 196.5 + 62 + 128 = 386.5 g Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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(a) Percentage composition of Zn 



Mass of Zn Mass of sample

 100

196.5 g  100  50.84% Zn 386.5 g

(b) Percentage composition of P 



Mass of P Mass of sample

 100

62 g  100  16.04% P 386.5 g

(c) Percentage composition of O 



Mass of O Mass of sample

128 g 386.5 g

 100

 100  33.12% O

36. Mass of organic compound, w = 0.90 g Mass of CO2, w 1 = 1.1 g Mass of water, H2O, w 2 = 0.3 g 1 mol C  1 mol CO2 12 g of C = 44 g CO2 (a) Percentage of carbon  

Mass of CO2 12 g   100 44 g Mass of sample 12 44



1.1 g 0.90 g

 100 

1320 39.6

 33.33% C

(b) 2 mol H = 1 mol H2O 2 g H = 18 H2O Percentage of hydrogen  

Mass of H2 O 2g   100 18 g Mass of sample 2 18



0.3 g 0.90 g

 100 

60 16.2

 3.70% H

(c) Percentage of nitrogen = 100 – (%C + %H) = 100 – (33.33 + 3.70) = 100 – 37.03 = 62.97% N 37.

∵

Molecular mass of CO2 = 44 g/mol 1 molecule of CO2 has 2 oxygen atoms 44 g of CO2 = 2 × 6.022 × 1023 oxygen atom




23 132 g of CO2 = 2  6.022  10 

9

132 g 44 mol

= 36.132 × 1023 atoms of oxygen Molecular mass of CO = 28 g/mol. It has 16 g O and one atom of O in one molecule of CO. ∵

In carbon monoxide, mole of CO = mole of C atom = mole of O atom 6.022 × 1023 O atoms = 1 CO = 28 g



36.132 × 10 atoms of O 23

28 g/mol  36.132  1023 atom



6.022  1023 atom/mol

 168 g CO

38. Element Percentage



Atomic mass

Relative no. of atoms

Divided by Simple lowest number ratio

S

47.4

32

47.4 = 1.48 32

1.48 =1 1.48

1

Cl

52.6

35.5

52.6 = 1.48 35.5

1.48 =1 1.48

1

The empirical formula of the compound is SCl

Calculation of molecular formula Empirical formula mass = 1 × 32 + 1 × 35.5 = 67.35 amu n 

Molecular mass Empirical formula mass



135 67.5

2

Molecular formula = n × empirical formula 2 × SCl = S 2Cl2

39. (a) Mass of Co + mass of O = mass of cobalt oxide Mass of O = mass of cobalt oxide – mass of Co = 0.2076 g – 0.1476 g = 0.06 g (b) Moles of Co in the oxide 

Mass of CO Atomic mass of cobalt

0.1476 g  0.002501 mol 59 g/mol



Moles of O in the oxide  

(c) Mole ratio 

Co O



Mass of O Atomic mass of oxygen 0.06 g 16 g/mol

0.002501 0.00375

 0.00375 mol

 0.67 

2 3

(d) Since, the mole ratio of atoms in a sample of a compound is equal to the ratio of atoms in its formula, the empirical formula of the oxide of cobalt is Co2O3. Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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40. Element Percentage




Atomic mass


Divided by lowest number

Simple ratio

24.7 = 0.62 40

0.62 =1 0.62

1

1.24

= 1.24

1.24 =2 0.62

2

= 1.23

1.23 =2 0.62

2

Ca

24.7

40

H

1.24

1

C

14.8

12

O

59.3

16

1 14.8 12 59.3 16

= 3.71

3.71 0.62

= 5.98

6

The empirical formula of the compound is CaH2C2O6 Calculation of molecular formula Empirical formula mass = 40 + 2 × 1 + 2 × 12 + 6 × 16 = 162 n






162 162

1

Molecular formula = n × empirical formula = 1 × CaH2C2O6 = CaH2C2O6 or Ca(HCO3)2

41. Hydrogen combines with oxygen of the oxide to give metal. Therefore, loss in mass during reduction of the oxide is equal to the mass of oxygen. (i) Let the fix the mass of oxygen with which metal combines = 1 g Mass of oxide = 3.45 g Mass of oxygen = 0.24 g Mass of lead = 3.45 g – 0.24 g = 3.21 g 0.24 g oxygen combines with 3.21 g lead 1 g oxygen combines with

3.21 0.24

= 13.4 g lead

(ii) Mass of oxide = 1.195 g Mass of oxygen = 0.156 g Mass of lead = 1.195 g – 0.156 g = 1.039 g 0.156 g oxygen combines with 1.039 g lead 1 g oxygen combines with

1.039 0.156

= 6.7 g lead

The ratio of the masses of lead that combines with a fixed mass of oxygen (1 g) is 6.7 : 13.4 or 1 : 2. This ratio is simple and therefore illustrates the law of multiple proportions. 42. Moles of H2SO4 in 25 mL of 0.2 M H2SO4 = M × V in litres  0.2 M ×

25

litres = 0.005 mol

1000




11

Moles of H2SO4 in 35 mL of 0.25 M H2SO4  0.25 M ×

35

litres = 0.00875 mol

1000

Moles of H2SO4 in 45 mL of 0.35 M H2SO4  0.35 M ×

45

litres = 0.01575 mol

1000

Total moles of H2SO4 = 0.005 + 0.00875 + 0.01575 = 0.0295 mol Total volume = 25 mL + 35 mL + 45 mL = 105 mL = 0.105 litre Final concentration 

0.0295 mole H2 SO4 0.105 litre of solution

43. Element Element ratio

Atomic mass

= 0.2809 M H2SO4


Simplest ratio

9 = 0.75 12

0.75 =3 0.25

Carbon

9

12

Hydrogen

1

1

1 =1 1

1 =4 0.25

3.5

14

3.5 = 0.25 14

0.25 =1 0.25

Nitrogen

Empirical formula = C3H4N Empirical formula mass = (3 × 12) + (4 × 1) + 14 = 54 n




108 54

2

Thus, molecular formula of the compound = 2 × empirical formula = 2 × C3H4N = C6H8N2 44.

Atomic Relative no. Element Percentage of atoms mass Carbon

50.0

12

4.166

Oxygen

50.0

16

3.125

Simplest ratio

Simplest whole number ratio

4.166 = 1.33 3.125

4

3.125 = 1 3.125

3

The empirical formula C4O3 Empirical formula mass = (4 × 12) + (3 × 16) = 96 Molecular mass = 290 n

Molecular mass Empirical mass



290 96

 3 approximately

Molecular formula = n × empirical formula = 3 × C4O3 = C12O9 Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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45.



Atomic Relative no. Element Percentage of atoms mass

Simplest ratio

Sodium

14.31

23

0.622

0.622 = 2 0.311

Sulphur

9.97

32

0.311

0.311 = 1 0.311

Hydrogen

6.22

1

6.22

6.22 = 20 0.311

Oxygen

69.50

16

4.34

4.34 = 14 0.311

The empirical formula = Na2SH20O14 Empirical formula mass = (2 × 23) + 32 + (20 × 1) + (14 × 16) = 322 Molecular mass = 322 Molecular formula = Na2SH20O14 All of the hydrogen is present in the form of water. Thus, 10 water molecules are present in the molecule. So, molecular formula = Na2SO4  10H2O

SECTION - B NCERT Questions

1.

(i) H2O : The molar mass of water, H2O = (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen) = [2(1.008) + 1(16.00 u)] = 2.016 u + 16.00 u = 18.016 = 18.02 u (ii) CO2 : The molar mass of carbon dioxide, CO2 = (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen) = [1(12.01 u) + 2(16.00 u)] = 12.01 u + 32.00 u = 44.01 u (iii) CH4 : The molar mass of methane, CH4 = (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen) = [1(12.01 u) + 4(1.008 u)] = 12.01 u + 4.032 u = 16.042 u



2.


13

Molar mass of Na2SO4 = [(2 × 23.0) + (32.06) + 4(16.00)] = 142.06 g Mass percent of an element  

Mass of that element in the compound Molar mass of the compound

 100

Mass percent of sodium  46.0 g  100 = 32.39 = 32.4% 142 g

Mass percent of sulphur  32 g  100 = 22.54 = 22.6% 142 g

Mass percent of oxygen  64.0 g  100 = 45.07 = 45.05% 142 g

3.

% of iron by mass = 69.9% [Given] % of oxygen by mass = 30.1% [Given] Relative moles of iron in iron oxide : 

% of iron by mass Atomic mass of iron



69.9 55.85

 1.25

Relative moles of oxygen in iron oxide : 

% of oxygen by mass 30.1   1.88 Atomic mass of oxygen 16.00

Simplest molar ratio of iron to oxygen = 1.25 : 1.88 = 1 : 1.5 

4.



2:3

The empirical formula of the iron oxide is Fe2O3.

The balanced reaction of combustion of carbon can be written as C(s)

 O2(g)   CO2(g)

1 mole

1 mole

1 mole

(12 g)

(32 g)

(44 g)

(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide. (ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant. (iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide. 5.

0.375 M aqueous solution of sodium acetate 1000 mL of solution containing 0.375 moles of sodium acetate 

Number of moles of sodium acetate in 500 mL 

0.375 1000

 500  1.1875 mol

Molar mass of sodium acetate = 82.0245 g mol–1 (Given) 

Required mass of sodium acetate = molar mass × no. of moles = (82.0245 g mol–1) (0.1875 mol) = 15.38 g


14

6.



Molarity  % by mass × Density × 10 mol. wt.



6 9  1.4 1 1 0 63

= 15.44 M 7.

1 mole of CuSO4 contains 1 mole of copper Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00) = 63.5 + 32.00 + 64.00 = 159.5 g 159.5 g of CuSO4 contains 63.5 g of copper 100 g of CuSO4 will contain 

8.

63.5  100 g of copper 159.5

Amount of copper that can be obtained from 100 g CuSO4 

63.5  100 159.5

= 39.81 g

Mass percent of iron (Fe) = 69.9% (Given) Mass percent of oxygen (O) = 30.1%

(Given)

Number of moles of iron present in the oxide 

69.90 55.85

Number of moles of oxygen present in the oxide 

 1.25

30.1 16.0

 1.88

Ratio of iron to oxygen in the oxide = 1.25 : 1.88 

1.25 1.88 : 1.25 1.25

= 1 : 1.5 =2:3 

The empirical formula of the oxide is Fe2O3

Empirical formula mass of Fe2O3 = [2(55.85) + 3(16.00)] g Molar mass of Fe2O3 = 159.69 g 

n

Molar mass 159.69 g   0.999  1 (approx.) Emprical formula mass 159.7 g

Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the empirical formula of the given oxide is Fe2O3 and n is 1. Hence, the molecular formula of the oxide is Fe2O3. 9.

The average atomic mass of chlorine = [(Fractional abundance of 35Cl) (Molar mass of 35Cl) + (Fractional abundance of 37Cl) (Molar mass of 37Cl)]   75.77    24.33   (34.9689 u) (36.9659 u)         =  100     100    

= 26.4959 + 8.9568 = 35.4527 u 

The average atomic mass of chlorine = 35.4527 u




15

10. (i) 1 mole of C2H6 contains 2 moles of carbon atoms 

Number of moles of carbon atoms in 3 moles of C2H6 = 2 × 3 = 6

(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms 

Number of moles of carbon atoms in 3 moles of C2H6 = 3 × 6 = 18

(iii) 1 mole of C2H6 contains 6.022 × 1023 molecules of ethane 

Number of molecules in 3 mole of C2H6 = 3 × 6.022 × 1023 = 18.066 × 1023

11. Molarity (M) of a solution is given by 

Number of moles of solute Volume of solution in litres



Mass of sugar/molar mass of sugar 2L



20 / [(12  12)  (1 22)  (11 16)] g 2L





20 g/342 g 2L 0.0585 mol 2L

= 0.02925 mol L–1 Molar concentration of sugar = 0.02925 mol L–1



12. Molar mass of methanol (CH3OH) = (1 × 12) + (4 × 1) + (1 × 16) = 32 g mol–1 = 0.032 kg mol–1 Molarity of methanol solution 



Density of methanol in kg L–1 Molar mass of methanol in kg mol –1 0.793 kg L1 1

0.032 kg mol

1

 24.78 mol L

(Since, density is mass per unit volume) Applying, M1V1 = M2V2 (Given solution) (Solution to be prepared) (24.78 mol L–1) V 1 = (2.5 L) (0.25 mol L–1) V1 = 0.0252 L V1 = 25.22 mL 13. Pressure is determined as force acting per unit area of the surface. Here, force is mass of air × acceleration due to gravity (9.8 m/s2) P

F A



1034 g × 9.8 ms 2 cm2



1 kg 1000 g



(100)2 cm 2 1 m2

= 1.01332 × 105 kg m –1 s –2 We know, 1 N = 1 kg ms–2 Then, 1 Pa = 1 Nm–2 = 1 kg m–1 s –2 1 Pa = 1 kg m–1 s–2 

Pressure = 1.01332 × 105 Pa


16



14. The SI unit of mass is kilogram (kg). 1 kilogram is defined as the mass equal to the mass of the international prototype of kilogram. 15.

Prefix

Multiples –6

(i)

micro

10

(ii)

deca

10

(iii)

mega

10

(iv)

giga

10

(v)

femto

10

6

9

–15

16. Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3. Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result. 17. (i) 1 ppm is equivalent to 1 part out of 1 million (106) parts. 

Mass percent of 15 ppm chloroform in water 

15 6

10

 100  1.5  10 3% 

(ii) 100 g of the sample contains 1.5 × 10–3 g of CHCl3 

1000 g of the sample contains 1.5 × 10–2 g of CHCl3



Molality of chloroform in water  

1.5  10 2 g Molar mass of CHCl3 1.5  10

–2

119.5

= 1.25 × 10–4 m (Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5) = 119.5 g mol–1) 18. (i) 0.0048 = 4.8 × 10–3 (ii) 234,000 = 2.34 × 105 (iii) 8008 = 8.008 × 103 (iv) 500.0 = 5.000 × 102 (v) 6.0012 = 6.0012 × 100 19. (i) 0.0025; has 2 significant figures. (ii) 208; has 3 significant figures. (iii) 5005; has 4 significant figures. (iv) 126,000; has 3 significant figures. (v) 500.0; has 4 significant figures. (vi) 2.0034; has 5 significant figures. Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



17

20. (i) 34.2 (ii) 10.4 (iii) 0.0460 (iv) 2810 21. (a) If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g and 80 g. The masses of dioxygen bear a whole number ratio of 1 : 2 : 1 : 5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers. (b) (i) 1 km = 

1 km 

1000 m 1 km

100 cm 1m



100 mm 1 cm

1 km = 106 mm 1 km  1 km 





1000 m 1 km



1 pm 1012 m

1 km = 1015 pm

Hence, 1 km = 106 mm = 1015 pm (ii) 1 mg  1 mg  

1 kg



1000 mg 1000 g

1 mg = 10–6 kg

1 mg  1 mg 

(iii)

1g

1g 1000 mg



1 ng (nano gram)



1 mg = 106 ng



1 mg = 10–6 kg = 106 ng

1 mL



 1 mL 

10 9 g

1L 1000 mL

1 mL = 10–3 L

1 mL

3

3

 1 cm  1 cm

1 dm  1 dm  1 dm 10 cm  10 cm  10 cm



1 mL = 10–3 dm 3



1 mL = 10–3 L = 10–3 dm 3

22. According to the question, Time taken to cover the distance = 2.00 ns = 2.00 × 10–9 s Speed of light = 3.0 × 108 ms –1 Distance travelled by light in 2.00 ns = Speed of light × time taken = (3.0 × 108 ms–1) (2.00 × 10–9 s) = 6.00 × 10–1 m = 0.600 m Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

18



23. A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed. (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B2. Thus, 200 molecules of B2 will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent. (ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent. (iii) According to the given reaction, 1 atom of A combines with 1 molecule of B2. Thus, all 100 atoms of A will combine with all 100 molecules of B2. Hence, the mixture is stoichiometric where no limiting reagent is present. (iv) 1 mol of atom A combines with 1 mol of B2. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B2 is the limiting reagent. (v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B2. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent. 24. (i) Balancing the given chemical equation, N2(g) + 3H2(g)  2NH3(g) From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g ) of dihydrogen to give 2 mole (34 g) of ammonia. 6g

2 2.00 × 103 g of dinitrogen will react with 28 g  2.00  10 g dihydrogen i.e., 428.6 g H2

So N2 is limiting reagent. 2.00 × 103 g of dinitrogen will react with 428.6 g of dihydrogen. Given, amount of dihydrogen = 1.00 × 103 g 

28 g of N2 produces 34 g of NH3

Hence, mass of ammonia produced by 2000 g of N2 

34 g 28 g

 2000 g = 2428.57 g

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted. (iii) Mass of dihydrogen left unreacted = 1.00 × 103 g – 428.6 g = 571.4 g 25. Molar mass of Na2CO3 = (2 × 23) + 12.00 + (3 × 16) = 106 g mol–1 Now, 1 mole of Na2CO3 means 106 g of Na2CO3 106 g



0.5 mol of Na2CO3 



0.50 M of Na2CO3 = 0.50 mol/L Na2CO3

1 mole

 0.5 mol Na2 CO3 =

53 g Na2CO3

Hence, 0.50 mol of Na2CO3 is present in 1 L of water or 53 g of Na 2CO3 is present in 1 L of water. 26. Reaction of dihydrogen with dioxygen can be written as 2H2(g) + O2(g)  2H2O(g) Now, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour. Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



19

27. (i) 28.7 pm 1 pm = 10–12 m 

28.7 pm = 28.7 × 10–12 m = 2.87 × 10–11 m

(ii) 15.15 pm 1 pm = 10–12 m 

15.15 pm = 15.15 × 10–12 m = 1.515 × 10–11 m

(iii) 25365 mg 1 mg = 10–3 g 25365 mg = 2.5365 × 104 × 10–3 g Since, 1 g = 10–3 kg 2.5365 × 101 g = 2.5365 × 10–1 × 10–3 kg 

2.5365 mg = 2.5365 × 10–2 kg

28. (i) 1 g Au (s) =

1 197

mol of Au (s) 23



6.022  10

atoms of Au (s) = 3.06 × 1021 atoms of Au (s)

197

1

(ii) 1 g Na (s) =

mol of Na (s)

23

23



6.022  10

atoms of Na (s) = 0.262 × 1023 atoms of Na (s)

23

= 26.2 × 1021 atoms of Na (s) (iii) 1 g of Li (s) =

1 7

mol of Li (s) 23



6.022  10 7

atoms of Li (s) = 0.86 × 1023 atoms of Li (s) = 86.0 × 1021 atoms of Li (s)

(iv) 1 g of Cl2(g) =

1 71

mol of Cl2 (g)

(Molar mass of Cl2 molecule = 35.5 × 2 = 71 g mol–1) 23



6.022  10 71

atoms of Cl2 (g) = 0.0848 × 1023 atoms of Cl2 (g) = 8.48 × 1021 atoms of Cl2 (g)

Hence, 1 g of Li (s) will have the largest number of atoms. Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

20


29. Mole fraction of C2H5OH  0.040


Number of moles of C2H5 OH Total moles in solution

nC

2H5OH



nC

2H5 OH

…(i)

 nH2O

Number of moles present in 1 L water 1000 g

nH O 

 55.55 mol

18 g mol1

2

Substituting the value of nC nC

2H5OH

 55.55

2H5OH

nH O 2

in equation (i)

 0.040

nC H OH  0.040nC H OH  (0.040)(55.55) 2

5

2

0.96nC nC

2H5OH



H5OH

2



5

 2.222 mol

2.222

mol

0.96

 2.314 mol

Molarity of solution 

2.314 mol 1L

 2.314 M

30. 1 mole of carbon atoms = 6.023 × 1023 atoms of carbon = 12 g of carbon 

31. (i)

Mass of one

12

C atom =

12 g 6.022  1023

= 1.993 × 10–23 g

0.02856  298.15  0.112 0.5785

Least precise number of calculation = 0.112 

Number of significant figures in the answer = Number of significant figures in the least precise number =3

(ii) 5 × 5.364 Least precise number of calculation = 5.364 

Number of significant figures in the answer = Number of significant figures in 5.364 =4

(iii) 0.0125 + 0.7864 + 0.0215 Since the least number of decimal places in the term is four, the number of significant figures in the answer is also 4. 32. Molar mass of argon 

   35.96755 



0.063   90.60      37.96272   39.9624 g mol1     100   100   100  

0.337 

= [0.121 + 0.024 + 39.802] g mol–1 = 39.947 g mol–1 Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



21

33. (i) 1 mole of Ar = 6.022 × 1023 atoms of Ar 

52 moles of Ar = 52 × 6.022 × 1023 atoms of Ar = 3.131 × 1025 atoms of Ar

(ii) 1 atom of He = 4 u of He or 4 u of He = 1 atom of He 1 u of He = 52 u of He =

1 4

atom of He

52 4

atom of He = 13 atoms of He

(iii) 4 g of He = 6.022 × 1023 atoms of He 

52 g of He =

6.022  10 4

23

 52

atoms of He = 7.8286 × 1024 atoms of He

34. (i) 1 mole (44 g) of CO2 contains 12 g of carbon 

3.38 g of CO2 will contain carbon =

12 g 44 g

 3.38 g =

0.9217 g

18 g of water contains 2 g of hydrogen 

2g

0.690 g of water will contain hydrogen =

18 g

 0.690 =

0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is = 0.9217 g + 0.0767 g = 0.9984 g 

0.9217 g

Percent of C in the compound = Percent of H in the compound =

0.9984 g 0.0767 g 0.9984 g

Moles of carbon in the compound =

92.32%

 100 =

7.68%

92.32 12.00

Moles of hydrogen in the compound = 

 100 =

= 7.69

7.68 1

= 7.68

Ratio of carbon to hydrogen in the compound = 7.69 : 7.68 = 1 : 1

Hence, the empirical formula of the gas is CH (ii) Given, weight of 10.0 L of the gas (at S.T.P.) = 11.6 g 

Weight of 22.4 L of gas at STP =

11.6 g 10.0 L

 22.4 L =

25.984 g  26 g

Hence, the molar mass of the gas is 26 g (iii) Empirical formula mass of CH = 12 + 1 = 13 g n

Molar mass of gas 26 g  Empirical formula mass of gas 13 g

n=2 

Molecular formula of gas = (CH)n = C2H2


22



35. 0.75 M of HCl  0.75 mol of HCl are present in 1 L of water  [(0.75

mol) × (36.5 g mol–1)] HCl is present in 1 L of water

 27.375

g of HCl is present in 1 L of water

Thus, 1000 mL of solution contains 27.375 g of HCl 

Amount of HCl present in 25 mL of solution 

27.375 g 1000 mL

 25 mL

= 0.6844 g From the given chemical equation, CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l) 2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO3 (100 g) 

Amount of CaCO3 that will react with 0.6844 g =

100 73

 0.6844 g =

0.938 g

36. 1 mol [55 + 2 × 16 = 87 g] MnO2 reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl 

5.0 g of MnO2 will react with 

146 g 87 g

 5.0 g of

HCl

= 8.4 g of HCl Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.

SECTION - C Model Test Paper Very Short Answer Type Questions :

1.

Precision means the closeness of various measurements for the same quantity. Accuracy is the agreement of a particular value to the true value of the result.

2.

When a substance does not contain discrete molecules of their constituent units and have a three dimensional structure, formula mass is used to calculate molecular mass which is the sum of the atomic masses of all atoms present in the formula.

3. 4.

1 mol = 6.022 × 1023 molecules One molal solution is that solution which contain 1 mole solute in 1 kg of solvent.

5.

The number of molecules present in 0.5 moles of CO2 is 6.022 × 1023 × 0.5 = 3.011 × 1023.

6.

Molarity 

Number of moles of solute Volume of solution in litres

Molality 

Number of moles of solute Mass of solvent in kg

7.

1947 = 1.947 × 103 0.00019 = 1.9 × 10–4 0.02601 = 2.601 × 10–2 2600.00 = 2.60 × 103

8.

Pure substance have fixed composition and cannot be separated by simple physical methods whereas mixtures do not have fixed composition or ratio and can be separated by physical methods.




23

Short Answer Type Questions :

9.

The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.

10. (i) 7.85 g of Fe 56 g of Fe contains 6.022 × 1023 atoms = 1 mol 56 g of Fe = 1 mol 9 g of Fe 

1 56

(ii) Mole of Ca 

 9 =

9  10

0.16 mol

–3

= 2.25 × 10–4

40

11. Empirical formula is a simplest whole number ratio of atoms in the molecule, therefore the empirical formula of given compounds are (i) CH (ii) CH2 (iii) HO (iv) H2O 12. Mass percent of A 

Mass of A Mass of solution

 100

4g

= 4 g of A + 18 g of water  100 

4 22

 100  18.18%

13. (i) The result of multiplication 3.24  0.08666 5.006

= 0.0560883

0.0561 As the least number of the significant figures is three, therefore the result is rounded off to 0.0561. The 0 is rounded off to 1 because the next digit is greater than 5. (ii) 5 significant figures (Answer should be given by least decimal place) (iii) 6 significant figures (iv) 3 significant figures 

Short Answer Type Questions :

14. Initial volume, V1 = 2 L Final volume, V2 = 3 L + 2 L = 5 L Initial molarity, M1 = 5 M Final molarity = M2 M1V1 = M2V2 5 M × 2 L = M2 × 5 L M2



5M×2L 5L

2M

Thus, the resulting solution is 2 M HCl 15. If two elements can combine to form more than one compounds, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. For example, hydrogen combines with oxygen to form two compounds, namely water and hydrogen peroxide. Hydrogen  Oxygen  W ater 2g

16 g

18 g


24



Hydrogen  Oxygen  Hydrogen peroxide 2g

32 g

34 g

Here, the masses of oxygen (i.e., 16 g and 32 g) which combine with the fixed mass of hydrogen (2 g) bear a simple ratio, i.e., 16 : 32 or 1 : 2. 16. Number of moles of solute (CH3OH)  (n A )

Number of moles of

water (nB )



120 18

Weight of CH3 OH Molecular mass of CH3OH



60 32

 1.875

 6.667

n A + nB = 1.875 + 6.667 = 8.542 B   A =

1.875 8.542

 0.220

1 – B = 0.780

17. Given, w = 32 gm Atomic mass of S = 32 Molar mass of S8 = 32 × 8 = 256 No. of molecules N = no. of moles × 6.022 × 1023 No. of moles 

w M



32 256

 0.125

So, number of molecules = 0.125 × 6.022 × 1023 = 7.52 × 1022 molecules 18. C2H6 = (2 × 12) + (6 × 1) = 30 C12H22O11 = (12 × 12) + (22 × 1) + (11 × 16) = 342 H2SO4 = (2 × 1) + 32 + (4 × 16) = 98 H3PO4 = (1 × 3) + 31 + (4 × 16) = 98 Long Answer Type Questions :

19. (i) 1 gram atom of calcium = gram atomic mass of calcium = 40 g 2.5 gram – atom of calcium = 40 × 2.5 = 100 g (ii) Molecular mass of water (H2O) 1 × 2 + 16 = 18 u 1 gram molecule of H2O = gram molecular mass of H2O = 18 g 1.5 gram molecule of H2O = 1.8 × 1.5 = 27 g 20. Molecular mass of sucrose = (12 × 12 + 1 × 22 + 16 × 11) = 342 342 g of sucrose contain = 6.022 × 1023 molecules 23



6.022  10 342

 6.84

= 12.04 × 1021 molecules Number of atoms of carbon in 6.84 g of sucrose 1 molecule of sucrose contains = 12 atoms of carbon Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



25

12.04 × 1021 molecules of sucrose contains = 12 × 12.04 × 1021 atoms of carbon = 144.48 × 1021 atoms of carbon Number of atoms of hydrogen in 6.84 g of sucrose 1 molecule of sucrose contains = 22 atoms of hydrogen 12.04 × 1021 molecule of sucrose contain = 22 × 12.04 × 1021 atoms of hydrogen = 264.88 × 1021 atoms of hydrogen Number of atoms of oxygen in 6.84 g of sucrose 1 molecule of sucrose contains = 11 atoms of oxygen 12.04 × 1021 molecules of sucrose contain = 11 × 12.04 × 1021 atoms of oxygen = 132.44 × 1021 atoms of oxygen OR

Mass of carbon present in 8.45 mg of CO2 

8.45  12 44

mg = 2.30 mg

Percentage of carbon 

2.30  100 4.24

 54.24%

Mass of hydrogen in 3.46 mg of H2O 

3.46  2 18

mg = 0.384 mg

Percentage of hydrogen 

0.384  100 4.24

 9.05%

Percentage of oxygen = 100 – 54.24 – 9.05 = 36.71% Calculation of empirical formula Element Percentage C

Atomic mass

54.24

12

Relative no. of moles

Simple Ratio

54.24

4.52

12 H

9.05

9.05

1

1 O



36.71

= 9.05

36.71

16

16

= 4.52

= 2.29

2.29 9.05 2.29 2.29 2.29

Simplest whole number ratio

= 1.97

2

= 3.95

4

=1

1

Empirical formula = C2H4O

Calculation of molecular formula : Molecular mass = 88 u n




88 44

2

Molecular formula = empirical formula × n = C 2H4O × 2 = C4H8O2 Aakash Educational Services Pvt. Ltd. -Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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21. The balanced equation for the reaction is AgNO3 + NaCl  AgCl + NaNO3 Number of mol of NaCl in 500 cm3 (0.05 L) of 0.300 M NaCl solution = 0.300 × 0.5 = 0.15 mol Number of moles AgNO3 in 100 cm3 (0.1 L) of 0.500 M AgNO3 solution = 0.500 × 0.1 = 0.05 mol The reaction equation shows that 1 mol of AgNO3 reacts with 1 mol of NaCl to give 1 mol of AgCl 0.05 mol of AgNO3 will react with 0.05 mol of NaCl to give 0.05 mol of AgCl NaCl is present in excess. AgNO3 is, therefore, the limiting reagent. Molar mass of AgCl = 143.4 g 

Mass of 0.05 mol of AgCl = 0.05 × 143.4 g of AgCl = 7.2 g of AgCl OR

(i) 1 mole atoms of nitrogen = 6.022 × 1023 atoms 

0.6 mole atoms of nitrogen = 6.022 × 1023 × 0.6

= 3.6132 × 1023 atoms (ii) 1 mole molecule of nitrogen = 6.022 × 1023 molecules 

0.3 mole molecules of nitrogen = 6.022 × 1023 × 0.3 = 1.8066 × 1023 molecules

1 molecule of nitrogen = 2 atoms 1.8066 × 1023 molecule of nitrogen = 1.8066 × 1023 × 2 atoms = 3.6132 × 1023 atoms (iii) 32 g sulphur contain = 6.022 × 1023 atoms 

3.2 g of sulphur contain 23



6.022  10

 3.2

32

= 6.022 × 1022 atoms 



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CLS Aipmt 14 15 XI Che Study Package 1 SET 1 Chapter 1

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