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Concentration questions
1.
What is the the molarity molarity (molar (molar concentratio concentration) n) of a solution solution made by by dissolving dissolving 2.355 2.355 g of sulfuri sulfuricc acid (H2SO4) in water and diluting to a final volume v olume of 50.0 mL?
2.355 g H2SO4 x
1 mol H2SO4 = 0.024 mol 98.08 g H2SO4
molar conc. = 0.024 mol 0.050 L
= 0.48 mol/L
2. Hydrochlor Hydrochloric ic acid is sold commerci commercially ally as a 12.0 mol/L mol/L solution. solution. How many many moles of HCl are in 300.0 300.0 mL solution solution??
12.0 mol/L =
# mol 0.300 L
# mol = 3.6 mol 3. The concentrati concentration on of cholestero cholesteroll in normal normal blood blood is (C27H46O) approximately 0.005 mol/L. How many grams of cholesterol are in 750 mL of blood?
0.005 mol/L =
# mol 0.750 L
0.00375 mol x
386.73 C27H46O = 1.45 g 1 mol C27H46O
# mol = 0.00375 mol 4. What is the final final concentra concentration tion if 75.0 mL of a 3.50 3.50 mol/L glucose glucose solution solution is diluted diluted to a volume of 400.0 400.0 mL?
ci•Vi = cf•Vf (3.5 mol/L) mol/L) • (75.0 mL) = cf • (400.0 mL) ci = 0.66 mol/L 5. Assume that you have a 5.75 5.75 % mass/mass solution of LiCl in water. water. What mass of solution (in grams) contains 1.60 g of of LiCl?
5.75 % m/m = 1.60 g LiCl mass sol’n mass sol’n = 1.60 g LiCl 5.75 % m/m
x 100%
x 100%
= 27.83 g
6. The legal limit limit for human exposure exposure to carbon carbon monoxide in the the workplace is is 35 ppm. Assuming that that the density of air is 1.3 g/L, how many grams of carbon monoxide are in 1.0 L of air at the maximum allowable concentration?
35 ppm = mass solute 1.3 g air 35 ppm
x 106
x 1.3 g air = mass solute 106
= 4.55 x 10-5 g
7. How many grams of NaOH would be required to prepare 800 grams of a 40% by mass NaOH solution? How many grams of water is required?
40 % m/m = x g NaOH 800 g sol’n
x 100%
40 % m/m · 800 g sol’n = x g NaOH = 320 g NaOH 100% 8.
Determine the molarity of a solution made by dissolving 20.0 g of NaOH in sufficient water to yield a 482 mL solution.
20.0 g NaOH x
1 mol NaOH = 0.50 mol 40.01 g NaOH
molar conc. = 0.50 mol 0.482 L
= 1.04 mol/L
9. How would you prepare 500 ml of 3 mol/L HCl using 6 mol/L HCl from the stockroom. In other words how much water and how much 6 M HCl would you mix to accomplish this dilution?
ci•Vi = cf•Vf (6 mol/L) • Vi = (3 mol/L) (500 mL) Vi = 250 mL VH2O = Vf – Vi VH2O = 500 mL – 250 mL = 250 mL of water 10. A meteorologist indicates the level of a given pollutant in the air is 244.5 ppm. According to this value, what is the mass of pollutants in 234.56 kg of air? (3)
244.5 ppm = mass solute x 106 234.56 kg air 244.5 ppm
x
234.56 kg air = mass solute 106
= 0.057 kg
11. What is the molar concentration (mol/L) of single-single coffee from Tim’s Horton’s (what I would enjoy), if 5.0 g of sugar (C6H12O6) are dissolved in 630 mL (extra-large) of coffee? (4)
5.0 g C6H12O6 x
1 mol C6H12O6 = 0.028 mol 180.16 g C6H12O6
molar conc. = 0.028 mol 0.630 L
= 0.044 mol/L
12. A saline solution (contact cleanser) contains 0.90 g of sodium chloride, dissolved to make a 100.0 mL solution. What is the molar concentration (mol/L) of this solution? (4)
0.90 g NaCl x
1 mol NaCl = 0.015 mol 58.44 g NaCl
molar conc. = 0.015 mol 0.100 L
= 0.15 mol/L
13. At 20ºC, a solution of KClO3 will permit the dissolution of 10 g of solute and the solution obtained has a molar concentration of 0.150 mol/L. What must be the volume of this solution? (4)
10 g KClO3 x
1 mol KClO3 = 0.082 mol 122.55 g KClO3
0.150 mol/L = 0.082 mol xL
xL =
0.082 mol = 0.54 L or 544 mL 0.150 mol/L
14. What volume of ethanol (in wine) is necessary for preparing a 800 mL ethanol solution (wine) of 12% v/v? If more than 72 mL of alcohol passes the legal limit for consumption behind the wheel, is this person legally drunk? (4)
12.0 % v/v =
12.0% v/v
vol. solute 800 mL sol’n
x 100%
x 800 mL sol’n = vol. solute 100%
= 96 mL
Legally drunk, over the 72 mL limit. 15. Balsamic vinegar is sold as a solution of 6% v/v acetic acid in water. What quantity of balsamic vinegar contains exactly 20.0 mL of pure acetic acid? (3)
6 % v/v = 20.0 ml acetic acid vol. sol’n vol. sol’n = 20.0 mL acetic acid 6 % v/v
x 100%
x 100%
= 333 mL
16. Hair dye can contain a solution that is 10.2% v/v hydrogen peroxide in water. What volume of the peroxide is to be found in 250.5 mL of the chemical hair treatment? (3)
10.2 % v/v =
10.2 % v/v
vol. peroxide x 100% 250.5 mL treatment
x 250.5 mL sol’n = vol. solute 100%
= 25.5 mL
17. A boric acid solution is used in ophthalmic drops (for eyes). What mass of boric acid is present in 250.0 mL of a solution that is 2.25 % m/v of acid in water? (3)
2.25 % m/v = x g boric acid 250.0 mL sol’n 2.25 % m/v
x 100%
x 250.0 mL sol’n = x g boric acid 100%
= 5.62 mL
18. Household chemical cleaners often contain ammonia. Industrial strength ammonia is 14.0 mol/L. If 3.0 L of an ammonia solution are needed to clean the house at a concentration of 0.10 mol/L, what would be the volume needed of the original solution that would be diluted ? What volume of water needs to be added to dilute the ammonia? (4)
ci•Vi = cf•Vf (14.0 mol/L) • Vi = (0.10 mol/L) (3.0 L) Vi = 0.0214 L or 21.4 mL VH2O = Vf – Vi VH2O = 3.0 L – 0.0214 L = 2.9786 L of water