solutions

Solutions

Q1 Calculate (i) molality (ii) molarity (iii) mole fraction of KI if the density of 20% (mass/ mass) aqueous KI is 1.202 gm / mL

a. 1.45m, 1.51M, 0.0263

b. 1.51m, 1.45M, 0.0263

c. 1.15m, 1.54M, 2.63

d. 11.5m, 15.4M, 26.3

Ans. b.1.51m, 1.45M, 0.0263

Explanation: Molar mass of KI = 39 + 127 = 166 g mol-1

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 - 20) g of water = 80 g of water

Therefore, molality of the solution

= moles of KI/mass of water in Kg = (20/166)/ 0.08

= 1.506 m

= 1.51 m (approximately)

(b) It is given that the density of the solution = 1.202 g mL-1

mass/ density =Volume of 100 g solution

100g/ 1.202 gml-1 = 83.19 mL = 83.19 × 10-3 L

Therefore, molarity of the solution = (20/166) / 83.19 × 10-3 L =1.45 M

Moles of water = 80/18 = 4.44

Therefore, mole fraction of KI = moles of KI/ moles of KI + moles of water = 0.12/0.12 + 4.44= 0.0263

Q2A. Assertion: Mole fraction depends on moles of solution

Reason: Mole fraction is the ratio of moles of either solute or solvent to the total no. of moles of the solution

Ans. a

Q2 dependent on temperature

a. Molarity

b. Mole fraction

c. Weight percentage

d. Molality

Ans. a. Molarity is a function of temperature as volume depends on temperature

Q2). what is the mole fraction of the solute in a 1.00m aqueous solution

a. 1.770

b. 0.0354

c. 0.0177

d. 0.177

Ans. c. 0.0177

Explanation: 1.00 m aqueous solution means 1 mole of solute in 1 kg of water.

1 kg of water corresponds to 181000=55.56 moles.

The mole fraction of solute is = 1/ 1+55.561=0.0177.

Xm =1mole/ 1000g = 0.0177

Q3. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO3? The concentrated acid is 70% HNO3.

a. 70.0g conc. HNO3

b. 54.0g conc. HNO3

c. 45.0g conc. HNO3

d. 90.0g conc. HNO3

Ans. c. 45.0g of conc. HNO3

Explanation: Molar Mass of HNO3 = 1+14+48 = 63u

Molarity = Moles of solute/ volume of solution in Litres

2.0 M =2 Moles HNO3/1 L of solution = 0.5 moles HNO3/ 250mL

2= w x 1000/63 x 250 = w x 63/2

Mass of acid x 70/100 = 63/2

mass of acid = 45g

Q4 can be used as antifreeze in automobile radiators

a. methyl alcohol

b. glycol

c. nitrophenol

d. ethyl alcohol

Ans. b. glycol

Explanation: a 35% (v/v) solution of ethylene glycol is used as an antifreeze in cars for cooling the engine. At this concentration, the antifreeze lowers the freezing point of water to 255.4 K (-17.6 'C)

Q5. Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g /mL. Volume of acid required to make one litre of 0.1 M H2SO4 solution is

a. 16.65 mL

b. 22.20 mL

c. 5.55 mL

d. 11.10 mL

Ans.

Q6. How many mL of 0.1M HCl are required to react completely with 1g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

2.1 Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

2.2 Calculate the mole fraction of benzene in solution containing 30% by mass in CCl4?

2.3 Calculate the molarity of each of the following solutions: (a) 30g of Co(NO3)2.6H2O in 4.3L of solution (b) 30mL of 0.5M H2SO4 diluted to 500mL.

2.4 Calculate the mass of urea NH2CONH2 required in making 2.5 kg of 0.25 molal aqueous solution.

2.5 Calculate

(a) molality

(b) molarity and

Ans. Mass% of KI Solution = 20%, density = 1.2 g/ mL

(a) Molality, m:

wA = 100-20=80g

2.6 H2 S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.

2.7 Henry’s law constant for CO2 in water is 1.67x108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

2.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase

2.9

2.10

2.11

2.12