# UNIT 9: Solutions - Online Test

Q1. The Henry's law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B is 0.2. The ratio of mole fraction of B and A dissolved in water will be
Explaination / Solution:

Given, (KH)A = x

(KH)B = y

xA/xB =  0.2

(xA/xB)in solution =  ?

PA = x (xA) in solution – (1)

PB = y (xB) in solution – (2)

(xA/xB)in solution = (PB/PA) . (x/y) = (xB/xA) . (x/y) = (1/0.2) . (x/y) = 5x/y

Q2. At 100oC the vapour pressure of a solution containing 6.5g a solute in 100g water is 732mm. If Kb = 0.52, the boiling point of this solution will be
Explaination / Solution:

Tb – 100 = 1.06

Tb = 100 + 1.06

= 101.06 ≈ 101o C

Q3. According to Raoults law, the relative lowering of vapour pressure for a solution is equal to
Explaination / Solution:

∆P/Pº= x2 (mole fraction of the solute)

Q4. At same temperature, which pair of the following solutions are isotonic ?
Explaination / Solution:

0.1 × 3 ion [Ba2+, 2NO3]

= 0.1 × 3 ion [2 Na+, SO4]

Q5. The empirical formula of a non-electrolyte(X) is CH2O. A solution containing six gram of X exerts the same osmotic pressure as that of 0.025M glucose solution at the same temperature. The molecular formula of X is
Explaination / Solution:

1)non electrolyte = (π2)glucose

C1RT = C2RT

[CH2O  => 12+2+16=30]

W1/M1 = W1/M2

6 / n(30) = 0.025

n = 8

molecular formula = C8H16O8

Q6. The KH for the solution of oxygen dissolved in water is 4 × 104 atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is
Explaination / Solution:

Q7. Normality of 1.25M sulphuric acid is
Explaination / Solution:

Normality of H2SO4 = (no.of replacable H+)×M

= 2 × 1.25

= 2.5 N

Q8. Two liquids X and Y on mixing gives a warm solution. The solution is
Explaination / Solution:

ΔHmix is negative and show negative deviation from Raoults law.

Q9. The relative lowering of vapour pressure of a sugar solution in water is 3.5 × 10–3. The mole fraction of water in that solution is
Explaination / Solution:

/ Pº = Xsugar

3.5 × 10–3 = Xsugar

Xsugar + XH2o = 1

XH2o = 1 – 0.0035 = 0.9965

Q10. The mass of a non-voltaile solute (molar mass 80 g mol–1) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90%
Explaination / Solution: