A substance ‘X’ (1.5 g) dissolved in 150 g of a solvent ‘Y’ (molar mass = 300 g ) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent ‘Y’ is ______ . (nearest integer) [Given : of the solvent =

Question

A substance ‘X’ (1.5 g) dissolved in 150 g of a solvent ‘Y’ (molar mass = 300 g ${mol}^{- 1}$ ) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent ‘Y’ is ______ $\times 10^{- 2}$ . (nearest integer)

[Given : $K_{b}$ of the solvent = 5.0 K kg ${mol}^{- 1}$ ]

Assume the solution to be dilute and no association or dissociation of X takes place in solution. [2026]

Smart Achievers · Accepted Answer

(3)

$Δ T_{b} = i \times K_{b} \times m$

$0.5 = i \times m \times 5$

$i \times m = \frac{0.5}{5} = 0.1$

$i \times a = \frac{15}{1000}$

(where a = moles of solute)

Now,

$\frac{P ° - P_{s}}{P °} = i X_{solute} = i \times \frac{a}{a + \frac{150}{300}}$

$= i \times \frac{a}{\frac{1}{2}} = \frac{15 / 1000}{1 / 2} = \frac{30}{1000} = 3 \times 10^{- 2} = 3$

Solution

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Solution

Ans. (3) ΔTb=i×Kb×m 0.5=i×m×5 i×m=0.55=0.1 i×a=151000 (where a = moles of solute) Now, P°-PsP°=iXsolute=i×aa+150300 =i×a12=15/10001/2=301000=3×10-2=3

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