Question

Vapour pressure of solution containing 6g of a non-volatile solute in 180 g of water at 20.0 torr. If 1 mole water is further added, vapour pressure increases by 0.02 torr. Which of the following is true?

• A. The molecular weight of solute is 54 g $mo{l}^{-1}$
• B. The vapour pressure of pure water is 20.22 torr
• C. Addition of more water in the solution will further raise the vapour pressure of solution.
• D. The vapour pressure of pure water is 22.22 torr.

Answer 1

Answer: (A), (B), (C)

The molecular weight of solute is 54 g $mo{l}^{-1}$
The vapour pressure of pure water is 20.22 torr
Addition of more water in the solution will further raise the vapour pressure of solution.

The relationship between the relative lowering of vapour pressure and the molecular weight of solute and solvent is as shown below :

$\frac{\Delta P}{P}=\frac{W}{M}\times \frac{m}{w}$

Vapour pressure of solution containing 6g of a non-volatile solute in 180 g of water us 20.0 torr. Substitute values in the above expression.

$\frac{P-20}{20}=\frac{6}{M}\times \frac{18}{180}=\frac{6}{M}\times \frac{1}{10}$.....(1)

1-mole water is further added vapour pressure increases by 0.02 torr.

Hence, $\frac{P-20.02}{20.02}=\frac{6}{M}\times \frac{18}{198}=\frac{6}{M}\times \frac{1}{11}.....\left(2\right)$

Thus, $M\simeq 54g/mole$

Substitute this in equation (1).

$\frac{P-20}{20}=\frac{6}{54}\times \frac{18}{180}=\frac{6}{M}\times \frac{1}{10}$

$P=20.22torr$

When more water is added in the solution the vapour pressure of solution further increases.

As dilution increases, the lowering in vapour pressure $\left(\Delta P\right)$ decreases. Hence, the vapour pressure of solution increases.

Therefore, option A, B and C are correct.