Question 16

NUMERICALMEDIUM

Vapour Pressure and Molar Volume of an Ideal Solution of Volatile Liquids A and B

Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of $100 \text{ mm Hg}$ at $300 \text{ K}$ . The vapour pressure of pure A at $300 \text{ K}$ is $105 \text{ mm Hg}$ . Assume that A and B behave as ideal gases in the vapour phase.

Given: The gas constant $R = 0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$ Molar mass of A is $50 \text{ g mol}^{-1}$ Molar mass of B is $57 \text{ g mol}^{-1}$ Density of liquid B at $300 \text{ K}$ is $0.5 \text{ g/mL}$ $1 \text{ atm} = 760 \text{ mm Hg}$

The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.

Correct Answer: 0.16

Detailed Solution

From previous calculation: Mole fraction of B in solution ( $x_B$ ) = 0.2. Vapour pressure of pure B ( $P_B^\circ$ ) = $80 \text{ mm Hg}$ . Total vapour pressure ( $P_T$ ) = $100 \text{ mm Hg}$ .
Calculate partial pressure of B ( $P_B$ ): $P_B = P_B^\circ x_B = 80 \times 0.2 = 16 \text{ mm Hg}$ .
Calculate mole fraction of B in vapour phase ( $y_B$ ): According to Dalton's Law: $P_B = y_B P_T$ $y_B = \frac{P_B}{P_T} = \frac{16}{100} = 0.16$ .

Question 16

Vapour Pressure and Molar Volume of an Ideal Solution of Volatile Liquids A and B

Detailed Solution

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