Question Stem (Q15 & Q16): 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 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase.

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

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

For an ideal liquid solution, Raoult’s law gives the partial vapour pressures:

$$P_A = x_A P_A^{*}, \qquad P_B = x_B P_B^{*}$$
and Dalton’s law gives the total pressure:

$$P_{\text{total}} = P_A + P_B = x_A P_A^{*} + x_B P_B^{*}$$

Step 1: Mole fractions in the liquid
“5 molal B in A” means 5 mol of B are dissolved in 1 kg (1000 g) of A.
Moles of A in 1 kg: $$\frac{1000\;\text{g}}{50\;\text{g mol}^{-1}} = 20\;\text{mol}$$
Total moles in the solution: $$n_{\text{tot}} = 20 + 5 = 25$$
Therefore,

$$x_B = \frac{5}{25} = 0.20, \qquad x_A = 1 - x_B = 0.80$$

Step 2: Vapour pressure of pure B
Given $$P_{\text{total}} = 100\;\text{mm Hg}, \; P_A^{*} = 105\;\text{mm Hg}$$

Using Raoult’s law:

$$100 = 0.80 \times 105 + 0.20 \times P_B^{*}$$
$$100 = 84 + 0.20\,P_B^{*}$$
$$0.20\,P_B^{*} = 16 \Longrightarrow P_B^{*} = 80\;\text{mm Hg}$$

Step 3: Partial pressures
$$P_B = x_B P_B^{*} = 0.20 \times 80 = 16\;\text{mm Hg}$$
$$P_A = x_A P_A^{*} = 0.80 \times 105 = 84\;\text{mm Hg}$$

Step 4: Mole fraction of B in the vapour
For ideal gases, $$y_B = \frac{P_B}{P_{\text{total}}} = \frac{16}{100} = 0.16$$

Hence, the mole fraction of B in the vapour phase in equilibrium with the solution is 0.16.