Raoult's Law (Liquid-Liquid Mixture)

For the ratio we need the molar volume of pure $$B$$ in its vapour phase $$\left(V_{m}^{\text{vap}}\right)$$ and in its liquid phase $$\left(V_{m}^{\text{liq}}\right)$$ at 300 K.

Step 1: Mole fractions in the given 5 molal solution
Take 1 kg of solvent $$A$$.
Moles of $$A$$ $$= \dfrac{1000\ \text{g}}{50\ \text{g mol}^{-1}} = 20\ \text{mol}$$.
5 molal means 5 mol of $$B$$ are present, so moles of $$B = 5$$.
Total moles $$= 20+5 = 25$$.
$$x_A = \dfrac{20}{25} = 0.8,\qquad x_B = \dfrac{5}{25} = 0.2$$.

Step 2: Vapour pressures above the solution
Raoult’s law for an ideal solution: $$P_A = x_A P_A^{*},\; P_B = x_B P_B^{*}$$.
Given $$P_A^{*}=105\ \text{mm Hg},\; P_{\text{total}} = 100\ \text{mm Hg}$$.
$$P_A = 0.8 \times 105 = 84\ \text{mm Hg}$$.
Hence $$P_B = P_{\text{total}} - P_A = 100 - 84 = 16\ \text{mm Hg}$$.
Therefore $$P_B^{*}=\dfrac{P_B}{x_B}= \dfrac{16}{0.2}=80\ \text{mm Hg}$$.

Step 3: Molar volume of pure $$B$$ in vapour phase
Convert $$P_B^{*}$$ to atm: $$P_B^{*}= \dfrac{80}{760}=0.1053\ \text{atm}$$.
For an ideal gas, $$V_{m}^{\text{vap}} = \dfrac{RT}{P}$$.
Thus $$V_{m}^{\text{vap}} = \dfrac{0.08\ \text{L atm K}^{-1}\ \times 300\ \text{K}}{0.1053\ \text{atm}} \approx 228\ \text{L}$$.

Step 4: Molar volume of pure $$B$$ in liquid phase
Density of liquid $$B$$ at 300 K is 0.5 g mL$$^{-1} = 500\ \text{g L}^{-1}$$.
Molar mass of $$B = 57\ \text{g mol}^{-1}$$.
$$V_{m}^{\text{liq}} = \dfrac{57\ \text{g mol}^{-1}}{500\ \text{g L}^{-1}} = 0.114\ \text{L}$$.

Step 5: Required ratio
$$\dfrac{V_{m}^{\text{vap}}}{V_{m}^{\text{liq}}}= \dfrac{228}{0.114}\approx 2.0\times10^{3}$$.

Hence, the ratio of the molar volume of pure $$B$$ in vapour phase to that in liquid phase at 300 K is
2000.00

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.