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.

At 300 K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is ___.

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
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