Liquid A and B form an ideal solution. The vapour pressure of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If $$x_A$$ and $$x_B$$ are the mole fraction of A and B in solution while $$y_A$$ and $$y_B$$ are the mole fraction of A and B in vapour phase then :

For an ideal liquid mixture Raoult’s law is valid.
If $$P_A^0$$ and $$P_B^0$$ are the vapour-pressures of the pure liquids, then the partial pressures over the solution are

$$p_A = x_A P_A^0$$ and $$p_B = x_B P_B^0$$ $$-(1)$$

The total pressure is $$P = p_A + p_B$$. The mole fractions of the components in the vapour phase are defined as

$$y_A = \frac{p_A}{P}$$ and $$y_B = \frac{p_B}{P}$$ $$-(2)$$

Divide the two expressions in $$(2)$$:

$$\frac{y_A}{y_B} = \frac{p_A/P}{p_B/P} = \frac{p_A}{p_B}$$

Now use $$(1)$$:

$$\frac{y_A}{y_B} = \frac{x_A P_A^0}{x_B P_B^0} = \frac{x_A}{x_B}\,\frac{P_A^0}{P_B^0}$$ $$-(3)$$

The data given are $$P_A^0 = 350\ \text{mm Hg}$$ and $$P_B^0 = 750\ \text{mm Hg}$$, so

$$\frac{P_A^0}{P_B^0} = \frac{350}{750} = 0.467 \lt 1$$

Substituting into $$(3)$$:

$$\frac{y_A}{y_B} = \frac{x_A}{x_B}\,(0.467)$$

Because the multiplier $$0.467$$ is less than $$1$$, we obtain

$$\frac{y_A}{y_B} \lt \frac{x_A}{x_B} \quad\Longrightarrow\quad \frac{x_A}{x_B} \gt \frac{y_A}{y_B}$$

Hence the correct relation is given in Option C.

Answer - Option C