Consider a binary solution of two volatile liquid components 1 and 2.$$x_{1}$$ and $$y_{1}$$ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of $$\frac{1}{x_{1}}$$ vs $$\frac{1}{y_{1}}$$ are given respectively as:

For a binary solution of two volatile liquids, find the slope and intercept of $$\frac{1}{x_1}$$ vs $$\frac{1}{y_1}$$.

Apply Raoult's law

Total pressure: $$P = x_1P_1° + x_2P_2° = x_1P_1° + (1-x_1)P_2° = P_2° + x_1(P_1° - P_2°)$$

Mole fraction in vapor: $$y_1 = \frac{x_1P_1°}{P} = \frac{x_1P_1°}{P_2° + x_1(P_1° - P_2°)}$$

Take reciprocal

$$\frac{1}{y_1} = \frac{P_2° + x_1(P_1° - P_2°)}{x_1P_1°} = \frac{P_2°}{x_1P_1°} + \frac{P_1° - P_2°}{P_1°}$$

$$\frac{1}{y_1} = \frac{P_2°}{P_1°} \cdot \frac{1}{x_1} + \frac{P_1° - P_2°}{P_1°}$$

Compare with y = mx + c format

Plotting $$\frac{1}{y_1}$$ (y-axis) vs $$\frac{1}{x_1}$$ (x-axis) -- but the question asks for $$\frac{1}{x_1}$$ vs $$\frac{1}{y_1}$$. Let me rearrange:

$$\frac{1}{x_1} = \frac{P_1°}{P_2°} \cdot \frac{1}{y_1} - \frac{P_1° - P_2°}{P_2°}$$

$$= \frac{P_1°}{P_2°} \cdot \frac{1}{y_1} + \frac{P_2° - P_1°}{P_2°}$$

Slope = $$\frac{P_1°}{P_2°}$$, Intercept = $$\frac{P_2° - P_1°}{P_2°}$$

The correct answer is Option 2: $$\frac{P_1°}{P_2°}$$ and $$\frac{P_2° - P_1°}{P_2°}$$.