1. Use Raoult’s law for an ideal binary volatile solution

For components $1$ and $2$:

$p_1 = x_1 P_1^0, \qquad p_2 = x_2 P_2^0$

where

$x_2 = 1-x_1$

Total pressure:

$P = p_1 + p_2 = x_1 P_1^0 + (1-x_1)P_2^0$

2. Relate vapour-phase mole fraction $y_1$ to liquid-phase mole fraction $x_1$

By definition,

$y_1 = \frac{p_1}{P} = \frac{x_1 P_1^0}{x_1 P_1^0 + (1-x_1)P_2^0}$

Now take reciprocal:

$\frac{1}{y_1} = \frac{x_1 P_1^0 + (1-x_1)P_2^0}{x_1 P_1^0}$

Simplify:

$\frac{1}{y_1} = \frac{x_1 P_1^0}{x_1 P_1^0} + \frac{(1-x_1)P_2^0}{x_1 P_1^0}$

$\frac{1}{y_1} = 1 + \frac{P_2^0}{P_1^0}\left(\frac{1-x_1}{x_1}\right)$

Since

$\frac{1-x_1}{x_1} = \frac{1}{x_1}-1$

we get

$\frac{1}{y_1} = 1 + \frac{P_2^0}{P_1^0}\left(\frac{1}{x_1}-1\right)$

$\frac{1}{y_1} = 1 + \frac{P_2^0}{P_1^0}\frac{1}{x_1} - \frac{P_2^0}{P_1^0}$

$\frac{1}{y_1} = \frac{P_2^0}{P_1^0}\frac{1}{x_1} + \left(1-\frac{P_2^0}{P_1^0}\right)$

3. Convert to the form of a linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$

The question asks for the slope and intercept of the plot of

$\frac{1}{x_1} \text{ vs } \frac{1}{y_1}$

So we must write $\dfrac{1}{x_1}$ as a linear function of $\dfrac{1}{y_1}$.

From

$\frac{1}{y_1} = \frac{P_2^0}{P_1^0}\frac{1}{x_1} + \left(1-\frac{P_2^0}{P_1^0}\right)$

rearrange:

$\frac{P_2^0}{P_1^0}\frac{1}{x_1} = \frac{1}{y_1} - 1 + \frac{P_2^0}{P_1^0}$

$\frac{1}{x_1} = \frac{P_1^0}{P_2^0}\frac{1}{y_1} + \frac{P_1^0}{P_2^0}\left(-1 + \frac{P_2^0}{P_1^0}\right)$

Simplify the constant term:

$\frac{1}{x_1} = \frac{P_1^0}{P_2^0}\frac{1}{y_1} + \frac{P_2^0-P_1^0}{P_2^0}$

4. Identify slope and intercept

Comparing with

$Y = mX + c$

where

$Y = \frac{1}{x_1}, \qquad X = \frac{1}{y_1}$

we get:

• Slope: $m = \frac{P_1^0}{P_2^0}$

• Intercept: $c = \frac{P_2^0-P_1^0}{P_2^0}$

5. Match with options

This corresponds to Option D.

Comparison with stored answer: Stored correct answer is D, which matches the derived result.