The plots of $\frac{1}{X_{A}}$ (on y-axis) vs $\frac{1}{Y_{A}}$ (on x-axis)(where X_A and Y_A are the mole fraction of liquid A in liquid and vapour phase respectively) is linear with slope and y-intercept respectively.

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$\frac{P_{A}^{0}}{P_{B}^{0}} and \frac{(P_{B}^{0} - P_{A}^{0})}{P_{B}^{0}}$

$\frac{P_{B}^{0}}{P_{A}^{0}} and \frac{(P_{A}^{0} - P_{B}^{0})}{P_{B}^{0}}$

$\frac{P_{B}^{0}}{P_{A}^{0}} and \frac{(P_{B}^{0} - P_{A}^{0})}{P_{B}^{0}}$

$\frac{P_{A}^{0}}{P_{B}^{0}} and \frac{(P_{A}^{0} - P_{B}^{0})}{P_{B}^{0}}$

answer is C.

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Detailed Solution

$P = P_{A}^{0} X_{A} + P_{B}^{0} (1 - X_{A})$ and $P_{A}^{0} X_{A} = Y_{A} P = Y_{A} [P_{A}^{0} X_{A} + P_{B}^{0} (1 - X_{A})]$
So, $\frac{1}{Y_{A}} = 1 + \frac{P_{B}^{0}}{P_{A}^{0}} (\frac{1}{X_{A}} - 1)$
So $, x = 1 + \frac{P_{B}^{0}}{P_{A}^{0}} (y - 1)$
Hence, $\Rightarrow (x - 1) \frac{P_{A}^{0}}{P_{B}^{0}} + 1 = y$
So, $y = mx + C$ gives the result