JEE Chemistry 2025 Question with Solution

Given: A binary solution contains two volatile liquid components 1 and 2. $x_1$ and $y_1$ are the mole fractions of component 1 in the liquid and vapor phases, respectively.

Find: The slope and intercept for the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$.

Using Raoult's law and Dalton's law:

$$p_1 = x_1 p_1^0$$ $$p_2 = x_2 p_2^0$$ $$P_{\text{total}} = p_1 + p_2$$

Therefore,

$$y_1 = \frac{p_1}{P_{\text{total}}} = \frac{x_1 p_1^0}{x_1 p_1^0 + x_2 p_2^0}$$

Now rearrange this relation to express the given variables in linear form. The provided the solution states that the correct option is A and concludes that the slope and intercept correspond to

$$\frac{p_1^0}{p_2^0} - \frac{p_1^0}{p_2^0}$$

Hence, the correct option is A.

Note: The algebra shown in the solution contains internal inconsistency between intermediate expressions and the final stated option, but the source explicitly marks A as correct.

Given: $x_1$ is the mole fraction of component 1 in liquid phase and $y_1$ is its mole fraction in vapor phase.

Find: The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$.

The source solution writes:

$$p_1 = x_1 \cdot p_1^0$$ $$p_2 = x_2 \cdot p_2^0$$ $$P_{\text{total}} = p_1 + p_2$$

So,

$$y_1 = \frac{p_1}{P_{\text{total}}} = \frac{x_1 \cdot p_1^0}{x_1 \cdot p_1^0 + x_2 \cdot p_2^0}$$

Then it rewrites the expression as

$$\frac{y_1}{x_1} = \frac{p_1^0}{p_1^0 + \frac{x_2}{x_1} \cdot p_2^0}$$

and after inversion states

$$\frac{1}{y_1} = \frac{1}{x_1} \cdot \frac{p_1^0 + x_2/x_1 \cdot p_2^0}{p_1^0}$$

The page then mentions the slope as $\frac{p_2^0}{p_1^0}$ and the intercept as $1 - \frac{p_1^0}{p_2^0}$, but finally concludes that the correct answer is option A.

Since the instruction is to treat the solution, the final marked answer is A.