QJEEM23Ch9dd9: JEE Main 2023 Chemistry - Class 12 JEEM23Ch9dd9 Solution: If the boiling points of two solvents X and Y (having same…

The boiling point elevation constant, also known as the ebullioscopic constant ( $K_{b}$ ), can be determined using the formula:

$$K_b=\frac{R\cdot T_b^2}{1000\cdot \mathrm{\Delta }{H}_{vap}}$$

where:

- $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)

- $T_{b}$ is the boiling point of the solvent (in K)

- $\Delta H_{vap}$ is the enthalpy of vaporization (in J/mol)

According to the problem, the ratio of boiling points for X and Y is 2:1, and the ratio of their enthalpy of vaporization is 1:2.

Let's denote the boiling point of Y as $T_{b}(Y)$ and the boiling point of X as $T_{b}(X)$ , and likewise for the enthalpy of vaporization $\Delta H_{vap}(Y)$ and $\Delta H_{vap}(X)$ .

We then have:

$T_{b}(X) = 2T_{b}(Y)$ and $\Delta H_{vap}(X) = 0.5\Delta H_{vap}(Y)$

We can substitute these values into the equation for $K_{b}$ :

$K_{b}(X) = R*(2T_{b}(Y))^2 / (1000 * 0.5\Delta H_{vap}(Y)) = 4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))$

and

$K_{b}(Y) = R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))$

Comparing these two equations:

$K_{b}(X) / K_{b}(Y) = [4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))] / [R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))]$

This simplifies to:

$K_{b}(X) / K_{b}(Y) = 4 / 0.5 = 8$

So, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.