JEE Main 2023 — Solutions Question with Solution

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

$$K_b=\frac{R\cdot T_b^2}{1000\cdot \Delta H_{vap}}$$

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 $K_b$:

$K_b(X)=R\ast (2T_b(Y){)}^2/(1000\ast 0.5\Delta H_{vap}(Y))=4\ast R\ast T_b(Y{)}^2/(500\ast \Delta H_{vap}(Y))$

and

$K_b(Y)=R\ast T_b(Y{)}^2/(1000\ast \Delta H_{vap}(Y))$

Comparing these two equations:

$K_b(X)/K_b(Y)=[4\ast R\ast T_b(Y{)}^2/(500\ast \Delta H_{vap}(Y))]/[R\ast T_b(Y{)}^2/(1000\ast \Delta H_{vap}(Y))]$

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.