If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio $2: 1$ and their enthalpy of vaporizations are in the ratio $1: 2$, then the boiling point elevation constant of $\mathrm{X}$ is $\underline{\mathrm{m}}$ times the boiling point elevation constant of Y. The value of m is ____________ (nearest integer)

The boiling point elevation constant, also known as the ebullioscopic constant ($K_{b}$), can be determined using the formula: <br/><br/> $K_{b} = \frac{R \cdot T_{b}^{2}}{1000 \cdot \Delta H_{vap}}$ <br/><br/> where:<br/><br/> - $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)<br/><br/> - $T_{b}$ is the boiling point of the solvent (in K)<br/><br/> - $\Delta H_{vap}$ is the enthalpy of vaporization (in J/mol) <br/><br/> 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. <br/><br/> 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 $\Delta H_{vap}(X)$. <br/><br/> We then have: <br/><br/> $T_{b}(X) = 2T_{b}(Y)$ and $\Delta H_{vap}(X) = 0.5\Delta H_{vap}(Y)$ <br/><br/> We can substitute these values into the equation for $K_{b}$: <br/><br/> $K_{b}(X) = R*(2T_{b}(Y))^2 / (1000 * 0.5\Delta H_{vap}(Y)) = 4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))$ <br/><br/> and <br/><br/> $K_{b}(Y) = R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))$ <br/><br/> Comparing these two equations: <br/><br/> $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))] $ <br/><br/> This simplifies to: <br/><br/> $K_{b}(X) / K_{b}(Y) = 4 / 0.5 = 8$ <br/><br/> So, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.