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 X is \underlinem times the boiling point elevation constant of Y. The value of m is \underline\hspace2cm (nearest integer)

Solutions PYQ — JEE Main 2023 · 8 Apr · Shift 2 · Q17 with Solution

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Correct answer: 8

For a solvent, the boiling point elevation constant is

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

where:

Since X and Y have the same molecular weight, $M_X = M_Y$ , so this factor cancels.

Thus,

$\frac{K_{b,X}}{K_{b,Y}} = \frac{T_{b,X}^2}{T_{b,Y}^2}\cdot \frac{\Delta H_{vap,Y}}{\Delta H_{vap,X}}$

Given:

$T_{b,X} : T_{b,Y} = 2:1$

So,

$\frac{T_{b,X}^2}{T_{b,Y}^2} = \left(\frac{2}{1}\right)^2 = 4$

Also,

$\Delta H_{vap,X} : \Delta H_{vap,Y} = 1:2$

Hence,

$\frac{\Delta H_{vap,Y}}{\Delta H_{vap,X}} = \frac{2}{1} = 2$

$\frac{K_{b,X}}{K_{b,Y}} = 4 \times 2 = 8$

So, if the boiling point elevation constant of X is $m$ times that of Y, then

$m = 8$

Stored correct answer = 8

Our derived answer also = 8, so they agree.