If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio 2: 1 and…

The boiling point elevation constant $\left({\mathrm{K}}_{\mathrm{b}}\right)$ of a volatile liquid is given by the following expression
${\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{RT}}_{\mathrm{b}}^2\mathrm{M}}{{\mathrm{\Delta H}}_{\mathrm{v}}}$

Where ${\mathrm{T}}_{\mathrm{b}}=$boiling point of the liquid
            $\mathrm{M}=$molar mass of the liquid
            ${\mathrm{\Delta H}}_{\mathrm{v}}=$ Enthalpy of vaporisation

The boiling points of the two solvents are in the ratio $2:1$, which means that the boiling point elevation constants are also in the ratio $2:1$. Let's call the boiling point elevation constant of solvent $\mathrm{X}$ as $({\mathrm{T}}_{\mathrm{b}}{)}_{\mathrm{x}}$, and the boiling point elevation constant of solvent $\mathrm{Y}$as $({\mathrm{T}}_{\mathrm{b}}{)}_{\mathrm{y}}$.
Given: ${\mathrm{M}}_{\mathrm{x}}={\mathrm{M}}_{\mathrm{y}}$
${\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{x}}=2{\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{y}}$
$2\left({\mathrm{\Delta H}}_{\mathrm{v}}\right)\mathrm{x}={\left({\mathrm{\Delta H}}_{\mathrm{v}}\right)}_{\mathrm{y}}$
$\therefore \frac{{\left({\mathrm{K}}_{\mathrm{b}}\right)}_{\mathrm{x}}}{{\left({\mathrm{K}}_{\mathrm{b}}\right)}_{\mathrm{y}}}={\left(\frac{2}{1}\right)}^2\times \frac{2}{1}=\frac{8}{1}=\mathrm{m}$
$\therefore \mathrm{m}=8$

Therefore, the value of m is $8$(nearest integer).