If the boiling points of two solvents X ~and Y (having same...

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

Where ${T}_{b} =$boiling point of the liquid
$M =$molar mass of the liquid
${\Delta H}_{v} =$ 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 $X$ as $( {T}_{b}{)}_{x}$, and the boiling point elevation constant of solvent $Y$as $( {T}_{b}{)}_{y}$.
Given: ${M}_{x} = {M}_{y}$
${\left({T}_{b}\right)}_{x} = 2{\left({T}_{b}\right)}_{y}$
$2\left({\Delta H}_{v}\right)x = {\left({\Delta H}_{v}\right)}_{y}$
$∴ \dfrac{{\left({K}_{b}\right)}_{x}}{{\left({K}_{b}\right)}_{y}} = {\left(\dfrac{2}{1}\right)}^{2} \times \dfrac{2}{1} = \dfrac{8}{1} = m$
$∴ m = 8$

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