JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Solutions

The molal boiling point elevation constant is $K_b=\frac{R{T}_b^{2}M}{\Delta H_{vap}}$, and with equal molar masses the ratio depends only on $T_b^2/\Delta H_{vap}$. The boiling points of X and Y are in the ratio two to one, so the square of this ratio contributes a factor of four, and the enthalpies of vaporisation are in the ratio one to two, so their reciprocal ratio contributes a further factor of two. Multiplying gives $4\times 2=8$, so $K_b$ of X is eight times $K_b$ of Y and $m=8$.