Chemistry · 2018
JEE Main · 15 April 2018, Shift 2 · Q36
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of…
Two $\displaystyle 5$ molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are $\displaystyle \mathrm{M}_{\mathrm{X}}$ and $\displaystyle \mathrm{M}_{\mathrm{Y}}$, respectively where $\displaystyle \mathrm{M}_{\mathrm{X}}=\frac{3}{4} \mathrm{M}_{\mathrm{Y}}$. The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of solvent, the value of " $\displaystyle m$ " is :
Official answer
(2)
$\displaystyle \frac{3}{4}$
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