Two 5 molal solutions are prepared by dissolving a non-electrolyte, non-volatile solute separately in the…
Two 5 molal solutions are prepared by dissolving a non-electrolyte, non-volatile solute separately in the solvents $\mathrm{X}$ and $\mathrm{Y}$. The molecular weights of the solvents are $\mathrm{M}_{\mathrm{X}}$ and $\mathrm{M}_{\mathrm{Y}}$, respectively where $\mathrm{M}_X=\frac{3}{4} \mathrm{M}_{\mathrm{Y}}$. The relative lowering of vapour pressure of the solution in $\mathrm{X}$ is " $\mathrm{m}$ " 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 "m" is:
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{4}{3}$
Solution
The relationship between molar masses of the two solvents is
$
\mathrm{M}_{\mathrm{X}}=\frac{3}{4} \mathrm{M}_{\mathrm{Y}}
$
The relative lowering of vapour pressure of the two solutions is
$
\left(\frac{\Delta \mathrm{P}}{\mathrm{P}}\right)_{\mathrm{X}}=\mathrm{m}\left(\frac{\Delta \mathrm{P}}{\mathrm{P}}\right)_{\mathrm{Y}}
$
But, the relative lowering of vapour pressure of solutions is directly proportional to the mole fraction of solute.
Given 5 molal solution, means 5 moles of solute are dissolved in $1 \mathrm{~kg}$ ( or $1000 \mathrm{~g}$ ) of solvent.
The number of moles of solvent $=\frac{1000 \mathrm{~g}}{\mathrm{M}}$
The mole fraction of solute $=\frac{5}{1000 / \mathrm{M}}$
$
=\mathrm{M} \times \frac{5}{1000}
$
hence $\mathrm{M}_{\mathrm{X}} \times \frac{5}{1000}=\mathrm{m} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000}$.
Substitute equation (i) in equation (ii)
$
\begin{aligned}
&\frac{3}{4} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000}=\mathrm{m} \times \mathrm{M}_{\mathrm{Y}} \times \frac{5}{1000} \\
&\mathrm{~m}=\frac{3}{4}
\end{aligned}
$