Question

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 the solvents are $M_X$ and $M_Y$, respectively where $M_X = \frac{3}{4} M_Y$ 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 "m" is :

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

We have $M_{ x }=\frac{3}{4} M_{ Y } $

Given that $\frac{\Delta p_{ x }}{p_{ x }^{\circ}}=m \times\left(\frac{\Delta p_{ Y }}{p_{ r }^{ o }}\right)$

$\Delta p / p^{\circ}$ represent relative lowering of vapour pressure.

Relative lowering of vapour pressure depends only on the mole fraction of the solute.

$M_{ x } \times \frac{5}{1000}=m \times M_{ Y } \times \frac{5}{1000}$

From Eq. (1) and Eq. (2), we get

$\frac{3}{4} \times M_{ Y } \times \frac{5}{1000} =m \times M_{ Y } \times \frac{5}{1000} $

$m =\frac{3}{4}$