Two 5 molal solutions are prepared by dissolving a non-electrolyt… - Vidyadip

MCQ

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

✓
$\frac{3}{4}$
B
$\frac{1}{2}$
C
$\frac{1}{4}$
D
$\frac{4}{3}$

✓

Answer

Correct option: A.

$\frac{3}{4}$

a
The relationship between molar masses of the two solvents is

$M_X\,=\,\frac {3}{4}M_Y$ .... $(i)$

The relative lowering of vapour pressure of two solution is

${\left( {\frac{{\Delta P}}{P}} \right)_X}\, = \,m{\left( {\frac{{\Delta P}}{P}} \right)_Y}$

But, the relative loering of vaoour pressure of solutin is directly proportional to the mole fraction of solution.

Given $5\,molal$ solution , mans $5\,molal$ of solute are dissolved in $1\,kg$ (or $1000\,g$) of solvent.

The number of moles of solvent $=\,\frac {1000\,g}{M}$

The mole fraction of solution $=\,\frac {5}{1000/M}$

$=M\times \,\frac {5}{1000}$

hence $M_X\times \frac {5}{1000}$ $=\,m\times M_Y\times \frac {5}{1000}$ .... $(ii)$

Substitute equation $(i)$ in equation $(ii)$

$\frac {3}{4}\times M_Y\times \frac {5}{1000}$ $=\,m\times M_Y\times \frac {5}{1000}$

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

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