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 $(ii)$

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

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

 

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free