The vapour pressure of a solution of a non- volatile solute $B$ in a solvent $A$ is $95 %$ of the vapour pressure of the pure solvent at the same temparature. If the molecular weight of the solvent is $0.3$ times the molecular weight of the solute, what is the ratio of the weight of solvent to solute?

0.15

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5.7

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0.2

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None of these

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Solution

The correct option is B $5.7$
As given, vapour pressure of the solution is $95 %$ of the pure solvent
$∴ P_{s} = 0.95 P_{0}$
According to Raoult's law,
$P_{s} = P_{0} χ_{A}$
where, $χA=mole fraction of solvent$
Given, $M_{A} = 0.3 M_{B}$
$M_{A} =$ molecular weight of solvent
$M_{B} =$ Molecular weight of solute
$W_{A} =$ mass of solvent
$W_{B} =$ mass of solute
$∴ 0.95 P_{0} = P_{0} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{W_{A}}{M_{A}}}{\frac{W_{A}}{M_{A}} + \frac{W_{B}}{M_{B}}} ⎞ ⎟ ⎟ ⎟ ⎠$
$\Rightarrow 0.95 = \frac{\frac{W_{A}}{0.3 M_{B}}}{\frac{W_{A}}{0.3 M_{B}} + \frac{W_{B}}{M_{B}}}$
$\Rightarrow 0.95 = \frac{W_{A}}{W_{A} + 0.3 W_{B}}$
$\Rightarrow 0.95 W_{A} + 0.285 W_{B} = W_{A}$
$\Rightarrow 0.05 W_{A} = 0.285 W_{B}$
$\Rightarrow \frac{W_{A}}{W_{B}} = \frac{0.285}{0.05}$
$\Rightarrow \frac{W_{A}}{W_{B}} = 5.7$

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