# The relative lowering of vaour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be Option 1) 18.0 Option 2) 342 Option 3) 160 Option 4) 180

D Divya Saini

As we learned in concept

Expression of relative lowering of vapour pressure -

$\frac{\Delta P}{ P^{0}}= x_{solute}$

$x_{solute}= \frac{ n_{solute}}{ n_{solute}+n_{solvent}}$

- wherein

$\Delta P \: is \: lowering \: o\! f \: v.p.$

$P^{0}\rightarrow \: vapour\: pressure\: of \: pure\: solvent$

$x_{solute}\rightarrow \: mole\: fraction \: of \:non\: volatile\: solute$

Suppose $\frac{\Lambda P}{P_0}=$ Relative lowering of vapour pressure

then, $\frac{\Lambda P}{P_0}= \frac{ n_{solute}}{ n_{solute}+n_{solvent}}$

$0.00713=\frac{71.5/m}{71.5/m +1000m/18}$

$m=\frac{71}{0.396}=179.3 \sim 180$

Option 1)

18.0

Incorrect

Option 2)

342

Incorrect

Option 3)

160

Incorrect

Option 4)

180

Correct

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