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

 

Answers (1)
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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