Discussion Forum

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^{0}C$. Assuming the concentration of solute is much lower than the concentration of solvent, the vapour Pressure (mm of Hg) of the solution is (take  $K_{b}=0.76 K kg mol^{-1})$

Answer:

Explanation:

The elevation in boiling Point is

 $\triangle T_{b}=K_{b},m:m=molality=\frac{n_{2}}{w_{1}} \times 1000$

[$n_{2}$ number of moles of solute,

$w_{1}$ = weight of solvent in gram ]

$\Rightarrow$    $2=0.76 \times \frac{n_{2}}{100}\times 1000$

$\Rightarrow$  $n_{2}=\frac{5}{19}$

Also, from Raoult's law of lowering of
vapour pressure :

$\frac{-\triangle p}{p_{0}}=x_{2}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{n_{2}}{n_{1}}[\because n_{1}>>n_{2}]$

$\Rightarrow$   $-\triangle p=760 \times \frac{5}{19} \times \frac{18}{100}$

=36 mm of Hg

 $\Rightarrow$     p=760-36=724 mm of Hg