Question

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 $2^{\circ} C$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure $( mm\, Hg )$ of the solution is (take $K_{b}=0.76 \,K\,kg\,mol ^{-1}$ )

• A. 724
• B. 740
• C. 736
• D. 718

Answer 1

Answer: (A)

When a non-volatile solute is added to pure solvent, there is a decrease in vapour pressure.
So, according to relative lowering in vapour pressure,
$\frac{p^{\circ}-P_{s}}{p^{\circ}}=x_{2}$
where $x_{2}$ is the number of moles of solute,
$p_{o}$ is the vapour pressure of the pure solvent
and $p_{s}$ is the vapour pressure of the solution.
Now, $x_{2}$ can also represented as $\frac{n_{2}}{\left(n_{1}+n_{2}\right)}$.
Thus, we have $\frac{p^{0}-p_{s}}{p^{\circ}}=n \frac{n_{2}}{n_{1}+n_{2}}$
Neglecting $n_{2}$ (as $n_{2}< n 1$ ) from the denominator, and substituting $p^{\circ}=1 \,atm =760 \,mm\, Hg$, we get
$\frac{760-p s}{760}=\frac{2.5 / M_{B}}{100 / 18} \times \frac{1000}{1000}$
$=\frac{18}{1000} \times m(1)$
According to elevation in boiling point:
$\Delta T_{b}=K_{b} \times m$
$\Rightarrow 2=0.76 \times m \Rightarrow m =\frac{2}{0.76}$
Putting this value in Eq. (1), we get
$\frac{760-p_{s}}{760}=\frac{8}{1000} \times \frac{2}{0.76} $
$\Rightarrow p_{s}=760-\frac{18 \times 2 \times 760}{1000 \times 0.76}$
$=760-36=724 \,mm \,Hg$