Solutions and Colligative Properties 2 Question 16
16. For a dilute solution containing $2.5 \mathrm{~g}$ of a non-volatile non-electrolyte solute in $100 \mathrm{~g}$ of water, the elevation in boiling point at $1 \mathrm{~atm}$ pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( $\mathrm{mm}$ of $\mathrm{Hg}$ ) of the solution is (take $K_{b}=0.76$ $\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$ ).
(2012)
(a) 724
(b) 740
(c) 736
(d) 718
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Solution:
- The elevation in boiling point is
$$ \Delta T_{b}=K_{b} \cdot m: m=\text { molality }=\frac{n_{2}}{w_{1}} \times 1000 $$
$\left[n_{2}=\right.$ Number of moles of solute, $w_{1}=$ Weight of solvent in gram]
$$ \begin{array}{ll} \Rightarrow & 2=0.76 \times \frac{n_{2}}{100} \times 1000 \ \Rightarrow & n_{2}=\frac{5}{19} \end{array} $$
Also, from Raoult’s law of lowering of vapour pressure :
$$ \begin{array}{rlrl} & \frac{-\Delta p}{p^{\circ}} & =x_{2}=\frac{n_{2}}{n_{1}+n_{2}} \approx \frac{n_{2}}{n_{1}} \quad\left[\because n_{1}>n_{2}\right] \ \Rightarrow \quad & -\Delta p & =760 \times \frac{5}{19} \times \frac{18}{100}=36 \mathrm{~mm} \text { of } \mathrm{Hg} & \ \Rightarrow \quad & p & =760-36=724 \mathrm{~mm} \text { of } \mathrm{Hg} & \end{array} $$
