Question

For a dilute solution containing $3.2g$ of a non-volatile nonelectrolyte solute in $100g$ water, the elevation in boiling point at $2$ atm pressure is $3^{\circ }C$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of $Hg$ ) of the solution is (Take, $K_b=0.76Kmo{l}^{-1}kg)$

• A. 724
• B. 740
• C. 742
• D. 718

Answer 1

Answer: (C)

$\Delta T_b=\frac{1000K_b{w}_1}{m_1{w}_2}$

$3=1000\times 0.76\frac{w_1}{m_1{w}_2}$

$\therefore \frac{w}{m_1}-=\frac{3}{1000\times 0.76}=\frac{3}{760}$

By Raoult's law,

$\frac{p^{\circ }-p_s}{p^{\circ }}={\chi }_{\text{solute }}=\frac{n_1}{n_1+n_2}$

But, $n_2>>n_1$

Thus, $\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{n_1}{n_2}=\frac{w_1{m}_2}{m_1{w}_2}$

$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{18}{760}$

$(m_2$ molar mass of solvent $=18)$

$p^{\circ }-p_s=\frac{18}{760}\times 760=18mmHg$

$\therefore p_s=p^{\circ }-18=760-18=742mmHg$