Question

0.5 g of a nonvolatile organic compound (molecular weight 65) is dissolved in $100mLofCC{l}_4$. If the vapour pressure of pure $CC{l}_{4}is140mmHg$, then calculate the vapour pressure of the solution.
$\text{(Density of}CC{l}_4\text{solution is}1.6gm{L}^{-1}$

• A. $145.5mmHg$
• B. $138.9mmHg$
• C. $165.2mmHg$
• D. $94.7mmHg$

Answer 1

The correct option is B $138.9mmHg$

$P_{CC{l}_4}^{\circ }=140mmHg\text{Weight of}CC{l}_4=V_{CC{l}_4}\times {\rho }_{CC{l}_4}=100\times 1.6=160g$

$\therefore n_{CC{l}_4}=\frac{160}{12+(4\times 36.5)}=\frac{160}{154}\approx 1.04mol\text{Weight of non volatile solute}=0.5g$

$n_{solute}=\frac{0.5}{65}=7.7\times {10}^{-3}mol$
$\therefore {\chi }_{solute}=\frac{n_{solute}}{n_{total}}=\frac{7.7\times {10}^{-3}}{1.04+7.7\times {10}^{-3}}{\chi }_{solute}=7.35\times {10}^{-3}$

For dilute solution:

${\chi }_{solute}=\frac{140-P_{Solution}}{140}7.35\times {10}^{-3}=\frac{140-P_{Solution}}{140}$

$P_{solution}=140\times (1-7.45\times {10}^{-3})=138.96mmHg$