Question

The boiling point of a solution containing 0.61 g of benzoic acid in 50 g of carbon disulphide assuming $84\%$ dimerization of the acid, where boiling point and Kb of $C{S}_2$ being ${46.2}^{\circ }C$ and $2.3Kkgmo{l}^{-1}$ is :

• A. ${46.3444}^{\circ }C$
• B. ${46.3434}^{\circ }C$
• C. ${46.3334}^{\circ }C$
• D. ${46.3344}^{\circ }C$

Answer 1

The correct option is C ${46.3334}^{\circ }C$

We have, $\Delta T=\frac{(1000\times K_b\times w)}{m\times W}$
$\Delta T_N=\frac{(1000\times 2.3\times 0.61)}{(122\times 50)}[m.wt.of{C}_6{H}_{5}COOH=122]\Delta T_N=0.23$
$\Delta T_{exp}=\Delta T_N(1-\alpha +\frac{\alpha }{n})=0.23\times \{1-0.84+\left(\frac{0.84}{2}\right)\}=0.23\times (1-\frac{0.84}{2})={46.3334}^{\circ }C\left(C\right)$