Question

Calculate the boiling point of a solution containing 0.60 g of benzoic acid in 50 g of $C{S}_2\left(l\right)$ assuming 90% association of the acid.
The boiling point and $K_b$ for $C{S}_2$ are ${46.2}^{\circ }C$ and $2.44{}^{\circ }\text{C kg/mol}$ respectively.

• A. $46.33{}^{\circ }C$
• B. $47.31{}^{\circ }C$
• C. $48.94{}^{\circ }C$
• D. $46.20{}^{\circ }C$

Answer 1

The correct option is A $46.33{}^{\circ }C$

$△T=i\times K_b\times \text{Molality}△T=i\times K_b\times \frac{W_{C_6{H}_{5}COOH}\times 1000}{M_{C_6{H}_{5}COOH}\times W_{C{S}_2}}$

If out of every mole of solute, $\alpha$ moles have associated, $\alpha /2$ moles of dimer are present in the solution.
$\Rightarrow$ Total moles of solute $n=1-\alpha /2$
$\Rightarrow i=1-\alpha /2=0.55$
$△T_b=i{K}_{b}m$
$=0.55\times 2.44\times \frac{0.60\times 1000}{122\times 50}=0.132{}^{\circ }C$

Final boiling point $=0.132+46.2=46.33{}^{\circ }C$