Question

The freezing point of a solution prepared by dissolving $20.5g$ of a non- volatile non- electrolyte with empherical formula $\left(C_3{H}_2\right)$ in $400g$ of benzene is $4.33^{\circ }C$. The benzene used to prepare the solution freezes at $5.48^{\circ }C$, using the same thermometer. (Given $:Kf$ of benzene $=5.1kgKmo{l}^{-1},Kb$ of benzene $=2.53kgKmo{l}^{-1}$ and boiling point of benzene $80.2^{\circ }C$ ).

• A. The boiling point of the solution will be $80.17^{\circ }C$.
• B. The boiling point of the solution will be $80.77^{\circ }C$
• C. The molecular formula of non electrolyte is $C_{18}{H}_{12}$.
• D. The molecular formula of non- electrolyte is $C_{15}{H}_{12}$

Answer 1

Answer: (B), (C)

$\Delta T_f=K_f\times m$
$(5.48-4.33)=5.1\times \frac{20.5}{38n}\times \frac{1000}{400}$
$n\approx 6.0$
Moleculár formula of non- electrolyte $={\left(C_3{H}_2\right)}_6$
$=C_{18}{H}_{12}.$
Molality of solution $(m)-\frac{20.5}{38\times 6}\times \frac{1000}{400}$
$\Delta Tb=Kb\times m$
$T_b^{\prime }=T_b+2.53\times \frac{20.5\times 5}{38\times 12}$
$\simeq 0.57$
$T_b^{\prime }=T_b+0.57=80.2+0.57=80.77^{\circ }C$
Alternatively, the carbons and hydrogens in the compound must be in the ratio of $3:2$, which is true in $C_{18}{H}_{12}$. The boiling of solution will be higher than $80.2^{\circ }C$, so it would mean that (d) cannot be the right choice and calculation suggest that the boiling point of solution is $80.77^{\circ }C$.