Question

The freezing point of a solution prepared by dissolving $20.5 \,g$ of a non- volatile non- electrolyte with empherical formula $\left( C _{3} H _{2}\right)$ in $400 \,g$ 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 $=5.1 \,kg \,K \,mol ^{-1}, Kb$ of benzene $=2.53\,kg \,K \,mol ^{-1}$ 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}{38 n } \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 $80.77^{\circ} C$.