Question

Ammonia undergoes self-dissociation according to the reaction,
$2N{H}_3(l)\rightleftharpoons N{H}_4^{+}{}_{(am)}^{+}+N{H}_{2(am)}^{-}$
where am, stands for ammoniated. When $1$ mole of $N{H}_{4}Cl$ is dissolved in $1kg$ of liquid ammonia, the b. p. at $760$ tour is observed at $-32^{\circ }C$ (Normal b.p. of $N{H}_{3(l)}$ is $-33.4^{\circ }C$ ). What conclusions are reached about the nature of the solution?

• A. $N{H}_{4}Cl$ is completely dissociated in $N{H}_3$
• B. $N{H}_{4}Cl$ is partially dissociated in $N{H}_3$
• C. $N{H}_{4}Cl$ is not dissociated in $N{H}_3$
• D. Boiling point is not raised

Answer 1

Answer: (A)

Vapour pressurs, $\pi \propto bp(T)$
$\therefore \frac{{\pi }_1}{{\pi }_2}=\frac{T_1}{T_2}$
$\Rightarrow \frac{760}{{\pi }_2}=\frac{240.3K}{306.4K}$
${\pi }_2=\frac{760\times 306.4}{240.3}=969$ torr
$\because {\pi }_2>{\pi }_1$ that means more gases are present
or $N{H}_{4}Cl$ gets completely dissociated into ammonia gas.