Question

Ammonia undergoes self-dissociation according to the reaction, $2 NH _{3}(l) \rightleftharpoons NH _{4}^{+}{ }_{(a m)}^{+}+ NH _{2(a m)}^{-}$ where am, stands for ammoniated. When $1$ mole of $NH _{4} Cl$ is dissolved in $1\, kg$ 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. $NH_{4}Cl$ is completely dissociated in $NH_{3}$
• B. $NH_{4}Cl$ is partially dissociated in $NH_{3}$
• C. $NH_{4}Cl$ is not dissociated in $NH_{3}$
• D. Boiling point is not raised

Answer 1

Answer: (A)

$NH_{4}Cl$ is completely dissociated in $NH_{3}$

Answer 2

Vapour pressurs, $\pi \propto b p(T)$
$\therefore \frac{\pi_{1}}{\pi_{2}}=\frac{T_{1}}{T_{2}}$
$\Rightarrow \frac{760}{\pi_{2}}=\frac{240.3\, K }{306.4\, K }$
$\pi_{2}=\frac{760 \times 306.4}{240.3}=969$ torr
$\because \pi_{2}>\pi_{1}$ that means more gases are present
or $NH _{4} Cl$ gets completely dissociated into ammonia gas.