Question

Ammonium chloride dissociates as, $N{H}_{4}Cl\left(g\right)\rightleftharpoons N{H}_3\left(g\right)+HCl\left(g\right)$. The vapour density becomes half the initial value when degree of dissociation is $0.5$.

• A. True
• B. False

Answer 1

The correct option is B False

Ammonium chloride dissociates as, $N{H}_{4}Cl\left(g\right)\rightleftharpoons N{H}_3\left(g\right)+HCl\left(g\right)$ The vapour density becomes two third the initial value when degree of dissociation is 0.5. Hence, the given statement is false.
Let us assume initially 1 mole of ammonia is present. Since the degree of dissociation is 0.5, the equilibrium number of moles of ammonium chloride, ammonia and $HCl$ will be 0.5 each.
The molar masses of ammonium chloride, ammonia and $HCl$ are $53.5g/mol$, $17g/mol$ and $36.5g/mol$ respectively.
The average molar mass of the mixture is $\frac{0.5\left(53.5\right)+0.5\left(17\right)+0.5\left(36.5\right)}{0.5+0.5+0.5}=\frac{2}{3}\times 53.5$ g/mol
Thus, the average molar mass becomes two third of the initial molar mass of ammonium chloride.
Vapour density is equal to one-half the molar mass. Hence, the vapour density also becomes two third of the initial value.