Question

The vapour density of fully dissociated ${NH}_{4}Cl$ would be:

• A. Less than half of the vapour density of pure ${NH}_{4}Cl$
• B. Double of the vapour density of pure ${NH}_{4}Cl$
• C. Half of the vapour density of pure ${NH}_{4}Cl$
• D. One-third of the vapour density of pure ${NH}_{4}Cl$

Answer 1

The correct option is C Half of the vapour density of pure ${NH}_{4}Cl$

The vapour density of fully dissociated ${NH}_{4}Cl$ would be half of the vapour density of pure ${NH}_{4}Cl$

During thermal dissociation of gas, the vapour density decreases as the molecular weight also decreases.

$N{H}_{4}Cl\to N{H}_3+HCl$

Molecular weight $=2\times$ vapour density.

The molecular weights of $N{H}_{4}Cl$, $N{H}_3$ and $HCl$ are 53.5 g/mol, 17 g/mol and 36.5 g/mol respectively. The average molecular weight of fully dissociated $N{H}_{4}Cl$ will be $\frac{17+36.5}{2}=26.75$ g/mol which is one half of 53.5 g/mol (the molecular weight of pure undissociated $N{H}_{4}Cl$).

Hence, option $C$ is correct.