Question

$105$ mL of pure water at $4$${}^{o}C$ is saturated with $N{H}_3$ gas yielding a solution of density $0.9g{mL}^{-1}$ and containing $30\%$ $N{H}_3$ by mass. The volume (in litres) of $N{H}_3$ gas at $4$${}^{o}C$ and $775$ mm of Hg, which was used to saturate the water is (as the nearest integer) :

Answer 1

$\because$ Solution has $30\%$ $N{H}_3$ by mass.

Therefore, $100$ g solution has $30$ g $N{H}_3$ dissolved in it.
or $70$ g $H_{2}O$ is saturated by $N{H}_3=$ $30$ g
$\therefore$ $105$ g $H_{2}O$ will be saturated by $N{H}_3$ $=\frac{30\times 105}{70}=45$ g $N{H}_3$
$\therefore$ Mass of saturated solution $=$ $100+45$ $=$ $150$ g
Volume of saturated solution $=\frac{150}{\text{density}}=\frac{150}{0.9}=166.67$ mL
Now, volume of $N{H}_3$ ($45$ g) at $4^o$ and $775$ mm pressure,
$PV=\frac{w}{m}RT$

$⟹$$\frac{775}{760}\times V=\frac{45}{17}\times 0.0821\times 277$

$⟹$ $V=59.03=59$ L