Question

$10g$ sample

of gas liquor $\left(N{H}_4^{+}salt\right)$ is boiled with NaOH and the resulting $N{H}_3$
is passed into $60ml$ of $0.90$ $N{H}_{2}S{O}_4$ . Excess $H_{2}S{O}_4$ required

$10c{m}^3$ of $0.40N$ $NaOH$ . What is the of $N{H}_3$ in gas liquor ?

• A. $8.5$
• B. $9.5$
• C. $7.9$
• D. $10.5$

Answer 1

The correct option is C $7.9$

Me of $H_{2}S{O}_4$ reacted with gas liquor $=(60\times 0.9)-(10\times 0.40)=1.4$

Me = me of $N{H}_3$ in the gas liquor $=w/17/3$

hence $w=1.4\times 17/3=7.93$ g