Question

$2$ litres of $N{H}_3$ at ${30}^{0}C$ and at $0.20$ atm is neutralised by $134$ mL of a solution of $H_{2}S{O}_4$. The normality of $H_{2}S{O}_4$ is (in multiples of $100$)

Answer 1

Given for $N{H}_3$: $V=2$ L, $T=303$ K, $P=0.20$ atm.

$PV=\frac{w}{m}RT$

$\frac{w}{m}$ or moles of $N{H}_3=\frac{PV}{RT}$ $=\frac{(0.2\times 2)}{(0.082\times 303)}=0.01609$

$N{H}_3$ is monoacidic base.
Moles of $N{H}_3=$Equivalent of $N{H}_3=0.01609$
Meq. of $N{H}_3=16.09$
Meq. of $H_{2}S{O}_4=$ Meq. of $N{H}_3$
$N\times 134=16.09$
Therefore, normality of $H_{2}S{O}_4=0.12N$
So, the answer is $12$.