Question

The amount of ammonia gas at NTP required to be passed into a $30mL$ of $N{H}_{2}S{O}_4$ solution to bring down the acid normality to $0.2$ N is (to the nearest integer) :

Answer 1

Initial Meq.of $H_{2}S{O}_4=30\times 1=30$
Meq.of $H_{2}S{O}_4$ after passing $N{H}_3=30\times 0.2=6$
$\therefore$ Meq.of $H_{2}S{O}_{4}lost=30-6=24$
Since, Meq. of $N{H}_3$ $=$ Meq.of $H_{2}S{O}_{4}lost=24$
$\therefore \left(w/17\right)\times 1000=24$
or, $w_{N{H}_3}=0.408$
$\therefore$ Volume of $N{H}_3$ at STP $=$ $(22.4\times 0.408)/17$
$=0.5376L=537.6mL$