Question

Two litres of $N{H}_3$ at $3^{0}C$ and $0.20$ atmosphere is neutralised by $134$ mL of a solution of $H_{2}S{O}_4$. If the normality of $H_{2}S{O}_4$ is $x$ N then, find the value of $100x$.

• A. $12$
• B. $16$
• C. $20$
• D. None of these

Answer 1

The correct option is A $12$

$PV=\frac{W}{Mw}RT$
$Moles\left(n\right)=\frac{W}{Mw}=\frac{PV}{RT}=\frac{0.2\times 2}{0.0821\times 303}=0.01608$ mol $=16.08$ mmol or mEq.
($1$ mmol of $N{H}_3=1$g Eq. of $N{H}_3$)
($1$ mmol of $N{H}_3=1mEq$ of $N{H}_3$)
$N\times V\left(mL\right)=mEq$
$N\times 134mL=16.08$
$\therefore N=0.12$