Question

$\text{0.80 g}$ of impure $(N{H}_4{)}_{2}S{O}_4$ was boiled with $100ml$. of a $\text{0.2 N NaOH}$ solution till all the $N{H}_3\left(g\right)$ evolved.The remaining solution was diluted to $\text{250 mL.25 mL}$ of this solution was neutralized using $5mL$ of a $\text{0.2 n H\_2SO\_4}$ solution.The percentage purity of the $(N{H}_4{)}_{2}S{O}_4$ sample is

• A. $82.5$
• B. $72.5$
• C. $62.5$
• D. $17.5$

Answer 1

Answer: (A)

$82.5$

Total m-eq. of $NaOH$ taken $=20$
m-eq. of $H_{2}S{O}_4$ = m-eq. of $NaOH$ reacted
$=\frac{5\times 0.2}{25}\times 250=10$
m-eq. of $NaOH$ reacted $=20-10=10$
$(N{H}_4{)}_{2}S{O}_4+2NaOH$
$\Rightarrow N{a}_{2}S{O}_4+2N{H}_3+2H_{2}O$
m-moles of $(N{H}_4{)}_{2}S{O}_4$ reacted $=5$
wt. of $(N{H}_4{)}_{2}S{O}_4$ in sample
$=\frac{0.66}{0.80}\times 100=82.5$