Question

$0.70g$ of mixture (NH ${}_4{)}_2{SO}_4$ was boiled with $100ml$ of $0.2N$ NaOH solution till all the $N{H}_3$ (g) evolved and get dissolved in solution itself. The remaining solution was diluted to $250mL.25mL$ of this solution was neutralized using $10mL$ of $a$ $0.1N{H}_2{SO}_4$ solution. The percentage purity of the ${\left({NH}_6\right)}_2{SO}_4$ sample is:

• A. $94.3$
• B. $50.8$
• C. $47.4$
• D. $79.8$

Answer 1

The correct option is A $94.3$

meq. of $N{H}_3=10$

m.mole of $N{H}_3=10$

m.mole of ${N{H}_4}_{2}S{O}_4=5$

Wt. of ${N{H}_4}_{2}S{O}_4=\frac{5}{1000}\times 132=0.66gm$

% of ${\left(N{H}_4\right)}_{2}S{O}_4=\frac{0.66}{0.7}\times 100=94.28$%