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Q.

0.80 gram of impure ${(N H_{4})}_{2} S O_{4}$ was boiled with 100 ml of a 0.2 N NaOH solution till all the $N H_{3}$ is evolved. The remaining solution was diluted to 250 ml, 25 ml of this solution was neutralized by using 5ml of 0.2 $N H_{2} S O_{4}$ solution. The percentage purity of the sample is ${(N H_{4})}_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O + 2 N H_{3}$

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a

76.5

b

66.5

c

33.5

d

82.5

answer is D.

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Detailed Solution

Meq of $H_{2} S O_{4} = M e q$ of NaOH remaining
$\begin{array}{l} = \frac{5 \times 0.2 \times 250}{25} \\ = 10 \end{array}$
Meq of NaOH consumed = 20-10
=10
M mole of ${(N H_{4})}_{2} S O_{4}$ reacted = 5
Wt of ${(N H_{4})}_{2} S O_{4} = 5 \times 10^{- 3} \times 132 = 0.66 g$
% purity $= \frac{0.66}{0.80} \times 100 = 82.5$

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