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$6 g$ of a mixture of ammonium sulphate and ammonium chloride was made up to $1000 c c$ with water. $25 c c$ of this solution was boiled with $50 c c$ of $\frac{M}{10}$ sodium hydroxide until no more ammonia was evolved and it was then found that the excess of sodium hydroxide required $24.3 c c$ of $\frac{M}{10}$ hydrochloric acid for neutralization. What was 10 mass percentage of ammonium chloride in the mixture?

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answer is 56.

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Detailed Solution

Here, the complete reaction is given as,

$\underset{a mol}{{({NH}_{4})}_{2} {SO}_{4}} + 2 NaOH ⟶ 2 {NH}_{3} + {Na}_{2} {SO}_{4} + H_{2} O$ $\underset{b m o l}{{NH}_{4} Cl + NaOH} ⟶ {NH}_{3} + NaCl + H_{2} O$

milli equivalent of NaOH used $= 50 \times \frac{1}{10} - 24.3 \times \frac{1}{10}$

$\begin{array}{l} = 2.57 \\ 2 a + b = \frac{2.57}{1000} \times \frac{1000}{25} \end{array}$ ...(i)

$a \times 132 + b \times 53.5 = 6$ ...(ii)

hus, the mass percentage of ammonium chloride in this mixture is 56%.

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