Question

$0.5g$ of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in $150mL$ of $N/5H_{2}S{O}_4$ solution. Excess sulphuric acid required $20mL$ of $1NNaOH$ for complete neutralization. The percentage of $N{H}_3$ in the ammonium chloride is:

• A. $68$%
• B. $34$%
• C. $48$%
• D. $17$%

Answer 1

Answer: (B)

$34$%

$\left(N{H}_3\right)Cl+NaOH\to +N{H}_3\uparrow +H_{2}O$

excess $H_{2}S{O}_4$ reg. $20ml$ of $1NNaOH$

$\Rightarrow$ moles of $H_{2}S{O}_4=(20\times 1)milimoles$

$=20mmol$

added amount of $H_{2}S{O}_4=\frac{N}{5}\times 150ml=30mmol$

amount of $h_{2}s{o}_4$ reacted with $n{h}_3=0.01MOLE$

$\Rightarrow 0.01mole$ of $N{H}_3$ present in $\left(N{H}_4\right)Cl$

So, $\%purity=\frac{\left(0.01\right)\times 17}{0.5}\times 100=34\%$