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Q.

$0.303 g$ of sample was analysed for nitrogen by Kjeldahl’s method. The ammonia gas evolved was absorbed in $50 m l o f 0.05 M H_{2} S O_{4}$ . The excess of acid required $25 m l o f 0.1 M N a O H$ for neutralization. The percentage of Nitrogen in the compound.

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a

11.5

b

23

c

12.5

d

14.5

answer is A.

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Detailed Solution

$\begin{array}{l} N_{1} V_{1} = N_{1} V_{2} \\ 25 \times 0.1 = 0.05 \times 2 \times V_{2} \\ V_{2} = 25 m l \\ % N = \frac{1.4 \times V o l a c i d \times N o f a c i d}{w t o f c o m p o u n t} \\ = \frac{1.4 \times 25 \times 0.05 \times 2}{0.303} \\ = 11.55 \end{array}$

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