0.56 g of an organic compound was kjeldahlised. Ammonia evolved was passed in $50 c m^{3}$ of $0.5 M H_{2} S O_{4}$ . Residual acid needed $60 c m^{3}$ of $\frac{M}{2} N a O H$ solution. The percentage composition of $^{'} N^{'}$ in this compound is:

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Solution

The correct option is D 56
Kjeldahl's method formula is
% $N = \frac{1.4 \times (N_{A} V_{A} - N_{B} V_{B})}{W t . o f o r g a n i c c o m p o u n d}$
Wt. of organic compound $= 0.56 g$
$N_{A} = (0.5) 2$ $N = 1 N$ $N_{B} = 0.5 (M) = 0.5 N$
$V_{A} = 50 {c m}^{3} = 50 m l$ $V_{B} = 60 {c m}^{3} = 60 m l$
% $N = \frac{(1.4) (50 - \frac{60}{2})}{0.56} = \frac{140}{56} (20) = 50$
$∴$ percentage of $N$ in compound is $56$ .

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