0.303 gr of an organic compound was analysed for Nitrogen by kjeldahl’s method. The $N H_{3}$ evolved was obsorbed in 50 ml of 0.1 N $H_{2} S O_{4}$ . The excess acid requires 25 ml of 0.1 N $N a O H$ for neutralization. The percentage of Nitrogen is

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20.3%

11.55%

46.6%

31.2%

answer is B.

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Detailed Solution

$% o f N$ $\begin{array}{l} = \frac{1.4 X N_{1} (V_{1} - \frac{N_{2} V_{2}}{N_{1}})}{u t o f o . c} \\ = \frac{1.4 X 0.1 (50 - \frac{0.1 X 25}{0.1})}{0.303} \\ = 11.55 % \end{array}$

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