$0.2 g$ of an organic compound was analysed by Kjeldahl’s method. The ammonia gas evolved was absorbed by $60 mL of N / 5 H_{2} S O_{4}$ . The unused acid required $40 mL of N / 10 N a O H$ for complete neutralisation. Find the percentage of $N_{2}$ in the sample.

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Solution

This can be solved on the basis of law of chemical equivalance
According to which;
$g r a m e q u i v a l e n t s o f N H_{3} + g r a m e q u i v a l e n t s o f N a O H = g r a m e q u i v a l e n t s o f H_{2} S O_{4}$
$m o l e s \times n f + \frac{N_{1} \times V_{1}}{1000} = \frac{N_{2} \times V_{2}}{1000}$
$\to n \times 1 = \frac{8}{1000}$
$\to n = \frac{8}{1000} = 8 \times 10^{- 3}$
Now,
1 mole of $N H_{3}$ contains 1 mole of `N`.
$∴$ $8 \times 10^{- 3}$ moles of $N H_{3}$ contains $8 \times 10^{- 3}$ moles of nitrogen.
Hence,
mass of `N `= moles $\times$ molar mass
= $8 \times 10^{- 3} \times 14$
= 0.112 g
$∴$ % of `N` = $\frac{0.112}{0.2} \times 100$

= `56 %`

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