The ammonia evolved from the treatment of $0.30 g$ of an organic compound for the estimation of nitrogen was passed in $100 m L$ of $0.1 M$ sulphuric acid. The excess of acid required $20 m L$ of $0.5 M$ sodium hydroxide solution for complete neutralization. The organic compound is:

acetamide

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benzamide

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urea

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thiourea

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Solution

The correct option is C urea
$H_{2} S O_{4} + 2 K O H \to K_{2} S O_{4} + 2 H_{2} O$
$\frac{M_{1} V_{1}}{1}$ (excess $H_{2} S O_{4}) = \frac{M_{2} V_{2}}{2} (K O H)$
$\frac{0.1 \times V_{1}}{1} = \frac{0.5 \times 20}{2}$
$V_{1} = 50 m L$
Volume of sulphuric acid used up to absorb $N H_{3} = 50 m L$
% $N = \frac{2.8 \times M V}{W} = \frac{2.8 \times 0.1 \times 50}{0.3} = 46.6$
Thus, the organic compound will be urea, which has $46.6$ % nitrogen.

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