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Q. $0.25\, g$ of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize $ 10\text{ }c{{m}^{3}} $ of $ 0.5\text{ }M\text{ }{{H}_{2}}S{{O}_{4}} $ . The percentage of nitrogen in the compound is

KEAMKEAM 2009Organic Chemistry – Some Basic Principles and Techniques

Solution:

From Kjeldahl's method,
Per cent of nitrogen
$ =\frac{1.4\times N\times V}{W} $ $ =\frac{1.4\times 0.5\times 2\times 10}{0.25} $ $ =56% $