Question

A sample of 0.25 g of an organic compound was
treated according to Kjeldahl's method. The
ammonia evolved was absorbed in 100 mL of 0.5 N
H2S04. The remaining acid after neutralisation by
ammonia consumed 80 mL of 0.5 N NaOH. The
percentage of nitrogen in the organic compound is
​

Answer 1

Given:

sample of the organic compound used = 0.25g

volume of acid taken= 100ml of 0.5N H2S04 = 50ml of 1N H2S04

volume of alkali used for neutralisation of excess acid

= 80ml of 0.5N NaOH = 40ml of 1N NaOH.

To Find:

The  percentage of nitrogen in the organic compound =?

Solution:

Now, through the reaction involved during Kjeldahl's method, we know that

1 mol H2S04 neutralise 2mol NaOH

thus, 40 mL of 1N  NaOH = 20 ml of 1 N H2S04

the volume of acid used by ammonia = 50-20 = 30ml

% Nitrogen = $\dfrac{1.4 \times N1 \times V}{w}$

here N1 = normality of standard acid = 0.5

        w = sample of the organic compound used = 0.25g

        v = volume of acid neutralised by ammonia = 30ml

On putting values,

% Nitrogen =$\dfrac{1.4 \times0.5 \times 30}{0.25}$ = 84%

Thus, the percentage of nitrogen in an organic compound is 84%