Ncert Exercises & Solutions

Question 12.33:
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method.
The ammonia evolved was absorbed in 50 mL of 0.5 M H $_{2}$ SO $_{4}$ . The residual acid required
60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of
nitrogen in the compound.

Given that, total mass of organic compound = 0.50 g
60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.
60 mL of 0.5 M NaOH solution H $_{2}$ S $= \frac{60}{2} m L o f 0.5 M$ H $_{2}$ O $_{4}$ = 30 mL of 0.5 M H $_{2}$ SO $_{4}$

Acid consumed in absorption of evolved ammonia is (50–30) mL = 20 mL
Again, 20 mL of 0.5 M H $_{2}$ SO $_{4}$ = 40 mL of 0.5 MNH $_{3}$
Also, since 1000 mL of 1 MNH $_{3}$ contains 14 g of nitrogen,
40 mL of 0.5 M NH $_{3}$ will contain $\frac{14 \times 40}{1000} \times 0.5$ = 0.28 g of N
Therefore, percentage of nitrogen in 0.50 g of organic compound = $\frac{0.28}{0.50} \times 100$ 56 %

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