Question

$1.4$ g of an organic compound was digested according to Kjeldahl's method and the ammonia evolved was absorbed in $60$ mL of $M/10H_{2}S{O}_4$ solution. The excess sulphuric acid required $20$ mL of $M/10NaOH$ solution for neutralization. The percentage of nitrogen in the compound is :

• A. $3$
• B. $5$
• C. $24$
• D. $10$

Answer 1

The correct option is D $10$

Out of $60$ mL of $\frac{M}{10}$ $H_{2}S{O}_4$, $10$ mL will be neutralized with $20$ mL of $\frac{M}{10}$ $NaOH$.

Hence, $60-10=50$ mL of $\frac{M}{10}$ $H_{2}S{O}_4$ will be neutralized with ammonia. This will be equal to $100$ mL ammonia.

$1000$ mL of $1M$ ammonia corresponds to $14$ g.

$100$ mL of $\frac{M}{10}$ ammonia will correspond to $\frac{100}{1000}\times \frac{1}{10}\times 14=0.14$ g nitrogen

Thus, $1.4$ g of organic compound contains $0.14$ g nitrogen.

Hence, $100$ g of organic compound will contain $\frac{0.14}{1.4}\times 100=10$ g nitrogen

Hence, the per cent of nitrogen in the organic compound is $10\%$.

Hence, the correct option is $D$