Question

A sample of $0.50g$ of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in $50mL$ of $0.5M{H}_{2}S{O}_4$. The residual acid required $60mL$ of $0.5M$ solution of $NaOH$ for neutralisation. What would be the percentage composition of nitrogen in the compound?

• A. $50$
• B. $60$
• C. $56$
• D. $44$

Answer 1

Answer: (C)

Given :
Mass of compound taken = $0.50g$
Vol. of $H_{2}S{O}_4=50mL$
Molarity of $H_{2}S{O}_4=0.5M$
Vol. of $NaOH$ required = $60mL$
Molarity of $NaOH$ required = $0.5M$
Method adopted : Kjeldahl’s method
Formula Used : $\%$ of $N=\frac{1.4\times M\times 2\left[V-\frac{V_1}{2}\right]}{m}\dots \left(i\right)$
By substituting the values in the formula, we get,
$\%$ of $N=\frac{1.4\times 0.5\times 2\left(50-60/2\right)}{0.5}=56$
$\therefore \%$ of $N$ in the given compound $=56\%$