Question

$0.2\text{g}$ of an organic compound was analysed by Kjeldahl’s method. The ammonia gas evolved was absorbed by $60\text{mL of}N/5H_{2}S{O}_4$. The unused acid required $40\text{mL of}N/10NaOH$ for complete neutralisation. Find the percentage of $N_2$ in the sample.

Answer 1

This can be solved on the basis of law of chemical equivalance
According to which;
$gramequivalentsofN{H}_3+gramequivalentsofNaOH=gramequivalentsof{H}_{2}S{O}_4$
$moles\times nf+\frac{N_1\times V_1}{1000}=\frac{N_2\times V_2}{1000}$
$\to n\times 1=\frac{8}{1000}$
$\to n=\frac{8}{1000}=8\times {10}^{-3}$
Now,
1 mole of $N{H}_3$ contains 1 mole of `N`.
$\therefore$ $8\times {10}^{-3}$ moles of $N{H}_3$ contains $8\times {10}^{-3}$ moles of nitrogen.
Hence,
mass of `N `= moles$\times$ molar mass
= $8\times {10}^{-3}\times 14$
= 0.112 g
$\therefore$ % of `N` = $\frac{0.112}{0.2}\times 100$

= `56 %`