Question

A $0.60$ g nitrogen containing compound was boiled with $NaOH$, and $N{H}_3$ thus formed, required $100$ mL of $0.2$ N $H_{2}S{O}_4$ for neutralisation. The percentage of nitrogen in the compound is:

• A. $46.67\%$
• B. $23.34\%$
• C. $60.00\%$
• D. $20.00\%$

Answer 1

The correct option is A $46.67\%$

100 mL of 0.2 N $H_{2}S{O}_4\equiv$ 0.02 equivalents of ammonia.

Since, basicity of ammonia is 1, moles of ammonia is also equal to 0.02 mol.

Each mole of ammonia has 1 mole of nitrogen atom, so moles of nitrogen atom is also equal to 0.02 .

So, 0.60 g of organic compound contains $\frac{14\times 20}{1000}$ g of nitrogen.
Therefore, % of nitrogen in the compound $=\frac{Massofnitrogen}{Massofcompound}\times 100=\frac{14\times 20\times 100}{1000\times 0.6}=46.67$ % of nitrogen.

So, percentage of nitrogen is $46.67$ %