Question

$0.257g$ of an organic substance was heated with conc. $H_{2}S{O}_4$ and then distilled with excess of strong alkali. The ammonia gas evolved was absorbed in $50ml$ of $N/10HCl$ which required $23.2ml$ of $N/10NaOH$ for neutralization at the end of the process. The percentage of nitrogen in the compound is:

• A. 14.6
• B. 18.0
• C. 17.0
• D. 15.5

Answer 1

Answer: (A)

Volume of $\frac{N}{10}HCl$ taken $=50ml$
Volume of $\frac{N}{10}NaOH$ used for neutralization of unused acid $=23.2ml$
Now $N_1{V}_1(NaOH)=N_2{V}_2(HCl)$
$23.2ml$ of $\frac{N}{10}NaOH\equiv 23.2ml$ of $\frac{N}{10}HCl$
$\therefore$ Volume of $\frac{N}{10}HCl$ unused $=23.2ml$
$\therefore$ Volume of $N/10HCl$ required for neutralization of $N{H}_3$
$=50-23.2=26.8ml$
$26.8ml$ of $\frac{N}{10}HCl=26.8ml$ of $N/10N{H}_3$
$1000ml$ of $1NN{H}_3$ solution contains nitrogen $=14g$
$26.8ml$ of $N{H}_3$ solution contains nitrogen
$=\frac{14\times 26.8}{10\times 1000}$
Percentage of nitrogen
$=\frac{14\times 26.8\times 100}{10\times 1000\times 0.257}=14.6\%$