Question

$0.257\, g$ of an organic substance was heated with conc. $H _{2} SO _{4}$ and then distilled with excess of strong alkali. The ammonia gas evolved was absorbed in $50\, ml$ of $N / 10\, HCl$ which required $23.2\, ml$ of $N / 10\, NaOH$ 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 $=50\, ml$
Volume of $\frac{N}{10} NaOH$ used for neutralization of unused acid $=23.2\, ml$
Now $N_{1} V_{1}( NaOH )=N_{2} V_{2}( HCl )$
$23.2\, ml$ of $\frac{N}{10} NaOH \equiv 23.2\, ml$ of $\frac{N}{10} HCl$
$\therefore$ Volume of $\frac{N}{10} HCl$ unused $=23.2\, ml$
$\therefore$ Volume of $N / 10\, HCl$ required for neutralization of $NH _{3}$
$=50-23.2=26.8\, ml$
$26.8\, ml$ of $\frac{N}{10} HCl =26.8\, ml$ of $N / 10\, NH _{3}$
$1000\, ml$ of $1\, N\, NH _{3}$ solution contains nitrogen $=14\, g$
$26.8\, ml$ of $NH _{3}$ solution contains nitrogen
$=\frac{14 \times 26.8}{10 \times 1000}$
Percentage of nitrogen
$=\frac{14 \times 26.8 \times 100}{10 \times 1000 \times 0.257}=14.6 \%$