Question

The $N{H_3}$ evolved due to the complete conversion of N from \[1.12\] g of sample of protein absorbed in 45 mL of $0.4$ N $HN{O_{\text{3}}}$ . The excess acid required 20 mL of $0.1$ N $NaOH$. The $\% {\text{N}}$ in the sample is:
A.8
B.16
C.20
D.25

Answer 1

Hint: Knowledge of different concentration factors is essential to solve this question. The normality of a solution is defined as the number of gram equivalents of a solute dissolved in one litre of the solvent. We shall find the amount of acid required to neutralise the given sodium hydroxide solution. Then, calculate the amount of ammonia required to neutralize the acid and thus calculate the percentage nitrogen present in the ammonia required.

Complete step by step solution:
$NaOH$ is a monoacidic base and $HN{O_{\text{3}}}$is a monobasic acid. Hence one molecule of$NaOH$ can neutralize one molecule o f$HN{O_{\text{3}}}$. So, 20 mL of $0.1$ N $NaOH$can neutralize 20 mL of$0.1$ N $HN{O_{\text{3}}}$.
Therefore, the amount of acid required to neutralize\[1.12\]g of N from a protein = \[\left[ {\left( {45 \times 0.4} \right) - \left( {20 \times 0.1} \right)} \right]\]
$ \Rightarrow \left[ {18 - 2} \right]$ = 16 milliequivalent of $HN{O_{\text{3}}}$
Again, $N{H_3}$being a monoacidic base can neutralize $HN{O_{\text{3}}}$.
So, 16 milliequivalent of $N{H_3}$ is required to neutralize 16 milliequivalent of $HN{O_{\text{3}}}$ .
Therefore, weight of $N{H_3}$ being w,
$ \Rightarrow \dfrac{{\text{w}}}{{17}} \times 1000$ = 16
$ \Rightarrow {\text{w}} = 0.2728$ g
So, 16 milliequivalent of $N{H_3}$ = $0.2728$g of$ N{H_3}$
Now, 17 g of $N{H_3}$ has 14 g of N, therefore,
$0.2728$g contains $\dfrac{{14}}{{17}} \times 0.2728$g of N = $0.2248$g of N.
Therefore, $\% {\text{N}}$ in the sample =$\dfrac{{0.2248}}{{1.12}} \times 100 = 20\% $

Hence the correct option is option C.

Note: The unit “equivalent” was introduced to account for the fact that when solutes dissolve in solvents to create a solution, the number of particles dispersed depends on the valence of the solute. Thus one molecule of $KCl$ dissociates in water to form two particles one ${K^ + }$ and one $C{l^ - }$. This leads to the equation that 1 ${\text{m}}{\text{.Eq = }}\dfrac{{{mass \times valency}}}{{{\text{Mol}}{\text{.Wt}}{\text{.}}}}$ and ${\text{Eq}}{\text{.Wt}}{\text{. = }}\dfrac{{{\text{Mol}}{\text{.wt}}{\text{.}}}}{{{\text{valency}}}}$