Question

he $N{{H}_{3}}$evolved from a 1.40 gm sample of protein was absorbed in 45 ml of 0.4M $HN{{O}_{3}}$. The excess acid required 20 ml of 0.1 M NaOH.
The % N in the sample is:
A.8
B. 16
C. 19.42
D. none of these

Answer 1

Hint: The acid is neutralized using a base. While ammonia gas evolved is absorbed using an acid, to prevent its conversion into other compounds. This problem uses the relation between molarity and volume, which is ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$. We will find the acid to neutralize the NaOH, then find ammonia to neutralize the acid and thus the percentage of nitrogen.
Formula used:
Molarity = $\dfrac{no.of\,moles}{Vol.of\,solution(in\,L)}$

Complete answer:
We have been given ammonia gas evolved from a protein sample of 1.40 gm. This gas is absorbed in 45 ml of 0.4M $HN{{O}_{3}}$. The acid is neutralized by the base, which requires 20 ml of 0.1 M of NaOH. As we know, NaOH is monoacidic base and $HN{{O}_{3}}$ is a monobasic acid, so number of milli equivalents of excess of $HN{{O}_{3}}$ is equal to that of NaOH. Therefore, volume of nitric acid in excess is,
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
0.4$\times {{V}_{1}}$ = $0.1\times 20$
Volume of $HN{{O}_{3}}$ = 5 ml
So, from 45 ml of $HN{{O}_{3}}$ 5 ml was in excess and 40 ml utilized for ammonia absorption, which was evolved from 1.40 of protein sample.
So, number of moles in 40 ml of 0.4M of $HN{{O}_{3}}$ are calculated from the formula, molarity = $\dfrac{no.of\,moles}{Vol.of\,solution(in\,L)}$
0.4 = $\dfrac{number\,of\,moles}{0.04}$
Number of moles of $HN{{O}_{3}}=0.4\times 0.04$
Number of moles of $HN{{O}_{3}}=0.016$
As ammonia gas is absorbed by 0.016 moles of nitric acid, so, moles of $N{{H}_{3}}$ will be the same as that of $HN{{O}_{3}}$ which is 0.016.
Now we will calculate the mass of ammonia evolved from the protein sample from number of moles as, $moles=\dfrac{mass}{molar\,mass}$
So, $mass=No.\,of\,moles\times molar\,mass$,
Mass of ammonia = $0.016\times 17$
Mass of ammonia = 0.272g
As we know, that in 17 g of $N{{H}_{3}}$ there is 14 g of Nitrogen, now we will calculate nitrogen in 0.272g of ammonia, as,
$\dfrac{14\,g\,N}{17\,g\,N{{H}_{3}}}\times 0.272g\,N{{H}_{3}}$ = 0.224 g of Nitrogen.
Now the percentage of nitrogen is calculated by the mass of nitrogen in evolved $N{{H}_{3}}$ upon mass of protein sample multiplied by 100.
% of N = $\dfrac{0.224}{1.4}\times 100$
% of N = 16%

Hence, % of N in evolved ammonia is 16%, so option B is correct.

Note:
Mass of nitrogen in 0.272g of ammonia is calculated by unitary method. The term milli equivalent, defines the ions that are dissociated in the solution, which depends on the solute added, like NaOH will dissolve and give, $N{{a}^{+}}$and $O{{H}^{-}}$, hence it is monoacidic as it produce one hydroxide ion, which neutralizes one hydrogen produced from $HN{{O}_{3}}$, when dissolves as ${{H}^{+}}$ and $N{{O}_{3}}^{-}$, which is monobasic.