Question

The ammonia evolved from the treatment of $0.30{\text{ }}g$of an organic compound for the estimation of nitrogen was passed in $100{\text{ }}ml$of $0.1{\text{ }}M$sulphuric acid. The excess of acid required $20{\text{ }}ml$of $0.5{\text{ }}M$sodium hydroxide for complete neutralization. The organic compound is:
A.Acetamide
B.Benzamide
C.Urea
D.Thiourea

Answer 1

The question says estimation of nitrogen in an organic compound so for this question, one can use the kjeldahl method and its formula that is $\%$nitrogen$= \dfrac{{1.4 \times N \times V}}{W}$. And also for calculating the percentage of nitrogen in the above options you can use the formula of mass percent composition: $Mass\% = \dfrac{{Mass}}{{Total{\text{ }}mass}} \times 100$

Complete answer:
Given in the question, weight of the organic compound$= 0.30g$, molarity of sulphuric acid$= 0.1M$, volume of sulphuric acid$= 100{\text{ }}mL$, molarity of sodium hydroxide$= 0.5M$, and volume of sodium hydroxide used$= 20{\text{ }}mL$. We need to identify the organic compound that is undergoing the following treatment.
The question says that the organic compound is treated for the estimation of nitrogen and so we will calculate the percentage of nitrogen which will give us the compound. For this, we will use the Kjeldahl process.
Let us assume that the unreacted $0.1{\text{ }}M{\text{ }}\left( {{\text{ }} = {\text{ }}0.2{\text{ }}N{\text{ }}} \right){\text{ }}{H_2}S{O_4}\; = {\text{ }}V'{\text{ }}ml$
So, $20{\text{ }}ml$of $0.5{\text{ }}M{\text{ }}NaOH = V'{\text{ }}ml{\text{ }}of{\text{ }}0.2{\text{ }}N{\text{ }}{H_2}S{O_4}\;$
$\Rightarrow 20{\text{ }} \times {\text{ }}0.5{\text{ }} = {\text{ }}V'{\text{ }} \times {\text{ }}0.2$
$\Rightarrow V'{\text{ }} = {\text{ }}50{\text{ }}ml$
Thus used ${H_2}S{O_4} = 100 - 50$
$= 50{\text{ }}mL$
Now for calculating the percentage of nitrogen we will use the formula:
$\%$nitrogen$= \dfrac{{1.4 \times N \times V}}{W}$
Where ‘N’ is the normality of the acid used, ‘V’ is the volume of the acid used, and ‘W’ is the weight of the compound.
Here, $N = 0.2\,Normal$, $V = 50{\text{ }}mL$ and, $W = 0.30g$
Hence by substituting the values we get,
$\%$nitrogen$= \dfrac{{1.4 \times N \times V}}{W}$
$\Rightarrow \% {\text{ nitrogen = }}\dfrac{{1.4 \times 0.2 \times 50}}{{0.30}}$
$\Rightarrow 46.67\%$
Thus, we get the percentage of nitrogen as $46.67\%$
Now to know the organic compound used we will find out the $\%$ of nitrogen in all the given options and then match our above value with those percentages. The value that matches the above percentage will be the answer. We will use the formula:
$Mass\% = \dfrac{{Mass}}{{Total{\text{ }}mass}} \times 100$
The percentage of nitrogen in acetamide$= \dfrac{{Mass}}{{Total{\text{ }}mass}} \times 100$
$\Rightarrow \dfrac{{14}}{{59}} \times 100 = 23.73\%$
The percentage of nitrogen in benzamide$= \dfrac{{Mass}}{{Total{\text{ }}mass}} \times 100$
$\Rightarrow \dfrac{{14}}{{122}} \times 100 = 11.48\%$
The percentage of nitrogen in urea $= \dfrac{{Mass}}{{Total{\text{ }}mass}} \times 100$
$\Rightarrow \dfrac{{28}}{{59}} \times 100 = 46.67\%$
The percentage of nitrogen in thiourea $= \dfrac{{Mass}}{{Total{\text{ }}mass}} \times 100$
$\Rightarrow \dfrac{{28}}{{76}} \times 100 = 36.84\%$
Thus by looking at the above answers we can clearly say that the organic compound is urea.

Therefore the correct option is C.

Note:
The Kjeldahl method is the method that is used to quantitatively determine nitrogen in organic substances and inorganic substances. It works on the principle that when any strong acid is used it helps in the digestion of food so that it will release nitrogen which can be determined by performing a titration.