For the estimation of nitrogen, 1.4 g of organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 ml of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is:

Basics of Organic Chemistry PYQ — JEE Main 2014 · Shift 0 · Q3 with Solution

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Correct answer: D

Given data

Mass of organic compound $= 1.4\,\text{g}$
Volume of sulphuric acid used $= 60\,\text{mL}$
Strength of sulphuric acid $= \dfrac{M}{10} = 0.1\,\text{M}$
Unreacted acid neutralized by $20\,\text{mL}$ of $\dfrac{M}{10}$ NaOH

We use Kjeldahl’s method:

Nitrogen in the compound is converted to $\mathrm{NH_3}$ .
The evolved $\mathrm{NH_3}$ is absorbed by the acid.
From the acid consumed, we find moles of $\mathrm{NH_3}$ , hence nitrogen.

Initial moles of $\mathrm{H_2SO_4}$

\text{moles of } \mathrm{H_2SO_4} = M \times V = 0.1 \times \frac{60}{1000} = 0.006

So initially,

\mathrm{H_2SO_4} = 0.006\,\text{mol}

Unreacted sulphuric acid from NaOH back titration

Reaction:

\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}

Moles of NaOH used:

0.1 \times \frac{20}{1000} = 0.002\,\text{mol}

Since $1$ mole of $\mathrm{H_2SO_4}$ reacts with $2$ moles of NaOH, unreacted moles of $\mathrm{H_2SO_4}$ are

\frac{0.002}{2} = 0.001\,\text{mol}

Moles of sulphuric acid consumed by ammonia

\text{Consumed } \mathrm{H_2SO_4} = 0.006 - 0.001 = 0.005\,\text{mol}

Absorption reaction:

\mathrm{H_2SO_4 + 2NH_3 \rightarrow (NH_4)_2SO_4}

Thus,

1\,\text{mol } \mathrm{H_2SO_4} \text{ reacts with } 2\,\text{mol } \mathrm{NH_3}

So moles of ammonia formed:

0.005 \times 2 = 0.010\,\text{mol}

Therefore moles of nitrogen $= 0.010\,\text{mol}$ .

Mass of nitrogen

\text{Mass of N} = 0.010 \times 14 = 0.14\,\text{g}

Percentage of nitrogen

\%N = \frac{0.14}{1.4} \times 100 = 10\%

Option check

A: $3\%$ — incorrect
B: $5\%$ — incorrect
C: $6\%$ — incorrect
D: $10\%$ — correct

Hence, the correct answer is:

\boxed{10\%}

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View written solutionFree

Correct answer: D

Given data

Mass of organic compound $= 1.4\,\text{g}$
Volume of sulphuric acid used $= 60\,\text{mL}$
Strength of sulphuric acid $= \dfrac{M}{10} = 0.1\,\text{M}$
Unreacted acid neutralized by $20\,\text{mL}$ of $\dfrac{M}{10}$ NaOH

We use Kjeldahl’s method:

Nitrogen in the compound is converted to $\mathrm{NH_3}$ .
The evolved $\mathrm{NH_3}$ is absorbed by the acid.
From the acid consumed, we find moles of $\mathrm{NH_3}$ , hence nitrogen.

Initial moles of $\mathrm{H_2SO_4}$

\text{moles of } \mathrm{H_2SO_4} = M \times V = 0.1 \times \frac{60}{1000} = 0.006

So initially,

\mathrm{H_2SO_4} = 0.006\,\text{mol}

Unreacted sulphuric acid from NaOH back titration

Reaction:

\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}

Moles of NaOH used:

0.1 \times \frac{20}{1000} = 0.002\,\text{mol}

Since $1$ mole of $\mathrm{H_2SO_4}$ reacts with $2$ moles of NaOH, unreacted moles of $\mathrm{H_2SO_4}$ are

\frac{0.002}{2} = 0.001\,\text{mol}

Moles of sulphuric acid consumed by ammonia

\text{Consumed } \mathrm{H_2SO_4} = 0.006 - 0.001 = 0.005\,\text{mol}

Absorption reaction:

\mathrm{H_2SO_4 + 2NH_3 \rightarrow (NH_4)_2SO_4}

Thus,

1\,\text{mol } \mathrm{H_2SO_4} \text{ reacts with } 2\,\text{mol } \mathrm{NH_3}

So moles of ammonia formed:

0.005 \times 2 = 0.010\,\text{mol}

Therefore moles of nitrogen $= 0.010\,\text{mol}$ .

Mass of nitrogen

\text{Mass of N} = 0.010 \times 14 = 0.14\,\text{g}

Percentage of nitrogen

\%N = \frac{0.14}{1.4} \times 100 = 10\%

Option check

A: $3\%$ — incorrect
B: $5\%$ — incorrect
C: $6\%$ — incorrect
D: $10\%$ — correct

Hence, the correct answer is:

\boxed{10\%}

Basics of Organic Chemistry question

Basics of Organic Chemistry question

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