JEE Main 2014 — Chemistry Organic Chemistry

2014mcqhard

For the estimation of nitrogen, $1.4g$ of an organic compound was digested by the Kjeldahl method and the evolved ammonia was absorbed in $60\mathrm{ml}$ of $\frac{ \text{M} }{ 1 0 }$ sulphuric acid. The unreacted acid required $20\mathrm{ml}$ of $\frac{ \text{M} }{ 1 0 }$ sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is

Official previous-year question

Held on 6 Apr 2014 · Verified 6 Jul 2026.

Options

A
$6%$
B
$10%$
C
$3%$
D
$5%$

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Solution

In Kjeldahl method,

$\mathrm{Per}\mathrm{cent}N=\frac{1.4NV}{W}$

Where, $\mathrm{NV}$ = gram equivalent of ammonia

$W$ = weight of the organic compound

$\Rightarrow 2{\mathrm{NH}}_{3}+\underset{\mathrm{excess}}{{H}_{2}{\mathrm{SO}}_{4}}\rightarrow {({\mathrm{NH}}_{4})}_{2}{\mathrm{SO}}_{4}$

Total of ${H}_{2}{\mathrm{SO}}_{4}$ used $=60\times \frac{1}{10}=\text{6 mm = 6 x 2 =12 gm equivalent}$

$\Rightarrow 2\mathrm{NaOH}+{H}_{2}{\mathrm{SO}}_{4}\rightarrow {\mathrm{Na}}_{2}{\mathrm{SO}}_{4}+2{H}_{2}O$

$\mathrm{NaOH}$ for back titration $=20\times \frac{1}{10}=\text{2 gm equivalent}$

gm equivalent of remaining ${H}_{2}{\mathrm{SO}}_{4}=12-2=10$

${H}_{2}{\mathrm{SO}}_{4}$ used for ${\mathrm{NH}}_{3}$ neutralisation $=10\mathrm{gm}\mathrm{equivalent}$

${\mathrm{NH}}_{3}(\mathrm{NV})=10\mathrm{gm}\mathrm{equivalent}$

$\text{Per cent of N}=\frac{1.4x10}{\text{1.4}}=\text{10%}$

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Solution

In Kjeldahl method,

$\mathrm{Per}\mathrm{cent}N=\frac{1.4NV}{W}$

Where, $\mathrm{NV}$ = gram equivalent of ammonia

$W$ = weight of the organic compound

$\Rightarrow 2{\mathrm{NH}}_{3}+\underset{\mathrm{excess}}{{H}_{2}{\mathrm{SO}}_{4}}\rightarrow {({\mathrm{NH}}_{4})}_{2}{\mathrm{SO}}_{4}$

Total of ${H}_{2}{\mathrm{SO}}_{4}$ used $=60\times \frac{1}{10}=\text{6 mm = 6 x 2 =12 gm equivalent}$

$\Rightarrow 2\mathrm{NaOH}+{H}_{2}{\mathrm{SO}}_{4}\rightarrow {\mathrm{Na}}_{2}{\mathrm{SO}}_{4}+2{H}_{2}O$

$\mathrm{NaOH}$ for back titration $=20\times \frac{1}{10}=\text{2 gm equivalent}$

gm equivalent of remaining ${H}_{2}{\mathrm{SO}}_{4}=12-2=10$

${H}_{2}{\mathrm{SO}}_{4}$ used for ${\mathrm{NH}}_{3}$ neutralisation $=10\mathrm{gm}\mathrm{equivalent}$

${\mathrm{NH}}_{3}(\mathrm{NV})=10\mathrm{gm}\mathrm{equivalent}$

$\text{Per cent of N}=\frac{1.4x10}{\text{1.4}}=\text{10%}$

JEE Main 2014 — Chemistry Organic Chemistry

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Solution

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JEE Main 2014 — Chemistry Organic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Organic Chemistry in other years

More JEE Main Organic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Inorganic Chemistry

Keep practicing Organic Chemistry