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A sample of \(0.50 \mathrm{~g}\) of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in \(50 \mathrm{ml}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\). The residual acid required \(60 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) solution of \(\mathrm{NaOH}\) for neutralisation. Find the percentage composition of nitrogen in the compound.
ChemistryGeneral Organic Chemistry
Solution:
2148 Upvotes Verified Answer
Step 1. To determine the volume of \(\mathrm{H}_2 \mathrm{SO}_4\) used.
Volume of acid taken \(=50 \mathrm{ml}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4=25 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\)
Volume of alkali used for neutralization of excess acid \(=60 \mathrm{ml}\) of \(0.5 \mathrm{M} \mathrm{NaOH}=30 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{NaOH}\).
Now 1 mole of \(\mathrm{H}_2 \mathrm{SO}_4\) neutralizes 2 moles of \(\mathrm{NaOH}\) (i. e. \(\mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}\) )
\(30 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{NaOH}=15 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\)
\(\therefore \quad\) Volume of acid used by ammonia
\(=25-15=10 \mathrm{ml}\)
Step 2. To determine percentage of nitrogen.
Again 1 mole of \(\mathrm{H}_2 \mathrm{SO}_4\) neutralizes 2 moles of \(\mathrm{NH}_3\)
\(\left(\text { i.e. } 2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\right)\)
\(10 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4 \equiv 20 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{NH}_3\)
\(\because \quad 1000 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{NH}_3\) contain nitrogen \(=14 \mathrm{~g}\)
\(20 \mathrm{ml}\) of \(1 \mathrm{M} \mathrm{NH}_3\) will contain nitrogen
\(=14 / 1000 \times 20 \mathrm{~g}\)
But this much amount of nitrogen is present in \(0.5 \mathrm{~g}\) of the organic compound.
Percentage of nitrogen \(=14 / 1000 \times 20 / 0.5 \times 100=56 \%\).
Alternatively, \(\%\) of \(\mathrm{N}\) can be determined by applying the following equation,
\(\% \mathrm{~N}=\frac{\begin{array}{c}1.4 \times \text { Molarity of the acid } \times \text { Basicity of the acid } \\ \times \text { Vol. of the acid used }\end{array}}{\text { Mass of substance taken }}\)
Substituting the values of all the items in the above equation, we have,
\(\% \mathrm{~N}=\frac{1.4 \times 1 \times 2 \times 10}{0.5}=56.0\)

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