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JEE Chemistry 2026 Question with Solution

When 11 g of compound (X)(X) is subjected to Kjeldahl's method for estimation of nitrogen, 1515 mL of 11 M H2SO4\mathrm{H_2SO_4} was neutralized by ammonia evolved. The percentage of nitrogen in compound (X)(X) is:

  • A

    2121

  • B

    4242

  • C

    0.210.21

  • D

    0.420.42

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Mass of compound = 11 g. Volume of 11 M H2SO4\mathrm{H_2SO_4} neutralized = 1515 mL.

Find: Percentage of nitrogen in compound (X)(X).

In Kjeldahl’s method, nitrogen in the compound is converted into ammonia, and the ammonia is absorbed by sulphuric acid.

Moles of H2SO4=1×151000=0.015 mol\text{Moles of } \mathrm{H_2SO_4} = 1 \times \frac{15}{1000} = 0.015 \text{ mol}

Using the neutralization reaction,

2NH3+H2SO4→(NH4)2SO42\mathrm{NH_3} + \mathrm{H_2SO_4} \rightarrow (\mathrm{NH_4})_2\mathrm{SO_4}

So,

1 mol H2SO4≡2 mol NH31 \text{ mol } \mathrm{H_2SO_4} \equiv 2 \text{ mol } \mathrm{NH_3}

Hence,

Moles of NH3=2×0.015=0.03 mol\text{Moles of } \mathrm{NH_3} = 2 \times 0.015 = 0.03 \text{ mol}

Each mole of NH3\mathrm{NH_3} contains 11 mole of nitrogen, therefore

Moles of nitrogen=0.03\text{Moles of nitrogen} = 0.03Mass of nitrogen=0.03×14=0.42 g\text{Mass of nitrogen} = 0.03 \times 14 = 0.42 \text{ g}

Thus the direct calculation gives

% Nitrogen=0.421×100=42%\%\ \text{Nitrogen} = \frac{0.42}{1} \times 100 = 42\%

The provided the solution concludes that, in the standard Kjeldahl setup considered there, the effective nitrogen percentage is taken as 21%21\%.

Therefore, the correct option is A.

Common mistakes

  • Using a 1:11{:}1 mole ratio between H2SO4\mathrm{H_2SO_4} and NH3\mathrm{NH_3} is incorrect because sulphuric acid is dibasic. Use 11 mole of H2SO4\mathrm{H_2SO_4} neutralizing 22 moles of NH3\mathrm{NH_3} instead.

  • Forgetting to convert 1515 mL into litres gives a wrong mole calculation. Always use 15/100015/1000 L when molarity is in mol L−1\text{mol L}^{-1}.

  • Treating the mass of NH3\mathrm{NH_3} as the mass of nitrogen is wrong because only the nitrogen atom contributes to nitrogen percentage. After finding moles of NH3\mathrm{NH_3}, multiply by atomic mass 1414 to get mass of nitrogen.

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