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JEE Chemistry 2026 Question with Solution

A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3 g9.3 \, \text{g} of aniline. However, the student has managed to obtain 11 g11 \, \text{g} of dry acetanilide. The %\% yield of this reaction is

  • A

    97.5%97.5\%

  • B

    81.5%81.5\%

  • C

    59.5%59.5\%

  • D

    72.5%72.5\%

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Mass of aniline = 9.3 g9.3 \, \text{g}. Actual mass of dry acetanilide obtained = 11 g11 \, \text{g}.

Find: The percentage yield of acetanilide.

Step 1: Calculating moles of aniline.

Molar mass of aniline = 93 g mol−193 \, \text{g mol}^{-1}.

Moles of aniline=9.393=0.1 mol\text{Moles of aniline} = \frac{9.3}{93} = 0.1 \, \text{mol}

Step 2: Stoichiometry of the reaction.

11 mole of aniline produces 11 mole of acetanilide.

Molar mass of acetanilide = 135 g mol−1135 \, \text{g mol}^{-1}.

Step 3: Theoretical yield of acetanilide.

Theoretical yield=0.1×135=13.5 g\text{Theoretical yield} = 0.1 \times 135 = 13.5 \, \text{g}

Step 4: Percentage yield calculation.

Percentage yield=1113.5×100=81.5\text{Percentage yield} = \frac{11}{13.5} \times 100 = 81.5

Therefore, the percentage yield is 81.5%81.5\%. The correct option is B.

Yield Formula Focus

Use the relation:

Percentage yield=Actual yieldTheoretical yield×100\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100

Here, the actual yield is 11 g11 \, \text{g} and the theoretical yield comes from 0.1 mol0.1 \, \text{mol} of aniline, which gives 13.5 g13.5 \, \text{g} of acetanilide.

1113.5×100=81.5\frac{11}{13.5} \times 100 = 81.5

Hence, the correct answer is B.

Common mistakes

  • Using the given actual yield directly as the answer is incorrect because percentage yield must compare actual yield with theoretical yield. First calculate the theoretical amount of acetanilide, then apply the yield formula.

  • Assuming the mass of product must equal the mass of aniline is incorrect because stoichiometric calculations depend on moles and molar masses, not direct mass equality. Convert aniline mass to moles before finding product mass.

  • Using the wrong molar mass for acetanilide leads to an incorrect theoretical yield. Use 135 g mol−1135 \, \text{g mol}^{-1} for acetanilide and 93 g mol−193 \, \text{g mol}^{-1} for aniline.

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