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JEE Chemistry 2024 Question with Solution

9.3 g9.3 \, \text{g} of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100%100\% completed is x×10−1 gx \times 10^{-1} \, \text{g}.

Answer

Correct answer:135

Step-by-step solution

Standard Method

Given: Mass of aniline = 9.3 g9.3 \, \text{g}. Acetic anhydride is in excess. The required answer is in the form x×10−1 gx \times 10^{-1} \, \text{g}.

Find: The value of xx for the mass of acetanilide formed.

The reaction is:

C6H5NH2+CH3COOCOCH3→C6H5NHCOCH3+CH3COOHC_6H_5NH_2 + CH_3COOCOCH_3 \rightarrow C_6H_5NHCOCH_3 + CH_3COOH

So, aniline and acetanilide are in a 1:11:1 mole ratio.

Molar mass of aniline (C6H7N)\left(C_6H_7N\right) is:

6×12+7×1+14=93 g/mol6 \times 12 + 7 \times 1 + 14 = 93 \, \text{g/mol}

Hence, moles of aniline are:

naniline=9.393=0.1n_{\text{aniline}} = \frac{9.3}{93} = 0.1

Since the reaction ratio is 1:11:1, moles of acetanilide formed are also:

nacetanilide=0.1n_{\text{acetanilide}} = 0.1

Molar mass of acetanilide (C8H9NO)\left(C_8H_9NO\right) is:

8×12+9×1+14+16=135 g/mol8 \times 12 + 9 \times 1 + 14 + 16 = 135 \, \text{g/mol}

Therefore, mass of acetanilide produced is:

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

Now write 13.5 g13.5 \, \text{g} in the form x×10−1 gx \times 10^{-1} \, \text{g}:

13.5=135×10−113.5 = 135 \times 10^{-1}

Therefore, the value of xx is 135135.

Molar Mass Expansion

Given: Aniline = 9.3 g9.3 \, \text{g}, reaction completed fully, acetic anhydride in excess.

Find: The value of xx in x×10−1 gx \times 10^{-1} \, \text{g}.

First compute the molar mass of aniline from its formula C6H7NC_6H_7N:

C:6×12=72H:7×1=7N:1×14=14\begin{aligned} C &: 6 \times 12 = 72 \\ H &: 7 \times 1 = 7 \\ N &: 1 \times 14 = 14 \end{aligned}

So,

Molar mass of aniline=72+7+14=93 g/mol\text{Molar mass of aniline} = 72 + 7 + 14 = 93 \, \text{g/mol}

Now calculate moles of aniline:

moles of aniline=9.393=0.1  mol\text{moles of aniline} = \frac{9.3}{93} = 0.1 \; \text{mol}

The product acetanilide has formula C8H9NOC_8H_9NO. Its molar mass is:

C:8×12=96H:9×1=9N:1×14=14O:1×16=16\begin{aligned} C &: 8 \times 12 = 96 \\ H &: 9 \times 1 = 9 \\ N &: 1 \times 14 = 14 \\ O &: 1 \times 16 = 16 \end{aligned}

Hence,

Molar mass of acetanilide=96+9+14+16=135 g/mol\text{Molar mass of acetanilide} = 96 + 9 + 14 + 16 = 135 \, \text{g/mol}

Because the reaction is 1:11:1, 0.1  mol0.1 \; \text{mol} of aniline gives 0.1  mol0.1 \; \text{mol} of acetanilide. Thus,

mass of acetanilide=0.1×135=13.5 g\text{mass of acetanilide} = 0.1 \times 135 = 13.5 \, \text{g}

Expressing this in the asked form:

13.5 g=135×10−1 g13.5 \, \text{g} = 135 \times 10^{-1} \, \text{g}

Therefore, the correct answer is 135135.

Common mistakes

  • Using the molar mass of acetanilide directly to find moles from 9.3 g9.3 \, \text{g} is incorrect because the given mass is for aniline. First calculate moles of aniline, then use the 1:11:1 stoichiometric ratio.

  • Assuming a wrong mole ratio between aniline and acetanilide leads to an incorrect result. From the balanced reaction, 11 mole of aniline forms 11 mole of acetanilide.

  • Stopping at 13.5 g13.5 \, \text{g} is incomplete because the question asks for the value of xx in the form x×10−1 gx \times 10^{-1} \, \text{g}. Convert 13.513.5 to 135×10−1135 \times 10^{-1} and report x=135x = 135.

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