JEE Chemistry 2024 Question with Solution

Given: mass of aniline = $9.3 \, \text{g}$. Acetic anhydride is in excess, so aniline is the limiting reagent.

Find: the value of $x$ in $x \times 10^{-1} \, \text{g}$ for the mass of acetanilide formed.

The reaction is:

$$C_6H_5NH_2 + CH_3COOCOCH_3 \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$$

This shows a $1:1$ mole ratio between aniline and acetanilide.

Molar mass of aniline $(C_6H_7N)$:

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

Moles of aniline:

$$n_{\text{aniline}} = \frac{9.3}{93} = 0.1 \, \text{mol}$$

Hence, moles of acetanilide formed = $0.1 \, \text{mol}$.

Molar mass of acetanilide $(C_8H_9NO)$:

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

Mass of acetanilide formed:

$$m = 0.1 \times 135 = 13.5 \, \text{g}$$

Now,

$$13.5 = 135 \times 10^{-1} \, \text{g}$$

Therefore, $x = 135$.

Given: $9.3 \, \text{g}$ of aniline reacts completely with excess acetic anhydride.

Find: the numerical value $x$.

First calculate the molar mass of aniline:

• C: $6 \times 12 = 72$
• H: $7 \times 1 = 7$
• N: $1 \times 14 = 14$

So,

$$M_{\text{aniline}} = 72 + 7 + 14 = 93 \, \text{g/mol}$$

Then,

$$\text{moles of aniline} = \frac{9.3}{93} = 0.1$$

Since the reaction forms acetanilide in a $1:1$ ratio, moles of acetanilide = $0.1$.

Now calculate the molar mass of acetanilide:

• C: $8 \times 12 = 96$
• H: $9 \times 1 = 9$
• N: $14$
• O: $16$

Thus,

$$M_{\text{acetanilide}} = 96 + 9 + 14 + 16 = 135 \, \text{g/mol}$$

Mass produced:

$$0.1 \times 135 = 13.5 \, \text{g}$$

Writing this in the required form,

$$13.5 \, \text{g} = 135 \times 10^{-1} \, \text{g}$$

So, the required value is 135.