JEE Chemistry 2025 Question with Solution

Given: Benzoic acid reacts with aqueous $\text{NaHCO}_3$ and releases $\text{CO}_2$. The volume of $\text{CO}_2$ obtained is $11.2 \, \text{L}$ at STP.

Find: The mass $X$ of benzoic acid.

Write the balanced chemical equation:

$$\text{C}_6\text{H}_5\text{COOH(aq)} + \text{NaHCO}_3\text{(aq)} \rightarrow \text{C}_6\text{H}_5\text{COONa(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$$

At STP, $1$ mole of a gas occupies $22.4 \, \text{L}$. Therefore, moles of $\text{CO}_2$ released are

$$n_{\text{CO}_2} = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{mol}$$

From the balanced equation, benzoic acid and $\text{CO}_2$ are in a $1:1$ mole ratio. Hence,

$$n_{\text{benzoic acid}} = 0.5 \, \text{mol}$$

The molar mass of benzoic acid $\left(\text{C}_6\text{H}_5\text{COOH}\right)$ is

$$6(12) + 5(1) + 12 + 2(16) + 1 = 122 \, \text{g/mol}$$

So the mass of benzoic acid is

$$X = n \times M = 0.5 \, \text{mol} \times 122 \, \text{g/mol} = 61 \, \text{g}$$

Therefore, the mass of benzoic acid is $61 \, \text{g}$.

Given: $11.2 \, \text{L}$ of $\text{CO}_2$ is formed at STP.

Find: Mass of benzoic acid.

At STP, $11.2 \, \text{L}$ corresponds to $0.5$ mole of gas. Since benzoic acid produces $\text{CO}_2$ in a $1:1$ ratio, benzoic acid used is also $0.5$ mole.

Using molar mass of benzoic acid $122 \, \text{g/mol}$:

$$X = 0.5 \times 122 = 61 \, \text{g}$$

This shortcut works because the balanced equation directly relates moles of benzoic acid and moles of $\text{CO}_2$ without any additional conversion factor.

Therefore, the required mass is $61 \, \text{g}$.