JEE Chemistry 2025 Question with Solution

Given: $0.01$ mole of compound $X$, hydrogen content $10\%$ by mass, and water produced on combustion is $0.9 \, \text{g}$.

Find: Molar mass of $X$.

First calculate moles of water formed:

$$\text{moles of } H_2O = \frac{0.9}{18} = 0.05$$

Each mole of water contains $2$ moles of hydrogen atoms, so moles of hydrogen atoms obtained are:

$$\text{moles of H atoms} = 2 \times 0.05 = 0.1$$

Mass of hydrogen present in the sample is therefore:

$$\text{mass of H} = 0.1 \, \text{g}$$

Since hydrogen is $10\%$ by mass, if mass of $0.01$ mole of $X$ is $m$ grams, then:

$$0.10m = 0.1$$ $$m = 1 \, \text{g}$$

This $1 \, \text{g}$ corresponds to $0.01$ mole of $X$, so molar mass is:

$$M = \frac{1}{0.01} = 100 \, \text{g mol}^{-1}$$

Therefore, the molar mass of $X$ is $100 \, \text{g mol}^{-1}$.

Given: $0.01 \, \text{mol}$ of organic compound $X$ produces $0.9 \, \text{g}$ of water on complete combustion, and $X$ contains $10\%$ hydrogen by mass.

Find: Molar mass of $X$.

Molar mass of water is:

$$M(H_2O) = 18 \, \text{g mol}^{-1}$$

So moles of water formed are:

$$\frac{0.9}{18} = 0.05 \, \text{mol}$$

Now each mole of $H_2O$ contains $2$ grams of hydrogen per $18$ grams of water, so hydrogen mass in $0.05$ mole of water is:

$$0.05 \times 2 = 0.1 \, \text{g}$$

Thus, the $0.01$ mole sample of $X$ contains $0.1 \, \text{g}$ hydrogen.

Hydrogen is $10\%$ of the mass of the compound sample, hence:

$$\frac{0.1}{\text{mass of sample}} = 0.10$$

Therefore:

$$\text{mass of sample} = \frac{0.1}{0.10} = 1 \, \text{g}$$

This $1 \, \text{g}$ sample is $0.01$ mole, so molar mass becomes:

$$M = \frac{1 \, \text{g}}{0.01 \, \text{mol}} = 100 \, \text{g mol}^{-1}$$

Therefore, the required numerical value is $100$.