JEE Chemistry 2026 Question with Solution

Given: Mass of the organic compound = $0.75 \, \text{g}$, mass of barium sulphate formed = $1.2 \, \text{g}$, atomic mass of sulphur = $32 \, \text{g mol}^{-1}$, molar mass of barium sulphate = $233 \, \text{g mol}^{-1}$.

Find: Percentage of sulphur in the organic compound.

In Carius method, sulphur present in the organic compound is converted into barium sulphate, $\text{BaSO}_4$, and its mass is used to calculate the sulphur content.

Using stoichiometry,

$$\% \, \text{S} = \frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \times \frac{\text{Mass of BaSO}_4\text{ formed}}{\text{Mass of organic compound taken}} \times 100$$

Substituting the given values,

$$\% \, \text{S} = \frac{32}{233} \times \frac{1.2}{0.75} \times 100$$

Now,

$$\frac{32}{233} \approx 0.13734$$

and

$$\frac{1.2}{0.75} = 1.6$$

Therefore,

$$\% \, \text{S} = 0.13734 \times 1.6 \times 100$$ $$\% \, \text{S} = 0.219744 \times 100$$ $$\% \, \text{S} = 21.9744\%$$

Hence, the percentage of sulphur is approximately $21.97\%$. The correct option is C.

Given: Mass of compound = $0.75 \, \text{g}$, mass of $\text{BaSO}_4$ = $1.2 \, \text{g}$.

Find: Percentage of sulphur.

For Carius method, directly use the relation between sulphur and $\text{BaSO}_4$:

$$\% \, \text{S} = \frac{32}{233} \times \frac{\text{mass of } \text{BaSO}_4}{\text{mass of sample}} \times 100$$

So,

$$\% \, \text{S} = \frac{32}{233} \times \frac{1.2}{0.75} \times 100 = 21.97\%$$

This shortcut works because one mole of $\text{BaSO}_4$ contains exactly one mole of sulphur. Therefore, the correct option is C.