JEE Chemistry 2025 Question with Solution

Given: $160 \, \text{mg}$ of organic compound gives $466 \, \text{mg}$ of $BaSO_4$.

Find: Percentage of sulphur in the compound.

In the Carius method, sulphur present in the organic compound is converted to $BaSO_4$.

Molar mass calculation:

$$M(BaSO_4) = 137 + 32 + 4(16) = 233 \, \text{g mol}^{-1}$$

So, $233 \, \text{g}$ of $BaSO_4$ contains $32 \, \text{g}$ of sulphur.

Therefore, $466 \, \text{mg}$ of $BaSO_4$ contains sulphur mass

$$\frac{32}{233} \times 466 \, \text{mg} = 64 \, \text{mg}$$

Now calculate percentage of sulphur in the given $160 \, \text{mg}$ sample:

$$\%S = \frac{64}{160} \times 100 = 40$$

Therefore, the percentage of sulphur in the given compound is $40\%$.

Given: Mass of compound = $160 \, \text{mg}$, mass of $BaSO_4$ formed = $466 \, \text{mg}$.

Find: Percentage of sulphur.

One mole of $BaSO_4$ contains one mole of sulphur.

$$M(BaSO_4) = 137 + 32 + (16 \times 4) = 233 \, \text{g mol}^{-1}$$

Hence,

$$233 \, \text{g of } BaSO_4 \text{ contains } 32 \, \text{g of } S$$

So, for $466 \, \text{mg}$ of $BaSO_4$,

$$\text{Mass of } S = \frac{32}{233} \times 466 \, \text{mg} = 63.95 \, \text{mg} \approx 64 \, \text{mg}$$

Now,

$$\%S = \frac{63.95}{160} \times 100 = 39.97\% \approx 40\%$$

Therefore, the required percentage of sulphur is $40\%$.