JEE Chemistry 2023 Question with Solution

Given: Mass of compound $X$ = $0.400 \, \text{g}$, mass of AgBr formed = $0.376 \, \text{g}$.

Find: Percentage of bromine in compound $X$.

In Carius method, bromine is estimated as AgBr.

Moles of AgBr formed:

$$\text{Moles of AgBr} = \frac{\text{Mass of AgBr}}{\text{Molar mass of AgBr}} = \frac{0.376}{188} = 0.002 \, \text{mol}$$

Moles of Br:

$$\text{Moles of Br} = \text{Moles of AgBr} = 0.002 \, \text{mol}$$

Mass of Br:

$$\text{Mass of Br} = \text{Moles of Br} \times \text{Molar mass of Br} = 0.002 \times 80 = 0.16 \, \text{g}$$

Percentage of Br in compound $X$:

$$\% \, \text{of Br} = \frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100 = \frac{0.16}{0.400} \times 100 = 40\%$$

Therefore, the percentage of bromine is $40\%$. The correct option is A.

Given: $0.376 \, \text{g}$ of AgBr is obtained from $0.400 \, \text{g}$ of compound $X$.

Find: Percentage of bromine in the compound.

Since $188 \, \text{g}$ of AgBr contains $80 \, \text{g}$ of Br, the mass of bromine in $0.376 \, \text{g}$ of AgBr is:

$$\text{Mass of Br} = \frac{80}{188} \times 0.376 = 0.16 \, \text{g}$$

Now calculate percentage of bromine:

$$\% \, \text{Br} = \frac{0.16}{0.400} \times 100 = 40\%$$

This works because each mole of AgBr contains exactly one mole of bromine. Hence, the correct option is A.