0.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is____
(Given: Molar mass AgBr=188g mol-1)
(Given: Molar mass AgBr=188g mol-1)
Correct Answer: 40
Solution and Explanation
1. Moles of AgBr formed: \[ \text{Moles of AgBr} = \frac{\text{Mass of AgBr}}{\text{Molar mass of AgBr}} = \frac{0.376}{188} = 0.002 \, \text{mol}. \] 2. Moles of Br: \[ \text{Moles of Br} = \text{Moles of AgBr} = 0.002 \, \text{mol}. \] 3. Mass of Br: \[ \text{Mass of Br} = \text{Moles of Br} \times \text{Molar mass of Br} = 0.002 \times 80 = 0.16 \, \text{g}. \] 4. Percentage of Br in compound X: \[ \% \text{of Br} = \frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100 = \frac{0.16}{0.400} \times 100 = 40\%. \]
Final Answer: \( \boxed{40\%} \).
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