0.25 g of an organic compound, on treatment with fuming nitric acid followed by AgNO3, gave 0.22 g of AgBr. The percentage of bromine in the compound is ______.
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Correct answer: D
Explanation
The correct answer is Option 4 i.e. 37.45%.
To calculate the percentage of bromine:
Mass of AgBr = 0.22 g
Molar mass of AgBr = 187.77 g/mol
Moles of AgBr = 0.22 ÷ 187.77 ≈ 0.00117 mol
Each mole of AgBr contains 1 mole of Br.
Mass of Br = 0.00117 × 79.9 ≈ 0.0935 g
Percentage of bromine = (0.0935 ÷ 0.25) × 100 ≈ 37.4%
Thus, the organic compound contains about 37.4% bromine by mass.