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In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (at. mass Ag = 108; Br = 80)

1. 24
2. 36
3. 48
4. 60

Correct answer: 24

Solution

To find the percentage of bromine in the compound, we first determine the moles of AgBr produced, which is 141 mg divided by the molar mass of AgBr (188 mg/mol). This gives us the moles of bromine, which is then converted to mass and expressed as a percentage of the original 250 mg of the organic compound, resulting in 24%.

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