In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is ____________ × 10^–1% (Nearest integer).
(Given : Molar mass of Ag is 108 and Br is 80 g mol^–1)

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answer is 255.

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Detailed Solution

$% Bromine = \frac{Molar Mass of Bro mine}{Molar Mass of Silver bromide} \times \frac{Weight of AgBr}{Weight of sample} \times 100$
$= \frac{80}{188} \times \frac{0.165}{0.25} \times 100 = \frac{4800}{188} = 25.53 = 255 \times 10^{- 1}$

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