In Carius method of estimation of halogen, $1$ g of an organic compound gave $0.094$ g of $AgBr$ . The percentage of bromine in the compound is :

4 %

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5 %

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7 %

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6 %

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Solution

The correct option is D $4 %$
Mass of halogen (Br) in $AgBr$ $= 0.094 \times \frac{80}{188} = 0.04$ g.
$%$ of halogen $= \frac{Mass of halogen \times 100}{Mass of organic compound taken} = \frac{0.04 \times 100}{1} = 4$ %

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In Carius method of estimation of halogen, 1 g of an organic compound gave 0.094 g of AgBr. The percentage of bromine in the compound is :

The correct option is D 4%Mass of halogen (Br) in AgBr =0.094×80188=0.04 g.% of halogen =Mass of halogen×100Mass of organic compound taken=0.04×1001=4 %

In Carius method of estimation of halogen, $1$ g of an organic compound gave $0.094$ g of $AgBr$ . The percentage of bromine in the compound is :

The correct option is D $4 %$
Mass of halogen (Br) in $AgBr$ $= 0.094 \times \frac{80}{188} = 0.04$ g.
$%$ of halogen $= \frac{Mass of halogen \times 100}{Mass of organic compound taken} = \frac{0.04 \times 100}{1} = 4$ %