In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine (Nearest integer) in the compound is .

[Atomic mass: Ag = 108 u, Br = 80 u]

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40.00

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40.0

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Solution

Mass of organic compound = 0.2 g

Mass of AgBr = 0.188 g

Mass of Br = $\frac{0.188 \times 80}{188} = 0.08 g$

$Percentage of Br in the compound = \frac{0.08 \times 100}{0.2}$

$Percentage of Br in the compound = 40 %$

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In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine (Nearest integer) in the compound is . [Atomic mass: Ag = 108 u, Br = 80 u]

Mass of organic compound = 0.2 g Mass of AgBr = 0.188 g Mass of Br =0.188×80188=0.08 g Percentage of Br in the compound =0.08×1000.2 Percentage of Br in the compound =40%

In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine (Nearest integer) in the compound is .

[Atomic mass: Ag = 108 u, Br = 80 u]

Mass of organic compound = 0.2 g

Mass of AgBr = 0.188 g

Mass of Br = $\frac{0.188 \times 80}{188} = 0.08 g$

$Percentage of Br in the compound = \frac{0.08 \times 100}{0.2}$

$Percentage of Br in the compound = 40 %$