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$0.20 \mathrm{~g}$ of an organic compound gave $0.12 \mathrm{~g}$ of AgBr By using Carius method, the percentage of bromine in the compound will be
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$25 \%$
$25 \%$
Given,
Mass of an organic compound $=0.20 \mathrm{~g}$
Mass of $\mathrm{AgBr}=0.12 \mathrm{~g}$
Molecular mass of $\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1}$
$188 \mathrm{~g}$ of AgBr contains $80 \mathrm{~g}$ of bromine
$\therefore 0.12 \mathrm{~g}$ of $\mathrm{AgBr}$ will contain $=\frac{80}{188} \times 0.12$
$=0.05 \mathrm{~g}$ of bromine
$\therefore$ Percentage of bromine $=\frac{0.05}{0.20} \times 100$
$=25 \%$
Mass of an organic compound $=0.20 \mathrm{~g}$
Mass of $\mathrm{AgBr}=0.12 \mathrm{~g}$
Molecular mass of $\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1}$
$188 \mathrm{~g}$ of AgBr contains $80 \mathrm{~g}$ of bromine
$\therefore 0.12 \mathrm{~g}$ of $\mathrm{AgBr}$ will contain $=\frac{80}{188} \times 0.12$
$=0.05 \mathrm{~g}$ of bromine
$\therefore$ Percentage of bromine $=\frac{0.05}{0.20} \times 100$
$=25 \%$
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