0.20 g of an organic compound gave 0.12 g of AgBr By using Carius method, the percentage of bromine in the compound will be

Correct answer: D. 25 \% Given, Mass of an organic compound =0.20 g Mass of AgBr=0.12 g Molecular mass of AgBr=188 g mol^ -1 188 g of AgBr contains 80 g of bromine 0.12 g of AgBr will contain =(80)/(188) × 0.12=0.05 g of bromine Percentage of bromine =(0.05)/(0.20) × 100=25 \%

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Chemistry
Chapter: General Organic Chemistry

BITSAT 2022 · Chemistry · General Organic Chemistry

0.20 g of an organic compound gave 0.12 g of AgBr By using Carius...

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

Options

A $34.06 \%$
B $44.04 \%$
C $54 \%$
D $25 \%$

Correct answer

D. $25 \%$

Step-by-step solution

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}

\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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