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BITSAT 2022 Paper

Section: Chemistry

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Question : 60 of 130

Marks: +1, -0

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

$34.06\%$
$44.04\%$
$54\%$
$25\%$

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\,mol}^{-1}

188\ \mathrm{g}

of

\mathrm{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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