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JEE Main 2015 Paper

Section: Chemistry

Question : 23 of 90

Marks: +1, -0

Solution:

Let the organic compound is R – Br

\mathrm{R-Br}+\mathrm{AgNO}_3\rightarrow\mathrm{AgBr}\downarrow

So number of moles of AgBr ≡ Number of moles of R – Br ≡ Number of moles of Br

\frac{141 \times 10^{-3}}{188}=

number of moles of Br

=0.75 \times 10^{-3}

Mass of bromine

=0.75 \times 80 \times 10^{-3}\text{g}=60

Percentage of bromine

=\frac{60}{250}\times100=24\%

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