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An organic compound weighing $0.15 \mathrm{~g}$ gave on Carius estimation, $0.12$ gf $\mathrm{AgBr}$.
The percentage of Br in the compound will be close to (Atomic mass of Ag $=108, \mathrm{Br}=80$ )
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The percentage of Br in the compound will be close to (Atomic mass of Ag $=108, \mathrm{Br}=80$ )
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Verified Answer
The correct answer is:
$34.1 \%$
Percentage of bromine in the compound
$$
\begin{aligned}
&=\frac{80}{188} \times \frac{\text { Mass of } \mathrm{AgBr} \text { formed }}{\text { Mass of substance taken }} \times 100 \\
&=\frac{80}{188} \times \frac{0.12}{0.15} \times 100=34.04 \%
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{80}{188} \times \frac{\text { Mass of } \mathrm{AgBr} \text { formed }}{\text { Mass of substance taken }} \times 100 \\
&=\frac{80}{188} \times \frac{0.12}{0.15} \times 100=34.04 \%
\end{aligned}
$$
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