In Carius method of estimation of halogen. 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound. (Molar masses : $\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1}: \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Select the correct option:
A
34.04%
B
40.04%
C
36.03%
D
38.04%
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Solution
Mass of organic compound $=0.45 \mathrm{gm}$
Mass of AgBr obtained $=0.36 \mathrm{gm}$
∴ Moles of $\mathrm{AgBr}=\frac{0.36}{188}$
$$
\begin{aligned}
& \therefore \text { Mass of Bromine }=\frac{0.36}{188} \times 80=0.1532 \mathrm{gm} \\
& \therefore \mathrm{Br} \text { in compound }=\frac{0.1532}{0.45} \times 100=34.04 \%
\end{aligned}
$$
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