JEE MAIN - Chemistry (2022 - 27th July Morning Shift - No. 12)
In Carius method of estimation of halogen, $$0.45 \mathrm{~g}$$ of an organic compound gave $$0.36 \mathrm{~g}$$ of $$\mathrm{AgBr}$$. Find out the percentage of bromine in the compound.
(Molar masses : $$\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$$)
34.04%
40.04%
36.03%
38.04%
Explanation
Mass of organic compound $=0.45 \,\mathrm{gm}$
Mass of $\mathrm{AgBr}$ obtained $=0.36 \,\mathrm{gm}$
$\therefore$ Moles of $\mathrm{AgBr}=\frac{0.36}{188}$
$\therefore$ Mass of Bromine $=\frac{0.36}{188} \times 80=0.1532 \,\mathrm{gm}$
$\therefore \%\, \mathrm{Br}$ in compound $=\frac{0.1532}{0.45} \times 100=34.04 \,\%$
Mass of $\mathrm{AgBr}$ obtained $=0.36 \,\mathrm{gm}$
$\therefore$ Moles of $\mathrm{AgBr}=\frac{0.36}{188}$
$\therefore$ Mass of Bromine $=\frac{0.36}{188} \times 80=0.1532 \,\mathrm{gm}$
$\therefore \%\, \mathrm{Br}$ in compound $=\frac{0.1532}{0.45} \times 100=34.04 \,\%$
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