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 : $$\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$$)
Solution
Mass of organic compound $=0.45 \,\mathrm{gm}$<br/><br/>
Mass of $\mathrm{AgBr}$ obtained $=0.36 \,\mathrm{gm}$<br/><br/>
$\therefore$ Moles of $\mathrm{AgBr}=\frac{0.36}{188}$<br/><br/>
$\therefore$ Mass of Bromine $=\frac{0.36}{188} \times 80=0.1532 \,\mathrm{gm}$<br/><br/>
$\therefore \%\, \mathrm{Br}$ in compound $=\frac{0.1532}{0.45} \times 100=34.04 \,\%$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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