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When 0.15 g of an organic compound was analyzed using

Question:

When 0.15 g of an organic compound was analyzed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is _______. (Nearest integer)

[Atomic mass : Silver = 108, Bromine = 80]

Solution:

Moles of $\mathrm{Br}=$ Moles of $\mathrm{AgBr}$ obtained

$\Rightarrow$ Mass of $\mathrm{Br}=\frac{0.2397}{188} \times 80 \mathrm{~g}$

therefore $\% \mathrm{Br}$ in the organic compound

$=\frac{\mathrm{W}_{\mathrm{Br}}}{\mathrm{W}_{\mathrm{T}}} \times 100$

$=\frac{0.2397 \times 80}{188 \times 0.15} \times 100=0.85 \times 80$

= 68

$\Rightarrow$ Nearest integer is '68'