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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Chemistry
Chapter
General Organic Chemistry

JEE Main 2025 · 24 January · Evening Shift · Chemistry · General Organic Chemistry

In Carius method of estimation of halogen, 0.25 g of an organic...

In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr)(\mathrm{AgBr}). The percentage of Bromine in the organic compound is ……\ldots\ldots ×10−1%\times 10^{-1 \%} (Nearest integer). (Given : Molar mass of Ag is 108 and Br is 80 g mol−180 \mathrm{~g} \mathrm{~mol}^{-1} )

Correct answer

0

Step-by-step solution

Mass of organic compound =0.25 g=0.25 \mathrm{~g} Mass of AgBr=0.15 g\mathrm{AgBr}=0.15 \mathrm{~g} No. of moles of Br=\mathrm{Br}= No. of moles of AgBr=0.15188\mathrm{AgBr}=\frac{0.15}{188} Mass of Br=0.15×80188 g% of Br=0.15×80×100188×0.25=25.5%=255×10−1%\begin{aligned} & \text{ Mass of } \mathrm{Br}=\frac{0.15 \times 80}{188} \mathrm{~g} \\ & \% \text{ of } \mathrm{Br}=\frac{0.15 \times 80 \times 100}{188 \times 0.25} \\ & =25.5 \% \\ & =255 \times 10^{-1} \% \end{aligned}$$
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