JEE MAIN - Chemistry (2025 - 24th January Evening Shift - No. 23)
In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide ( AgBr ). The percentage of Bromine in the organic compound is ________ $\times 10^{-1} \%$ (Nearest integer).
(Given : Molar mass of Ag is 108 and Br is $80 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Answer
255
Explanation
$$\begin{aligned}
& \% \text { Bromine }= \frac{\text { Molar Mass of Bro mine }}{\text { Molar Mass of Silver bromide }} \\
& \times \frac{\text { Weight of } \mathrm{AgBr}}{\text { Weight of sample }} \times 100 \\
&=\frac{80}{188} \times \frac{0.165}{0.25} \times 100 \\
&= \frac{4800}{188}=25.53=255 \times 10^{-1}
\end{aligned}$$
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