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A current of 15.0 amperes is passed through a solution of $\mathrm{CrCl}_2$ for 45 minutes. The volume of $\mathrm{Cl}_2$ (in $\mathrm{L}$ ) obtained at the anode at $1 \mathrm{~atm}$ and $273 \mathrm{~K}$ is around $\left(1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right.$, At. wt. of $\mathrm{Cl}=35.5, \mathrm{R}=0.082$ $\mathrm{L}$-atmK ${ }^{-1} \mathrm{~mol}^{-1}$ )
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The correct answer is:
$4.7$
$\mathrm{Cl}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}$
$\begin{aligned} & \mathrm{m}=\frac{\mathrm{i} \times \mathrm{i} \times \mathrm{E}}{96500}=\frac{15.0 \times(45 \times 60) \times 35.5}{2 \times 96500} \\ & =7.445 \mathrm{~g} \\ & \Rightarrow \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{7.445 \times 0.082 \times 273}{35.5 \times 1} \\ & =4.7 \mathrm{~L} .\end{aligned}$
$\begin{aligned} & \mathrm{m}=\frac{\mathrm{i} \times \mathrm{i} \times \mathrm{E}}{96500}=\frac{15.0 \times(45 \times 60) \times 35.5}{2 \times 96500} \\ & =7.445 \mathrm{~g} \\ & \Rightarrow \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{7.445 \times 0.082 \times 273}{35.5 \times 1} \\ & =4.7 \mathrm{~L} .\end{aligned}$
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