A current of 15.0 amperes is passed through a solution of CrCl_2 for 45 minutes. The volume of Cl_2 (in L ) obtained at the anode at 1 ~atm and 273 ~K is around (1 ~F=96500 C mol^-1 . , At. wt. of Cl=35.5, R=0.082 L -atmK ^-1 ~mol^-1 )

Correct answer: (A) 4.7. Cl_2+2 e^- arrow 2 Cl^- m=(i × i × E)/96500=(15.0 ×(45 × 60) × 35.5)/(2 × 96500) =7.445 ~g ⇒ V=nRT/P=(7.445 × 0.082 × 273)/(35.5 × 1) =4.7 ~L .

Q22 / 160 · TS EAMCET 2023 (14 May Shift 1) · Chemistry

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 ${ }^{-1} \mathrm{~mol}^{-1}$ )

Correct answer
A
$4.7$
B
$3.7$
C
$2.7$
D
$5.7$

Solution

$\mathrm{Cl}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}$ $\begin{aligned} & \mathrm{m}=\frac{\mathrm{i} \times \mathrm{i} \times \mathrm{E}}{96500}=\frac{15.0 \times(45 \times 60) \times 35.5}{2 \times 96500} \\ & =7.445 \mathrm{~g} \\ & \Rightarrow \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{7.445 \times 0.082 \times 273}{35.5 \times 1} \\ & =4.7 \mathrm{~L} .\end{aligned}$

Hence, the correct answer is option: 'A'.

Solution

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