Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulfate solution for 100 seconds is: (Given : Molar mass of Cu : $63 g {mol}^{- 1} 1 F = 96487 C)$

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0.0315 g

3.15 g

0.315 g

31.5 g

answer is D.

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Detailed Solution

$\begin{aligned} W = ZQ \\ W = \frac{Gmw}{valency} \times \frac{i \cdot t}{F} \\ W = \frac{63}{2} \times \frac{9.6487 \times 100}{96487} \\ W = 0.315 g \end{aligned}$

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