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Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of $\mathrm{Cu}: 63 \mathrm{~g} \mathrm{~mol}^{-1}, 1 \mathrm{~F}=96487 \mathrm{C}$ )
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0.315 g
$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$
$\begin{aligned} \text { Mass of Cu deposited }(w) & =\frac{M \times i \times t}{n F} \\ & =\frac{63 \times 9.6487 \times 100}{2 \times 96487} \\ & =0.315 \mathrm{~g}\end{aligned}$
$\begin{aligned} \text { Mass of Cu deposited }(w) & =\frac{M \times i \times t}{n F} \\ & =\frac{63 \times 9.6487 \times 100}{2 \times 96487} \\ & =0.315 \mathrm{~g}\end{aligned}$
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