Mole ConceptMedium
Question
Mass in grams of copper deposited by passing 9.6487 A current through a voltameter containing copper sulphate solution for 100 seconds is (Given : Molar mass of $Cu:63\text{ }g{\text{ }mol}^{- 1},1\text{ }F = 96487C$ )
Options
A.31.5 g
B.0.0315 g
C.3.15 g
D.0.315 g
Solution
Sol. [Key Concept: Faraday law for copper deposition]
$${Cu}^{2 +}(aq) + 2e^{-} \rightarrow Cu(s)$$
$$\begin{matrix} \text{~Mass of~}Cu\text{~deposited~}(w) & \ = \frac{M \times i \times t}{nF} \\ & \ = \frac{63 \times 9.6487 \times 100}{2 \times 96487} \\ & \ = 0.315\text{ }g \end{matrix}$$
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