1 litre of $1 M$ $C u {S O}_{4}$ solution is electrolysed. After passing $2 F$ charge, the molarity of $C u {S O}_{4}$ will be:

M / 2

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M / 4

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M

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zero

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Solution

The correct option is D zero
$1 M C u S O_{4}$
Moles $\times$ Volume factor $= \frac{q}{F}$
Moles = Molarity $\times$ volume
$= 1 \times 1 = 1 m o l e s$
$1 \times 2 = \frac{q}{f}$
$q = 2 f$
Therefore for passing $2 F$ charge would consume the solution. Completely and hence molarity becomes zero.

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