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  4. Charge required to liberate 11.5 g sodium is

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Q. Charge required to liberate $11.5\, g$ sodium is

Electrochemistry

Solution:

$Na^{+} + e^{-} \to Na$
Charge (in F) = moles of $e^{-}$ used = moles of Na deposited =$\frac{11.5}{25}g=0.5 F$