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Charge required to liberated 11.5 g sodi...

Charge required to liberated `11.5 g` sodium is .

A

0.5 F

B

0.1 F

C

1.5 F

D

96500 coulombs

Text Solution

Verified by Experts

The correct Answer is:
A
`Na^(+) (aq) + e^(-) to Na(s)`
Charge (in F) = moles of `e^(-)` used = moles of Na deposited-
`= 11.5/23 g = 0.5` Faraday
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