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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^+ + e^- rarr N`
Charge (in F) = moles of `e^-)` used = moles of Na deposited
`= (11.5)/(23) g= 0.5 ` Faraday.
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