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When 0.04 F of electricity is passed thr...

When 0.04 F of electricity is passed through a solution of `CaSO_(4)`, then the weight of `Ca^(2+)` metal deposited at the cathode is

A

0.2 gm

B

0.4 gm

C

0.6 gm

D

0.8 gm

Text Solution

Verified by Experts

The correct Answer is:
D
`Ca^(++)+2e^(-)toCa`
`E_(Ca)=(40)/(2)=20`
`W_(Ca)=E_(Ca)xx`No.of faradays=`20xx0.04=0.8gm`
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