When 0.04 faraday of electrcity is passed through a solution of `CaSO_4`. Then the weight of `Ca^(2+)` metal deposited at the cathode isA. 0.2 gmB. 0.4 gmC. 0.6 gmD. 0.8 gm

When 0.04 faraday of electrcity is passed through a solution of `CaSO_

1.	When 0.04 faraday of electrcity is passed through a solution of `CaSO_4`. Then the weight of `Ca^(2+)` metal deposited at the cathode isA. 0.2 gmB. 0.4 gmC. 0.6 gmD. 0.8 gm
Answer» Correct Answer - 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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When 0.04 faraday of electrcity is passed through a solution of `CaSO_4`. Then the weight of `Ca^(2+)` metal deposited at the cathode isA. 0.2 gmB. 0.4 gmC. 0.6 gmD. 0.8 gm

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