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 is

Answer»

0.2 gm
0.4 gm
0.6 gm
0.8 gm

SOLUTION :`CA^(++)+2e^(-)toCa`
`E_(Ca)=(40)/(2)=20`
`W_(Ca)=E_(Ca)XX`No.of faradays=`20xx0.04=0.8gm`


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