1.

The atomicweight of Fe is 56 The weightof Fe deposited fromFeCI_(3)solutionby passing0.6 faraday of electricity is

Answer»

5.6 g
11.2 g
22.4 g
33.6 g

Solution :`FE^(3+) +3e^(-) rarr Fe`
`therefore` 3 MOLESOF electrondeposite = 1 moleof Fe
`therefore`0.6 molesof electron deposite `=1/3xx0.6` moleof Fe
`because` 1 moleof Fe =56 g (atomicweightof fe =56)
`therefore`0.2 moleof Fe `=(56)/(1) xx0.2 =11.2 g`


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