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A certain current liberates 0.504g of H(...

A certain current liberates `0.504g` of `H_(2)(g)` in 2 hours. The weight of `Cu` deposited by same current flowing for the same time in `CuSO_(4)` solution is `……………………..g`.

Text Solution

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The correct Answer is:
`(16g)`
`(16g)`
`(W_(H_(2)))/(Ew(H_(2)))=(W_(Cu))/(Ww(Cu))`
`implies(0.0504)/(1)=(W_(Cu))/(63.5//2)impliesW_(Cu)=16.0g`
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