When a quantity of electricity is passed through `CuSO_(4)` solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of `H_(2)` liberated at STP will be : (given atomic weight of Cu=64)

Options

A. `56 cm^(3)`
B. `550 cm^(3)`
C. `5.6 cm^(3)`
D. `11.2 cm^(3)`

Answer

C

Explanation

`W_(Ag)/W_(O) = E_(Ag)/E_(O)`
`0.108/W_(O) = 1-8/8`
`W_(o) = 8 xx 10^(-3) g`
`(22.4 xx 10^(3))/32 xx 8 xx 10^(-3) = 5.6 mL`