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. `4.0cm^(3)`
B. `224cm^(3)`
C. `604cm^(3)`
D. =`(8.0cm^3)`

Answer

B

Explanation

Amount of electricity to deposit 0.16g copper from `CUSO_(4)` solution = `2F = (0.1)/(64) = 0.005F`
Since 2F electricity liberates 1 mol `H_(2)` 0.005 F electricity will liberates
`= (1)/(2F)xx0.005Fmol_(2)= 2.5xx10^(-3)molH_(2)= 2.5xx10^(-3)xx22400mlH_(2)= 56ml` or `56 cm^(3)`