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 : at.wt. of Cu = 64].

Options

A. `4.0 cm^(3)`
B. `56 cm^(3)`
C. `604 cm^(3)`
D. `8.0 cm^(3)`

Answer

B

Explanation

`("wt. of Cu deposited")/("wt. of " H_(2) " produced")=("eq. wt. of Cu")/("eq. wt. of H")=(64//2)/(1)`
or, `(0.16)/("wt. of " H_(2))=(32)/(1)`
or, wt. of `H_(2) =(0.16)/(32)=5xx10^(-3)g`.
`therefore` Volume of `H_(2)` liberated at STP
`=(22400)/(2)xx5xx10^(-3)` cc = 56 cc.