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 acidullated water, then the volume of `H_(2)` gas liberated at S.T.P. will be (Given at wt. of Cu=64)

Options

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

Answer

B

Explanation

`("Mass of Cu deposited")/("Volume of" H_("liberated"))=("Eq. wt of Cu")/("Eq. wt. of H")`
`(0.18)/("Mass of" H_(2) "liberated")=(64//2)/(1)`
or mass of `H_(2)` liberated `=(0.16)/(32)g=5xx10^(-3)g`
`therefore` volume of `H_(2)` liberated at STP
`=(22400)/(2)xx5xx10^(-3)cm^( 3)=56cm ^(3)`