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. `4.0 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") =(64//2)/1`
or Wt. of `H_(2) = 0.16/3 = 5 xx 10^(-3)`g
`therefore` Volume of `H_(2)` liberated at STP
`=22400/2 xx 5 xx 10^(-3), cc = 56 cc`