Same amount of electric current is passed through solutions of `AgNO_(3)` and HCl. If 1.08 g of silver is obtained in the first case, the amount of hydrogen liberated at S.T .P. in the second case is:

Options

A. `112 cm^(3)`
B. `22400 cm^(3)`
C. `224 cm^(3)`
D. `1.008 g`

Answer

A

Explanation

`("Eq. wt. of Ag")/("Eq. wt. of" H_(2)) = ("Mass of Ag")/("Mass of" H_(2))`
`(108)/(1) = (1.08)/(Mass of H_(2))`
`therefore` Mass of `H_(2) = (1.08)/(108) = 10^(-2)` g
2 g of `H_(2)` occupies at STP = `22400 cm^(3)`
`10^(-2)` g of `H_(2)` occupies at STP = `112 cm^(3)`