`0.05 g` of a piece of metal in dilute acid gave `24.62 mL` of `H_(2)` at `27^(@)C` and `760mm` pressure. The `Ew` of metal is

Options

A. 25
B. `12.5`
C. 50
D. `37.5`

Answer

A

Explanation

Volume of `H_(2)` at STP `= 24.62 xx (273)/(300) = 22.40 mL`
22400 mL of `H_(2)` at STP = 1 mole = 2 Equivalent of `H_(2)`
22.4 mL of `H_(2) = (2)/(22400) xx 22.4 = 0.002` equivalent of `H_(2) = 0.002` equivalent of metal
`:. ("Weight")/("Equivalent weight") = 0.02 rArr (0.05)/(0.002) = "Equivalent weight" = 25`