0.24 gm of metal is dissolved completely in excess of dil. `H_(2)SO_(4)` solution. If 82.5ml dry hydrogen gas is liberated at `27^(@)C` and 95cm Hg. What is the equivalent weight of the metal ?

Explanation

From ideal gas equation, `PV =nRT`
`( 95)/( 76) xx( 82)/( 1000) = ( w)/( 2) = 0.082 xx 300`
Weight of hydrogen gas liberated out, w `=8.38 xx 10^(-3) gm` hydrogen is liberated from 0.24 gm metal
`:. ` 1 gm hydrogen will be liberated from `( 0.24)/( 8.38 xx 10^(-3)) xx 1 `
`= 28.64 gm` Equivalent weight of metal = 28.64