When dissolved in dilute `H_(2)SO_(4), 0.275 g` of metal evolved `119.7 mL` of `H_(2)` at `20^(@)C` and `780.4 mm` pressure. `H_(2)` was collected over water. Aqueous tension is `17.4` mm at `20^(@)C`. Calculate equivalent weight of metal.

Explanation

Mole of `H_(2(n))=(PV)/(RT)`
`=((780.4-17.4)xx119.7)/(760xx1000xx0.0821xx293)=5xx10^(-3)`
Let eq.wt. of metal be `E`.
Eq. of metal =Eq. of `H_(2)`
`0.275/E`= mole of `H_(2)xx2=2xx5xx10^(-3)`
`E=27.52`