0.3 g of a metal reacts with dilute acid and produce 110 mL of `H_(2)` which is collected above water at `17^(@)C` temperature and 755 mm Hg pressure. Find equivalent mass of the metal. [pressure of water vapour at `17^(@)C=14.4mmHg`].

Explanation

Actual pressure of `H_(2)` gas`=(755-14.4)=740.6m m Hg`
Let, the volume of `H_(2)` gas at STP be V mL, then
`(110xx740.6)/((273+17))=(Vxx760)/(273)" "therefore V=100.91mL`
At STP, mass of 100.91 mL `H_(2)`
`=(2xx1.008xx100.91)/(22400)=0.00908g`
`therefore 1.008g" "H_(2)` replaces`(0.3xx1.008)/(0.00908)=33.3g` metal,
`E_("metal")=33.3`