5.249g of metallic carbonate on being strongly heated liberates 1309.28 mL of `CO_(2)` at `27^(@)C` annd 755 mm pressure of Hg. Find the equivalent mass of the metal.

Explanation

let, at `27^(@)C` temperature and 755 mmHg pressure,
volume of 1309.28 mL `CO_(2)` gas at STP be V mL, then
`(755xx1309.28)/((273+27))=(760xxV)/(273)" "[because(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))]`
or, V=1183.6mL
At STP, mass of 22400 mL `CO_(2)` gas=44g
`therefore`At STP, 1183.6 mL `CO_(2)` gas `=(44)/(22400)xx1183.6=2.325g`
`("mass of metallic carbonate")/("mass of "CO_(2)" produced")`
`=("equivalent mass of metallic carbonate")/("equivalent mass of "CO_(2))`
or, `(5.249)/(2.325)=(E+30)/(22)` [E=equivalent mass of the metal]
Equivalent mass of `CO_(3)^(2-)=(60)/(2)=30` [Equivalent mass of `(44)/(2)=22`]
`thereforeE=19.66~~20`