`A` mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of `13.2%` by weight. `2.58g` of the mineral on heating lost `1.232g` of `CO_(2)`. Calculate the `%` by weight of the other metal.

Answer

`21.70`

Explanation

`M_(1)CO_(3)+M_(2)CO_(3),` mol `wt` of `M_(1)=x,M_(2)`
`{:("moles"," "n,n,),(,M_(1)CO_(3),rarrM_(1)O+CO_(2),),(," "n," "n,),(,M_(2)CO_(3),rarrM_(2)O+CO_(2),),(," "n," "n,):}`
`88n=1.232impliesn=0.014`
`(0.014)(60+x)+0.014(60+y)=2.58`
`x+y=64.285" "...(i)`
`(0.014xx x)/(2.58)=(13.2)/(100)`
`x=24.32, y~=40`
`%M_(2)=(0.014xx40)/(2.58)xx100=21.7`