`4.2 g` of a metallic carbonate `MCO_(3)` was heated in a hard glass tube and `CO_(2)` evolved was found to have `1120 mL` of volume at `STP`. The `Ew` of the metal is

Options

A. 12
B. 24
C. 18
D. 15

Answer

A

Explanation

`22400 mL = 1 "mol of" CO_(2) = 2 Eq "of" CO_(2)`
`11200 mL = 1//2 "mol of" CO_(2)` at `STP` would be is `Ew`.
`Ew` of metallic carbonate `= (4.2 xx 11200)/(1120) = 42 g`
`Ew` of metal `= Ew "of" MCO_(3) - Ew "of" CO_(3)^(2-)`
`= 42 - 30 = 12`
`[Ew "of" CO_(3)^(2-) = (60)/(2) = 30]`
Alternatively:
`22400 mL = 1 "mol of" CO_(2) = 2 Eq CO_(3)^(2-)`
`11200 mL implies 1 Eq "of" CO_(3)^(2-)`
`: 1120 mL implies (1)/(11200) xx 1120 = 0.1 Eq "of" CO_(3)^(2-)`
Eq of `MCO_(3) = Eq "of" CO_(3)^(2-)`
`("weight")/(Ew) = ("Weight")/(Ew) = 0.1 Eq`
`:. 0.1 = (4.2)/(Ew "of" MCO_(3))`
`:. 0.1 Eq "of" MCO_(3) = 42`
`:. Ew "of" M = Ew "of" MCO_(3) - Ew "of" CO_(3)^(2-)`
`= 42 - 30 = 12`