InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`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 isA. 12B. 24C. 18D. 15 |
|
Answer» Correct Answer - A `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` |
|