Two metallic oxides contain `27.6%` and `30%` oxygen, respectively. If the formula of the second oxide is `M_(2)O_(3)`, that of the first will be

Options

A. `M_(3) O_(4)`
B. `M_(2) O_(5)`
C. `MO_(3)`
D. `M_(2) O`

Answer

A

Explanation

In `M_(2) O_(3)`, 30 parts of `O` combine with 70 (parts of `M`. Thus, 70 parts of `M` in `M_(2)O` correspond to 2 atoms of `M`. In first oxide, `72.4` parts will correspond to `(2)/(70) xx 74.7 = 2.07 M` atoms. 30 parts of `O` in `M_(2) O_(3)` corrrespond to 3 atom of `O`. In first oxide, 27.6 parts of `O` will correspond to `(3)/(30) xx 27.6 = 2.76 O` atoms. Ratio of `M` to `O` in the first oxide `= 2.07 : 2.76 = (2.07)/(2.07) = (2.76)/(2.07) = 1 : 1 : 33 = 3 : 4`