1.

Two oxides of a metal contain 36.4% and 53.4% of oxygen by mass respectively. If the formula of first oxide is M_(2)O, then that of the second is

Answer»

`M_(2)O_(3)`
`MO`
`MO_(2)`
`M_(2)O_(5)`

Solution :In `M_(2)O`, 36.4 PARTS of O combine with M = 63.6
parts, i.e., 2 ATOMS of `M-=63.6` parts
`"or1 ATOM of M = 31.8 parts"`
1 atom of O = 36.4 parts
In 2nd oxide, oxygen present = 53.4 parts
`therefore"M present = 46.6 parts"`
31.8 parts of M = 1 atom of M
`"46.6 parts of M"=(1)/(31.8)xx46.6" parts = 1.5 atoms"`
36.4 parts of O = 1 atom of O
`"53.4 parts of O"=(1)/(36.4)xx53.4=1.5" atoms"`
`therefore" In 2nd oxide M : O" = 1.5:1.5=1:1`
Hence, formula is MO
Alternatively, if m is the atomic mass of M,
For `M_(2)O,(16)/(14+16X)xx100=53.4`
`or (16x)/(14+16x)=0.534 or 16x=.476+8.544x`
or `7.456x=7.476 or x=1`
Hence, the formula is MO.


Discussion

No Comment Found

Related InterviewSolutions