In an experiment, 4g of `M_(2)O_(x)` oxide was reduced to 2.8g of the metal. If the atomic mass of the metal is `56g"mol"^(-1)`, the number of oxygen atoms in the oxide is:

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

`M_(2)O_(x) overset("Reduvtion")toM`
Eq. of `M_(2)O_(x)` = eq. of Metal
`("Wt. of" M_(2)O_(x))/("Eq.wt of " M_(2)O_(x)) = ("Wt.of Metal")/("Eq. wt. of Metal")`
`(4)/((2 xx 56 + x + 16)/(2x)) = (2.8)/((56)/(x))" ".......(i)`
On solving we get,
`(4)/(56+8x) = (2.8)/(56)`
` (1)/(14+2x) = (1)/(20)`
2x = 6
x = 3
Hence , the oxide is `M_(2)O_(3)`.