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("Reduction")(rarr) M`
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/((2xx56+xxx16)/(2x))=2.8/(56/x)` ...(i)
On solving we get,
`4/(56+8x)=2.8/56 implies 1/(14+2x)=1/20 implies 2x=6 implies x=3`
Hence, the oxide is `M_(2)O_(3)`.