On reduction with hydrogen, `3.6 g` of an oxide of matel left `3.2 g` of metal. If the vapour density of metal is `32`, the simplest formula of the oxide would be

Options

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

Answer

C

Explanation

We know that,
Equivalent weight ` = ("Weight of Metal")/("Weight of Oxygen") xx 8`
` = (3.2)/(0.4) xx 8 = 64`
Vapour density ` = ("mol . Wt")/(2)`
Mol. `wt = 2 xx V.D = 2 xx 32 = 64`
As we know that `n = ("mol.wt")/("eq.wt") = (64)/(64) = 1`
Suppose, the formula of metal oxide be `M_(2)O_(n)`
Hence the formula of metal oxide ` = M_(2)O`