A given sample of pure compound cotains `9.81 gm` of `Zn,1.8 xx 10^23` atoms of chromium and `0.60` mole of oxygen atoms. What is the simplest formula ?

Answer

4

Explanation

`ZnCr_(2)O_(x), Zn : Cr : O = 1:2 :x`
So for `1.8 xx 10^(23)` atoms of Cr i.e. 0.3 mol of `Cr, (0.3)/(2) xx x` mol of O atoms must be present.
`(0.3)/(2) xx x = 0.6 rArr x = 4`