Vapour density of a metal chloride is 66 . Its oxide contains $53\%$ metal. The atomic weight of the metal is

The vapour density of metal chloride given is $=66$

The atomic weight of oxygen $=8$

Let the total weight of the metal oxide is $100 \mathrm{g}$.

The weight of metal given is $53 \mathrm{g}$. So, the weight of oxygen is $=100-53=47 \mathrm{g}$.

,Equivalent weight of metal $=($ weight of metal $)/($ weight of oxygen $)\times$ atomic weight of oxygen

Equivalent wt. $=\frac{53}{47}\times 8=9.02$

The valency $=\frac{2\times \mathrm{VD}}{\mathrm{E}+35.5}$

$\mathrm{Valency}=2\times 66/9.02+35.5=3$

Hence, the atomic weight of the metal $=$ equivalent weight $\times$ valency

Atomic weight of the metal $=9.02\times 3=27.06$