An oxide of a metal contains 60% metal. If vapour density of chloride of metal is 47.5. Find the atomic weight of metal.

In metal oxide, metal is 60% which means that in 100 g of metal oxide, 60 g is metal and 40 g of oxygen.

$$\begin{gathered} \frac{\mathrm{weight} \mathrm{of} \mathrm{metal}}{\mathrm{weight} \mathrm{of} \mathrm{oxygen}}=\frac{\mathrm{equivalent} \mathrm{weight} \mathrm{of} \mathrm{metal}}{\mathrm{equivalent} \mathrm{weight} \mathrm{of} \mathrm{oxygen}} \\ \frac{60}{40}=\frac{\mathrm{E}}{8} \\ \mathrm{E}=12 \end{gathered}$$

Molecular weight of metal chloride is equal to its equivalent weight multiplied by valency.

 $\mathrm{Molecular} \mathrm{weight}=\left(\mathrm{Equivalent} \mathrm{weight}\right)\mathrm{x}+\left(35.5\right)\mathrm{x}$ 

Here, x is valency. 

Molecular weight is twice of vapour density. So, the value of x is calculated as:

 $$\begin{gathered} \left(12\right)\mathrm{x}+\left(35.5\right)\mathrm{x}=2\times 47.5 \\ \mathrm{x}=2 \end{gathered}$$

Atomic weight of metal (At) is:

$\mathrm{At}=12\times 2=24$