Question

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

Answer 1

Step 1: Calculation of Equivalent weight of the metal

Let the weight of metal oxide be $100\mathrm{g}$

Given that its oxide contains $53\%$metal.

Then, the weight of the metal is$53\mathrm{g}$.

The weight of the oxygen is$47\mathrm{g}$.

The equivalent weight of the metal$=\frac{{\mathrm{w}}_2}{{\mathrm{w}}_1}\times {\mathrm{a}}_1$

Where, the ${\mathrm{w}}_2=$weight of the metal, ${\mathrm{w}}_1=$weight of the oxygen, and ${\mathrm{a}}_1=$atomic weight of the oxygen

The equivalent weight of the metal$=\frac{53}{47}\times 8$

The equivalent weight of the metal $=9.02$

Step 2: Calculation of Valency

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

Where VD is vapor density and E is equivalent weight.

Valency $=\frac{2\times 66}{9.02+35.5}$

Valency $=2.96\approx 3$

Step 3: Calculation of atomic weight of the metal

Atomic weight $=$Equivalent weight $\times$Valency

Atomic weight $=9.02\times 3$

Atomic weight $=27.06$

Therefore, the atomic weight of the metal is $\mathbf{27}\mathbf{.}\mathbf{06}$.