Question

he vapour density of a volatile chloride of a metal is 95 and the specific heat of the metal is 0.13cal/g.The equivalent weight of metal will be approximately?

Answer 1

24

Solution

For metals, atomic weight $$\times$$ specific heat $$=6.4 $$

Substituting value in the above expression,

atomic weight $$\times 0.13$$ $$=6.4 $$

atomic weight $$= 47.23$$ g/mol

The relationship between molecular weight and vapour density is molecular weight $$=2 \times$$vapour density $$= 2 \times95=190$$ g/mole.

Thus, 190 g of metal chloride contains 47.23 g of metal.

The amount of chlorine present is $$190-47.23=142.77 g$$. This is approximately equal to 4 moles of chlorine. Hence, the valency of metal is 4.

Therefore, equivalent mass $$=\dfrac{atomic \ mass}{valency} = \dfrac{47.23}{4} $$

So, approximately, the equivalent mass of metal $$=12$$ g/mol.