Question

$0.4686 g$ of a metal oxide yields on reduction $0.3486 g$ of the metal which forms an alkali double sulphate isomorphous with common alum. Find the atomic weight of the metal.

Answer 1

SOLUTION : First part of the problem gives the equivalent weight of the metal, and the last portion the valency of the metal. By knowing these two values atomic weight is casily worked out.

Wt. of metal oxide $=0.4686 g$ and wt. of metal $=0.3486 g$

$\therefore$ Wt. of metal $=(0.4686-0.3486) g =0 \cdot 120 g$

$\therefore \quad \text { Eq. wt. }=\frac{\text { wt. of metal }}{\text { wt. of oxygen }} \times 8=\frac{0 \cdot 3486}{0 \cdot 120} \times 8=23 \cdot 24$

Formula of common alum : $K _{2} SO _{4} \cdot Al _{2}\left( SO _{4}\right)_{3} \cdot 24 H _{2} O$

Formula of alkali double sulphate of $M$ :

$K _{2} SO _{4} \cdot M _{2}\left( SO _{4}\right)_{3} \cdot 24 H _{2} O \text { where } M \text { is the metal. }$

By isomorphism, valency of $M=3$ (valency of $A l$ )

$\therefore \quad$ Atomic weight of metal $=23 \cdot 24 \times 3= 6 9 \cdot 7 2$.