SOLUTION : Wt. of metal in hydrated sulphate $=8 \cdot 1 \%$
and wt. of $SO _{4}^{2-}$ in hydrated sulphate $=43 \cdot 2 \%$
$\therefore \quad$ Wt. of water present $=100-(8 \cdot 1+43 \cdot 2)=48 \cdot 7 \%$
We know that equivalent weight of $SO _{4}^{2-}=\frac{96}{2}=48$
Wt. of sulphate and metal in the hydrated compound $=8 \cdot 1+43 \cdot 2=51 \cdot 3 \%$
$\therefore 43 \cdot 2 g$ sulphate is present when metal is $8 \cdot 1 g$
$\therefore 48 g$ sulphate is present when metal is $\frac{8 \cdot 1 \times 48}{43 \cdot 2} g =9 g$
$\therefore$ Equivalent weight of metal $=9 \cdot 0$
By Dulong and Petit's law,
$\text { rough atomic weight }=\frac{6 \cdot 4}{ sp . \text { heat }}=\frac{6 \cdot 4}{0 \cdot 242}=26 \cdot 45$
and hence, valency $=\frac{26.45}{9}=2 \cdot 94 \approx 3$
and, therefore, atomic weight of metal $=9 \times 3=27$
Alse, formula of hydrated sulphate of metal $M=M_{2}\left( SO _{4}\right)_{3} \cdot x H_{2} O$
where $x$ is the no. of moles of water of crystallisation.
Mol. wt. of $M _{2}\left( SO _{4}\right)_{3} \cdot x H _{2} O =27 \times 2+3 \times 96+18 x=342+18 x$
$\because \quad 342 g$ anhydrous sulphate associated with $18 x g$ water
$\therefore 51.3 g$ anhydrous sulphate associated with $\frac{18 x \times 51 \cdot 3}{342} g$ water and this is equal to $48 \cdot 7 g$ ( given in the problem).
$\therefore \quad \frac{18 x \times 51 \cdot 3}{342}=48 \cdot 7, x=\frac{48 \cdot 7 \times 342}{18 \times 51 \cdot 3}=18 \cdot 0$
$\therefore$ The correct formula of hydrated sulphate of metal $= M _{2}\left( SO _{4}\right)_{3} \cdot 1 8 H _{2} O$.
