SOLUTION : Wt. of $C d C l_{2}=1.5276 g$ and wt. of $C d=0.9367 g$
$\therefore \quad$ Wt. of chlorine $=(1.5276-0.9367) g =0.5909 g$
We know that
$\text { eq. wt. }=\frac{\text { wt. of } C d}{\text { wt. of chlorine }} \times 35 \cdot 5=\frac{0 \cdot 9367 \times 35 \cdot 5}{0.5909}=56 \cdot 275$
Hence equivalent weight of cadmium $=56 \cdot 275$
$\begin{array}{ll}\because \quad \text { Atomic weight } \quad=\text { equivalent weight } \times \text { valency } \\\therefore \quad \text { Atomic weight } \quad=56 \cdot 275 \times 2=112 \cdot 550 . \\\quad\left(\because \text { Valency of } C d=2, \text { from } C d C l_{2}\right)\end{array}$
