SOLUTION : Wt. of metal $=20 \cdot 2 g$
and hence, wt. of chlorine $=(100-20 \cdot 2) g =79 \cdot 8 g$
Eq. wt. of metal $=\frac{w t \text {. of metal }}{\text { wt. of chlorine }} \times 35 \cdot 5=\frac{20 \cdot 2}{79 \cdot 8} \times 35 \cdot 5=8 \cdot 986$
At. wt. (approx.) $=\frac{6.4}{\text { sp. heat }}=\frac{6.4}{0 \cdot 224}=28 \cdot 57$
and, therefore, valency $=\frac{28 \cdot 57}{8 \cdot 986}=3 \cdot 18 \approx 3$
and the correct atomic weight $=8 \cdot 986 \times 3=26 \cdot 958$.
Empirical formula of the chloride $= MCl _{3}$ and molecular formula $=\left( MCl _{3}\right) \times n$
Its molecular weight (given $)=133 \times 2=266$
Calculated mol. wt. of $\left( MCl _{3}\right) \times n=(26 \cdot 958+3 \times 35 \cdot 5) n=133.458 \times n$
Now,
$133.458 \times n=266 \quad \therefore n=\frac{266}{133.458} \approx 2$
Correct molecular formula $=\left( MCl _{3}\right) \times 2= M _{2} Cl _{6}$.