SOLUTION : Calculation of eq. wt. of metal
Wt. of metal $=34 \cdot 46 g$
and hence, w. of chlorine in the chloride $=(100-34 \cdot 46) g =65 \cdot 54 g$ $\because$ Eqtwt. $=\frac{w t \text {. of metal }}{\text { wt. of chlorine }} \times 35 \cdot 5=\frac{34 \cdot 46 \times 35 \cdot 5}{65 \cdot 54}=18.66$
By Dulong and Petit's law,
rough at. wt. $=\frac{6 \cdot 4}{ sp . \text { heat }}=\frac{6 \cdot 4}{0 \cdot 115}=55 \cdot 65$
and hence, valency $=\frac{\text { rough at. wt. }}{\text { eq. wt. }}=\frac{55 \cdot 65}{18 \cdot 66} \approx 3$
and, therefore, exact at. wt. $=18 \cdot 60 \times 3= 5 5 \cdot 9 8$.
The empirical formula of the chloride $= MCl _{3}$
We suppose that mol. formula $=\left( MCl _{3}\right) \times n$ and mol. wt. according to this formula
$=(55 \cdot 98+35 \cdot 5 \times 3) \times n=162 \cdot 48 \times n$
Given V.D. $=81 \cdot 5$, and then mol. wt. $=81 \cdot 5 \times 2=163$
Equating them, $162 \cdot 48 \times n=163, \quad \therefore n=\frac{163}{162 \cdot 48} \approx 1$
$\therefore$ The correct formula $= MCl _{3}$.
Note : If V.D., molecular weight and sp. heat are given in a problem the correct molecular formula must be evaluated for such problems as deduced in the above example.
