The chloride of a metal M is to contain 20.2% metal. The specific heat of the metal is 0.224.

Question

The chloride of a metal $M$ is to contain $20 \cdot 2 \%$ metal. The specific heat of the metal is $0 \cdot 224$. The vapour density of the chloride is $133(H=1)$. What is the atomic weight of the metal and molecular formula of the chloride?

Answer 1

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 $=\left( MCl _{3}\right) \times 2= M _{2} Cl _{6}$.