Given, specific heat $=0.16$
Let metal chloride be $M Cl _{x}$ then,
$\frac{6.4}{\text { specific heat }}=$ Atomic weight of metal
$\frac{6.4}{0.16}=$ Atomic weight of metal
Atomic weight $=40$
$40$ is the atomic weight of calcium. According to question, metal chloride $\left(M Cl _{x}\right)$ have $\approx 65 \%$ chlorine present in it.
$\frac{x \times \text { Atomic weight of chlorine }}{40+x \times \text { Atomic weight of chlorine }} \times 100=65$
$\frac{x \times 35.5}{40+x \times 355} \times 100=65$
$x=209 \approx 2($ approx. $)$
So, the formula of metal chloride will be $M Cl _{2}$.