A metal $M$ (specific heat 0.16 ) forms a metal chloride with $65 \%$ chlorine present in it. The formula of the metal chloride will be

Given, specific heat $=0.16$ Let metal chloride be $M \mathrm{Cl}_{x}$ 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 $\left(M C l_{x}\right)$ have $\approx 65 \%$ chlorine present in it. $\frac{x \times \text { Atomic weight of chlorine }}{40+x \times \text { Atomic weight of chlorine }} \times 100=65$ $$ \begin{array}{l} \frac{x \times 355}{40+x \times 355} \times 100=65 \\ x=209 \approx 2(a p p r o x .) \end{array} $$ So, the formula of metal chloride will be $M \mathrm{Cl}_{2}$