Question

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

• A. $MCl$
• B. $MC{l}_2$
• C. $MC{l}_3$
• D. $MC{l}_4$

Answer 1

Answer: (B)

Given, specific heat $=0.16$
Let metal chloride be $MC{l}_x$ 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 $\left(MC{l}_x\right)$ have $\approx 65\%$ chlorine present in it.
$\frac{x\times \text{ Atomic weight of chlorine }}{40+x\times \text{ Atomic weight of chlorine }}\times 100=65$
$\frac{x\times 35.5}{40+x\times 355}\times 100=65$
$x=209\approx 2($ approx. $)$
So, the formula of metal chloride will be $MC{l}_2$.