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. $\ce{ MCl}$
• B. $\ce{ MCl_2}$
• C. $\ce{ MCl_3}$
• D. $\ce{ MCl_4}$

Answer 1

Answer: (B)

$\ce{ MCl_2}$

Answer 2

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 $M Cl _{2}$.