Question

A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be

• A. $MC{l}_3$
• B. $MC{l}_2$
• C. $MC{l}_4$
• D. $MC{l}_5$

Answer 1

Answer: (C)

74.75% of chlorine means 74.75g chlorine is present in lOOg of metal chloride. Weight of metal $=11Og-74.75g$
$=25.25g$
Equivalent weight
$=\frac{weightofmetal}{weightofchlorine}$
$=\frac{25.25}{74.75}\times 35.5=12$
Valency of metal
$=\frac{2\times V.D}{Equivalentwt.ofmetal+35.5}$
$=\frac{2\times 94.8}{12+35.5}=4$
$\therefore$ Formula of compound $=MC{l}_4$