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. $MCl_3$
• B. $MCl_2$
• C. $MCl_4$
• D. $MCl_5$

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
$= \frac{weight \,of \,metal}{weight \,of \,chlorine}$
$= \frac{25.25}{74.75}\times35.5 = 12$
Valency of metal
$= \frac{2\times V.D}{Equivalent \,wt. \,of\, metal\,+\,35.5}$
$= \frac{2\times94.8}{12+35.5} = 4$
$\therefore $ Formula of compound $= MCl_{4}$