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)

$MCl_4$

Answer 2

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 $= MCl_{4}$