A transition metal M forms a volatile chlorid | vedclass

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Question

(C) Given,vapour density $(V.D.)$ $= 94.8$.Molecular weight of metal chloride $= 2 	imes V.D. = 2 	imes 94.8 = 189.6 \ g/mol$.Mass of chlorine in $1

Vedclass · Accepted Answer

$MCl_4$

Vedclass · Answer

(C) Given,vapour density $(V.D.)$ $= 94.8$.Molecular weight of metal chloride $= 2 	imes V.D. = 2 	imes 94.8 = 189.6 \ g/mol$.Mass of chlorine in $189.6 \ g$ of metal chloride $= 74.75\% 	ext{ of } 189.6 = 0.7475 	imes 189.6 \approx 141.7 \ g$.Number of chlorine atoms $= \frac{141.7}{35.5} \approx 4$.Mass of metal $= 189.6 - 141.7 = 47.9 \ g$.Since the metal chloride is $MCl_4$,the valency of the metal $M$ is $4$.Therefore,the formula of the metal chloride is $MCl_4$.

$A$ 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:

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