Vapour density of a metal chloride is 66. Its | vedclass

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(C) Let the equivalent weight of the metal be $E$.In the metal oxide,the mass of metal is $53 \ g$ and the mass of oxygen is $(100 - 53) = 47 \ g$.The

Vedclass · Accepted Answer

$27.06$

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(C) Let the equivalent weight of the metal be $E$.In the metal oxide,the mass of metal is $53 \ g$ and the mass of oxygen is $(100 - 53) = 47 \ g$.The equivalent weight of the metal is calculated as: $E = \frac{	ext{mass of metal}}{	ext{mass of oxygen}} 	imes 8 = \frac{53}{47} 	imes 8 \approx 9.02$.The molecular weight of the metal chloride is $2 	imes 	ext{Vapour Density} = 2 	imes 66 = 132$.The formula for the molecular weight of a metal chloride is $M = n(E + 35.5)$,where $n$ is the valency of the metal.$132 = n(9.02 + 35.5) = n(44.52)$.$n = \frac{132}{44.52} \approx 2.96$,which rounds to $n = 3$.Atomic weight of the metal = $	ext{Equivalent weight} 	imes 	ext{Valency} = 9.02 	imes 3 = 27.06$.

Isotope	$\%$ Natural Abundance	Molar Mass $(g/mol)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

Vapour density of a metal chloride is $66$. Its oxide contains $53\%$ metal. The atomic weight of the metal is

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(in $u$)

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