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(B) Let the element be $M$ and its valency be $n$. The chloride formed is $MCl_n$.Mass of $Cl$ in $315.5 \ g$ of $MCl_n = 315.5 \ g - 209 \ g = 106.5

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$48$

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(B) Let the element be $M$ and its valency be $n$. The chloride formed is $MCl_n$.Mass of $Cl$ in $315.5 \ g$ of $MCl_n = 315.5 \ g - 209 \ g = 106.5 \ g$.Equivalent mass of $Cl = \frac{35.5}{1} = 35.5 \ g$.Number of equivalents of $Cl = \frac{106.5}{35.5} = 3$.Since equivalents of $M$ = equivalents of $Cl$,equivalents of $M = 3$.Equivalent mass of $M = \frac{	ext{Mass of } M}{	ext{Equivalents}} = \frac{209}{3} \ g$.Now,for the oxide $M_xO_y$,equivalents of $M$ = equivalents of $O$.For $418 \ g$ of $M$,equivalents of $M = \frac{418}{209/3} = 418 	imes \frac{3}{209} = 2 	imes 3 = 6$.Equivalent mass of $O = \frac{16}{2} = 8 \ g$.Mass of $O = 	ext{Equivalents} 	imes 	ext{Equivalent mass} = 6 	imes 8 = 48 \ g$.

$209 \ g$ of an element reacts with chlorine to form $315.5 \ g$ of its chloride. What is the weight (in $g$) of oxygen that reacts with $418 \ g$ of same element?
$(Cl = 35.5 \ u; O = 16 \ u)$

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$209 \ g$ of an element reacts with chlorine to form $315.5 \ g$ of its chloride. What is the weight (in $g$) of oxygen that reacts with $418 \ g$ of same element?$(Cl = 35.5 \ u; O = 16 \ u)$