1.5 g of CdCl_2 was found to contain 0.9 g | vedclass

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(C) Mass of $Cd = 0.9 \ g$. Mass of $Cl_2 = 1.5 \ g - 0.9 \ g = 0.6 \ g$. Atomic weight of $Cl = 35.5 \ g/mol$,so mass of $Cl_2 = 2 	imes 35.5 = 71 \

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$106.5$

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(C) Mass of $Cd = 0.9 \ g$. Mass of $Cl_2 = 1.5 \ g - 0.9 \ g = 0.6 \ g$. Atomic weight of $Cl = 35.5 \ g/mol$,so mass of $Cl_2 = 2 	imes 35.5 = 71 \ g/mol$. In $CdCl_2$,$0.6 \ g$ of $Cl_2$ corresponds to $71 \ g/mol$. Therefore,$0.9 \ g$ of $Cd$ corresponds to atomic weight $X$ of $Cd$. Using the ratio: $\frac{	ext{Mass of } Cd}{	ext{Mass of } Cl_2} = \frac{	ext{Atomic weight of } Cd}{	ext{Atomic weight of } Cl_2}$. $\frac{0.9}{0.6} = \frac{X}{71}$. $X = \frac{0.9 	imes 71}{0.6} = 1.5 	imes 71 = 106.5 \ g/mol$.

$1.5 \ g$ of $CdCl_2$ was found to contain $0.9 \ g$ of $Cd$. Calculate the atomic weight of $Cd$.

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