(C) Mass of $Cd = 0.9 \ g$. Mass of $Cl_2 = 1.5 \ g - 0.9 \ g = 0.6 \ g$. Atomic weight of $Cl = 35.5 \ g/mol$. Mass of $Cl_2$ in $CdCl_2 = 2 \times 3

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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 $Cl = 35.5 \ g/mol$. Mass of $Cl_2$ in $CdCl_2 = 2 \times 35.5 = 71 \ g/mol$. According to the law of chemical equivalence: $\frac{\text{Mass of } Cd}{\text{Atomic weight of } Cd} = \frac{\text{Mass of } Cl_2}{\text{Equivalent weight of } Cl_2}$. $\frac{0.9}{x} = \frac{0.6}{71}$. $x = \frac{0.9 \times 71}{0.6} = 1.5 \times 71 = 106.5 \ g/mol$.

Class 11 Chemistry Some Basic Concepts of Chemistry Atomic, Molecular and Equivalent masses

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

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A
$118$
B
$112$
C
$106.5$
D
$53.25$

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Some Basic Concepts of Chemistry

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Atomic, Molecular and Equivalent masses

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

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