Question

$1.5276g$ of $CdC{l}_2$ was found to contain $0.9367g$ of $Cd$. Atomic weight of $Cd$ is

• A. 112.5
• B. 122.5
• C. 144.5
• D. 245

Answer 1

The correct option is A

112.5

Weight of Cd = $0.9367g$

$\therefore$ Weight of $Cl$ = $1.5276-0.9367=0.5909g$

For $CdC{l}_2$; $n_{eq.}$ of cadmium = $n_{eq.}$ of chlorine

$\frac{0.9367}{E}=\frac{0.5909}{35.5}$

$\therefore$ $E=56.275$

$\therefore$ Atomic weight = $Eq.wt.\times Valency$

= $56.275\times 2=112.55$.