Question

$4.16g$ of $BaC{I}_2$ contains $2.74g$ of Ba. The atomic weight of Ba is nearly:

• A. $137$
• B. $208$
• C. $67.3$
• D. $97.27$

Answer 1

Answer: (A)

The correct option is A $137$
$B{a}^{2+}+2C{l}^{-}⟶BaC{l}_2$

Weight of $Ba=2.74g$

Weight of$BaC{l}_2=4.16g$

Formula :

$\frac{\text{Moles of}B{a}^{2+}}{\text{Stoichiometry of}B{a}^{2+}}=\frac{\text{Moles of}BaC{l}_2}{\text{stoichiometry of}BaC{l}_2}$

$\therefore \frac{\text{Moles of}B{a}^{2+}}{1}=\frac{\text{Moles of}BaC{l}_2}{1}$

$\therefore \frac{weight}{\text{GMM of}B{a}^{2+}}=\frac{weight}{\text{GMM of}BaC{l}_2}$

$\frac{2.74g}{GMMofB{a}^{2+}}=\frac{4.16g}{208.23gmo{l}^{-1}}$

GMM of $B{a}^{2+}=\frac{2.74g\times 208gmo{l}^{-1}}{4.16g}$

Atomic mas of $Ba=137.15$