Question

$1.00 \,g$ of $BaCl _{2}$ is treated with excess of aqueous $AgNO _{3}$ and all chlorine is recovered as $1.38 \,g$ of $AgCl$. What is the atomic weight of $Ba$ ? $( Cl =35.5, Ag =108)$

• A. 137
• B. 172.5
• C. 33
• D. 68.5

Answer 1

Answer: (A)

$\underset{1\,g}{BaCl_{2}}+2AgNO_{3} \to \underset{1.38\,g}{2AgCl}+Ba(NO_{3})_{2}$

All chlorine atoms are recovered as $AgCl$ hence number of moles of chlorine on both sides must be equal.

Thus; $2 \times$ number of moles of $BaCl _{2}=1 \times$ number of moles of $AgCl$

$2 \times \frac{1}{(x+35.5 \times 2)}=1 \times \frac{1.38}{(108+35.5)}\Rightarrow x \approx 137$

(Say atomic mass of $Ba =x )$