Question

50 ml of 20.8% (w/V) $BaC{l}_2$ (aq) and 100 ml of 9.8% (w/V) $H_{2}S{O}_4$ (aq) solutions are mixed. Molarity of $C{l}^{-}$ ions in the resulting solution is : (At mass of Ba =137)

• A. 0.333 M
• B. 0.666 M
• C. 0.1 M
• D. 1.33 M

Answer 1

The correct option is A 0.333 M

$50mLBaC{l}_2$ (aq) solution contain $10.4gBaC{l}_2$
$\therefore n_{BaC{l}_2}=\frac{10.4}{137+71}=0.05$
moles of $H_{2}S{O}_4$ in $100mL{H}_{2}S{O}_4$ (aq) solution $=\frac{9.8}{98}=0.1$
$BaC{l}_2+H_{2}S{O}_4\to BaS{O}_4+2HCl$;
moles of $HCl$ formed $=0.1$
$\left[C{l}^{-}\right]=\frac{0.1\times 1000}{50+100}=0.666M$