Question

Calculate the weight of $BaC{l}_2$ needed to prepare $250mL$ of a solution having the same concentration of $C{l}^{-}$ ions as in a solution of $KCl$ of concentration $80g/L.(Ba=137.4,Cl=35.5)$ -

Answer 1

Answer: 27.92

Concentration of $C{l}^{-}$ of $BaC{l}_2=C{l}^{-}$ of $KCl2C{l}^{-}=C{l}^{-}$
$S=80g/L$
$\therefore BaC{l}_2=74.5$
$\text{ Moles }=\frac{S}{M}=\frac{80}{74.5}$
$1$ litre solution contain $=\frac{80}{74.5}$
$\frac{1}{4}$ litre solution contain $=\frac{80}{74.5}\times \frac{1}{4}$
$\frac{80}{74.5\times 4}=\frac{x}{207}$
$x=27.93$