Question

Calculate the weight of $BaCl _{2}$ needed to prepare $250 \,mL$ of a solution having the same concentration of $Cl^-$ ions as in a solution of $KCl$ of concentration $80\, g / L. ( Ba =137.4, Cl =35.5)$ -

Answer 1

Concentration of $Cl^ -$ of $BaCl _{2}= Cl^ -$ of $KCl\, 2Cl^- =Cl^-$
$S =80\, g / L$
$\therefore BaCl _{2}=74.5$
$\text { Moles }=\frac{ S }{ M }=\frac{80}{74.5}$
$1$ 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} $
$x=27.93$