91Ó°ÊÓ

The solution volume and molar mass of potassium sulfate can be multiplied as,

$\begin{array}{l}Mass of K_{2}S{O}_4=475×{10}^{-3}L so\ln ×\frac{5.62×{10}^{-2}mol K_{2}S{O}_4}{L so\ln }×\frac{174.26 g K_{2}S{O}_4}{1 mol K_{2}S{O}_4}\\ Mass of K_{2}S{O}_4=4.65 g\end{array}$

On multiplying the given calcium carbonate mass by molar mass reciprocal to find mole number present in solution.

$\begin{array}{l}Moles of CaC{l}_2=7.25×{10}^{-3} g CaC{l}_2×\frac{1 mol CaC{l}_2}{110.98 g CaC{l}_2}\\ =6.53×{10}^{-5}mol CaC{l}_2\\ Molarity=\frac{6.53×{10}^{-5}mol CaC{l}_2}{1×{10}^{-3}Lso\ln }\\ Molarity=6.53×{10}^{-2}M\end{array}$

The given volume of solution can be multiplied by molarity, the ratio between ions, and the Avogadro number to find the number of magnesium ions.

$\begin{array}{l}M{g}^{2+}ions=1×{10}^{-3}L so\ln ×\frac{0.184 mol MgB{r}_2}{1 mol MgB{r}_2}×\frac{1 mol M{g}^{2+}ions}{1 mol MgB{r}_2}×\frac{6.022×{10}^{23} M{g}^{2+}ions}{1 mol M{g}^{2+}ions}\\ M{g}^{2+}ions=1.11×{10}^{20}M{g}^{2+}ions\end{array}$