How many total moles of ions are released when each of the following samples dissolves completely in water?

(a) 0.805 mol of ${\mathbf{\text{Rb}}}_{\mathbf{\text{2}}}{\mathbf{\text{SO}}}_{\mathbf{\text{4}}}$(b)${\mathbf{\text{3.85×10}}}^{\mathbf{\text{-3}}}$ g of$\mathbf{\text{Ca}}{\left({\text{NO}}_{\text{3}}\right)}_{\mathbf{\text{2}}}$

(c) ${\mathbf{\text{4.03×10}}}^{\mathbf{\text{19}}}$Formula units of$\mathbf{\text{Sr}}{\left({\text{HCO}}_{\text{3}}\right)}_{\mathbf{\text{2}}}$

The ionic compounds and the water molecules are polar in nature therefore both attract each other by the force of attraction.

Most ionic compounds dissociate and dissolve in water easily.

The dissociation reaction of ${\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4}}$is:

${\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{2Rb}}^{\text{+}}{{\text{+SO}}_{\text{4}}}^{\text{2-}}$

Since each mole of ${\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4}}$when dissolved in water gives 2 moles of ${\text{Rb}}^{\text{+}}$ ions and 1 mole of ${{\text{SO}}_{\text{4}}}^{\text{2-}}$ ions, therefore a total of 3 moles of ions are produced.

Therefore, number of moles of ions formed when $0.805\text{\hspace{0.17em}mol}$ is dissolved in water is:

$\left({\text{0.805molRb}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{x}\frac{\text{3molion}}{{\text{1molRb}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{=2.415molofions}$

The dissociation reaction of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$is:

$\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\to {\text{Ca}}^{\text{2+}}{{\text{+2NO}}_{\text{3}}}^{\text{-}}$

Since each mole of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ when dissolved in water gives 1 mole of ${\text{Ca}}^{\text{2+}}$ ions and 2 moles of ${{\text{NO}}_{\text{3}}}^{\text{-}}$ions, therefore a total of 3 moles of ions are produced.

Number of moles of ions formed when $3.85\times {10}^3\text{\hspace{0.17em}g}$of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$is dissolved in water is given below.

localid="1661942919121" $\begin{array}{l}\text{=}\left({\text{3.85x10}}^{\text{-3}}\text{gmolCa}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\text{x}\left(\frac{\text{1molCa}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{164.10gmCa}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\text{x}\left(\frac{\text{3molions}}{\text{1molCa}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\\ {\text{=7.038x10}}^{\text{-5}}\text{molofions}\end{array}$

The dissociation reaction of $\mathrm{Sr}{\left({\mathrm{HCO}}_3\right)}_2$is:

$Sr{\left(HC{O}_3\right)}_2\to S{r}^{2+}+2HC{O_3}^{-}$

Since each mole of $Sr{\left(HC{O}_3\right)}_2$when dissolved in water gives 1 mole of $S{r}^{2+}$ ions and 2 moles of ${{\mathrm{HCO}}_3}^{-}$ ions, therefore a total of 3 moles of ions are produced.

Number of moles of ions formed when $4.03\times {10}^{19}\text{\hspace{0.17em}g}$of $Sr{\left(HC{O}_3\right)}_2$ is dissolved in water is given below.

$\begin{array}{l}=4.03x{10}^{19}formulaunitsofSr{\left(HC{O}_3\right)}_{2}x\frac{1molSr{\left(HC{O}_3\right)}_2}{6.022x{10}^{23}formulaunitsofSr{\left(HC{O}_3\right)}_2}x\frac{3molions}{1molSr{\left(HC{O}_3\right)}_2}\\ =2.008x{10}^{-4}molofions\end{array}$