91Ó°ÊÓ

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 ${\mathrm{K}}_3{\mathrm{PO}}_4$ is:

${\mathrm{K}}_3\left({\mathrm{PO}}_4\right)→3{\mathrm{K}}^{+}+{{\mathrm{PO}}_4}^{-3}$

Since each mole of ${\mathrm{K}}_3{\mathrm{PO}}_4$ when dissolved in water gives 3 moles of${\text{K}}^{\text{+}}$ ions and 1 mole of localid="1656603512946" ${\text{PO}}_{\text{4}}^{\text{-3}}$. Therefore a total of 4 ions are formed from the dissociation of 1 mole of ${\mathrm{K}}_3{\mathrm{PO}}_4$. The total number of ions formed from the dissociation of ${\mathrm{K}}_3{\mathrm{PO}}_4$of is given below. $\left(0.75mol{K}_{3}P{O}_4\right)x\frac{4mol\text{ }ion}{1mol{K}_{3}P{O}_4}=3.00molofions$

The dissociation reaction of ${\mathrm{NiBr}}_2$ is:

${\mathrm{NiBr}}_2→{\mathrm{Ni}}^{+2}+2{\mathrm{Br}}^{-}$

Since each mole of ${\mathrm{NiBr}}_2.3{\mathrm{H}}_2\mathrm{O}$ when dissolved in water gives 1 mole of ${\mathrm{Ni}}^{+2}$ions and 2 moles of ${\mathrm{Br}}^{-}$ ions, therefore a total of 3 moles of ions are produced.

The total number of moles of ions formed when $6.88\text{ }×{10}^{−3}\text{ g}$ of ${\mathrm{NiBr}}_2$ is dissolved in water is given below.

$\begin{array}{l}=\left(6.88x{10}^{-3}gmolNiB{r}_2.3H_{2}O\right)x\frac{1molNiB{r}_2.3H_{2}O}{272.55gmNiB{r}_2.3H_{2}O}x\frac{3mol\text{ }ions}{1molNiB{r}_2.3H_{2}O}\\ =7.57x{10}^{-5}molofions\end{array}$

The dissociation reaction of ${\mathrm{FeCl}}_3$is:

${\mathrm{FeCl}}_3→{\mathrm{Fe}}^{+3}+3{\mathrm{Cl}}^{-}$

Since each mole of ${\mathrm{FeCl}}_3$when dissolved in water gives 1 mole of ${\mathrm{Fe}}^{+3}$ ions and 3 moles of ${\mathrm{Cl}}^{-}$ ions, therefore, a total of 4 moles of ions are produced.

The total number of moles of ions formed when localid="1661943071139" $2.23\text{ }×{10}^{22}\text{ f´Ç°ù³¾³Ü±ô²¹â€‰â¶Ä‰u²Ô¾±³Ù²õ}$of ${\mathrm{FeCl}}_3$is dissolved in water is given below.

$\begin{array}{l}=2.23x{10}^{22}formulaunitFeC{l}_{3}x\frac{1molFeC{l}_3}{6.022x{10}^{23}formulaunitFeC{l}_3}x\frac{4molions}{1molFeC{l}_3}\\ =0.418molofions\end{array}$