Q4.16 P - Chemistry: Molecular Nature Of Matter And Change

Question

How many total moles of ions are released when each of the following samples dissolves completely in water?(a) 0.75 mol of ${\mathbf{\text{K}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}$  (b)  ${\mathbf{\text{6.88×10}}}^{\mathbf{\text{-3}}}$ g of ${\mathbf{\text{NiBr}}}_{\mathbf{\text{2}}}{\mathbf{\text{.3H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}$ (c)  ${\mathbf{\text{2.23×10}}}^{\mathbf{\text{22}}}$ formula units of ${\mathbf{\text{FeCl}}}_{\mathbf{\text{3}}}$

Step-by-Step Solution

Answer

Total moles of ions are released when each of the following samples dissolves completely in the water:

(a) $3.00\text{\hspace{0.17em}\hspace{0.17em}mol}$ of ions

(b) $7.57\times {10}^{-5}mol$ of ions

(c) $0.418\text{\hspace{0.17em}mol}$ of ions

Determine the solubility of Ionic compounds in water

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.

Calculate total moles of ions released when K 3 PO 4 dissolve in water

The dissociation reaction of ${\mathrm{K}}_3{\mathrm{PO}}_4$  is:  

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

                                     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 ${\text{PO}}_{\text{4}}^{\text{-3}}$ .  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.      $\left(0.75 mol K_{3}P{O}_4\right)x\frac{4 molion}{1 mol K_{3}P{O}_4}=3.00 mol of ions$

Calculate total moles of ions released when NiBr 2 dissolve in water

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

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

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 $6.88\times {10}^{-3}\text{\hspace{0.17em}g}$  of ${\mathrm{NiBr}}_2$  is dissolved in water is given below.

$\begin{array}{l}=\left(6.88x{10}^{-3}gmol NiB{r}_2.3H_{2}O\right)x\frac{1mol NiB{r}_2.3H_{2}O}{272.55 gm NiB{r}_2.3H_{2}O}x\frac{3molions}{1 mol NiB{r}_2.3H_{2}O}\\ =7.57x{10}^{-5}mol of ions\end{array}$

Calculate total moles of ions released when NiBr 2 dissolve in water

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

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

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 $2.23\times {10}^{22}\text{\hspace{0.17em}formula\hspace{0.17em}\hspace{0.17em}units}$ of ${\mathrm{FeCl}}_3$ is dissolved in water is given below.

$\begin{array}{l}=2.23x{10}^{22} formula unit FeC{l}_{3}x\frac{1 mol FeC{l}_3}{6.022x{10}^{23}formula unit FeC{l}_3}x\frac{4 mol ions}{1 mol FeC{l}_3}\\ =0.418 mol of ions\end{array}$