91Ó°ÊÓ

The dissociation reaction of ${\mathrm{Na}}_2{\mathrm{HPO}}_4$is:

${\mathrm{Na}}_2{\mathrm{HPO}}_4→2{\mathrm{Na}}^{+}+{{\mathrm{HPO}}_4}^{2-}$

Since each mole of ${\mathrm{Na}}_2{\mathrm{HPO}}_4$ when dissolved in water gives 2 moles of ${\mathrm{Na}}^{+}$ ions and 1 mole of ${{\mathrm{HPO}}_4}^{2-}$ ions, therefore a total of 3 moles of ions are produced.

Total number of moles of ions produced when $0.734\text{ â¶Ä³¾´Ç±ô}$of ${\mathrm{Na}}_2{\mathrm{HPO}}_4$is dissolved in water.

$\left(0.734molN{a}_{2}HP{O}_4\right)x\frac{3mol\text{ }ion}{1molN{a}_{2}HP{O}_4}=2.202molofions$

The dissociation reaction of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$is:

${\mathrm{CuSO}}_4→{\mathrm{Cu}}^{2+}+{{\mathrm{SO}}_4}^{2-}$

Since each mole of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ when dissolved in water gives 2 moles of ${\mathrm{Cu}}^{2+}$ ions and 1 mole of ${{\mathrm{SO}}_4}^{2-}$ ions, therefore a total of 2 moles of ions are produced.

Total number of moles of ions formed when $3.86\text{ â¶Ä‰g}$ of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$is dissolved in water.

$\begin{array}{l}=\left(3.86gCuS{O}_4.5H_{2}O\right)x\frac{1molCuS{O}_4.5H_{2}O}{249.72gmCuS{O}_4.5H_{2}O}x\frac{2molions}{1molCuS{O}_4.5H_{2}O}\\ =3.09x{10}^{-2}molofions\end{array}$

The dissociation reaction of ${\text{NiCl}}_{\text{2}}$is:

${\mathrm{NiCl}}_2→{\mathrm{Ni}}^{+2}+2{\mathrm{Cl}}^{-}$

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

Total number of moles of ions formed when $8.66\text{ }×\text{ }{10}^{20}$ formula units of ${\text{NiCl}}_{\text{2}}$ is dissolved in water is given below.

$\begin{array}{l}=8.66x{10}^{20}formulaunitNiC{l}_{2}x\frac{1mol\text{ }NiC{l}_2}{6.022x{10}^{23}formulaunitsNiC{l}_2}x\frac{3mol\text{ }ions}{1molNiC{l}_2}\\ =4.31x{10}^{-3}molofions\end{array}$