Consider the ionization of H₂SO₄ as follow:

${\mathrm{H}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{H}}_3{\mathrm{O}}^{\oplus }+{\mathrm{SO}}_4^{2-}$

The total number of ions furnished by 100 mL of 0.1 M H₂SO₄ will be

Given reaction is balanced and according to the balanced chemical reaction, one mole of H₂SO₄ gives three moles of ions, two moles of ${\mathrm{H}}_3{\mathrm{O}}^{\oplus }$ and one mole of ${\mathrm{SO}}_4^{2-}.$

Number of moles present in 100 mL of 0.1 M of H₂SO₄ is calculated as:

$\begin{array}{ccc}\mathrm{Molarity}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} (\mathrm{in} \mathrm{L})}\\ 0.1 \mathrm{M}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{100 \mathrm{mL}\times {\displaystyle \frac{1 \mathrm{L}}{1000 \mathrm{mL}}}}\\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}& =& 0.01 \mathrm{mol}\end{array}$

$\begin{array}{ccc}1 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4& \mathrm{gives}& 2 \mathrm{mol} \mathrm{of} {\mathrm{H}}_3{\mathrm{O}}^{\oplus }\\ 0.01 \mathrm{mol} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4& =& 0.01 \mathrm{mol} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4\times 2\\ & =& 0.02 \mathrm{mol} \mathrm{of} {\mathrm{H}}_3{\mathrm{O}}^{\oplus }\end{array}$

$\begin{array}{ccc}1 \mathrm{mol} {\mathrm{H}}_2{\mathrm{SO}}_4& \mathrm{gives}& 1 \mathrm{mol} {\mathrm{SO}}_4^{2-}\\ 0.01 \mathrm{mol} {\mathrm{H}}_2{\mathrm{SO}}_4& =& 0.01 \mathrm{mol} {\mathrm{H}}_2{\mathrm{SO}}_4\\ & =& 0.01 \mathrm{mol} {\mathrm{SO}}_4^{2-}\end{array}$

Therefore, the total number of ions produced by 0.01 moles of H₂SO₄ is:

$\begin{array}{ccc}\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{ions} \mathrm{produced}& =& \mathrm{Number} \mathrm{of} {\mathrm{H}}_3{\mathrm{O}}^{\oplus }+\mathrm{Number} \mathrm{of} {\mathrm{SO}}_4^{2-}\\ & =& 0.02 \mathrm{mol}+0.01 \mathrm{mol}\\ & =& 0.03 \mathrm{mol}\end{array}$

Number of ions present in one mole is $6.023\times {10}^{23}.$ Number of ions present in 0.03 mol is calculated as:

$\begin{array}{ccc}1 \mathrm{mol}& =& 6.023\times {10}^{23} \mathrm{ions}\\ 0.03 \mathrm{mol}& =& 0.03 \mathrm{mol}\times \frac{6.023\times {10}^{23} \mathrm{ions}}{1 \mathrm{mol}}\\ & =& 0.18\times {10}^{23} \mathrm{ions}\end{array}$