Number of ions present in 1 ml of 0.1M barium nitrate solution are

$$\begin{gathered} \text{Barium} \text{nitrate} \text{formula}=\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=2 {\text{NO}}_3^{-}\text{ions} \\ \text{M}={\text{n}}_{\text{moles}}\times \frac{1000}{\text{V}\left(\text{mL}\right)} \\ 0.1={\text{n}}_{\text{moles}}\times \frac{1000}{1}\Rightarrow {\text{n}}_{\text{moles}}={10}^{-4}\text{moles} \\ \text{No.of} \text{moles} \text{of} {\text{NO}}_3^{-}\text{ions}=2\times {10}^{-4}\text{moles} {\text{NO}}_3^{-}\text{ions} \\ =6\times {10}^{23}\times 2\times {10}^{-4} \\ =1.2\times {10}^{19}{\text{NO}}_3^{-}\text{ions} \\ \text{Total} \text{no.of} \text{ions}={10}^{-4}\text{moles} \text{of} {\text{Ba}}^{+2}+2\times {10}^{-4}\text{mole} {\text{NO}}_3^{-}\text{ions} \\ =3\times {10}^{-4}\text{moles} \text{of} \text{total} \text{ions} \\ =3\times {10}^{-4}\times 6\times {10}^{23} \\ =1.8\times {10}^{19} \\ =1.8\times {10}^{19} \\ =1.8\times {10}^{20} \text{no.of} \text{ions} \end{gathered}$$