# Calcium chloride when dissolved in water

Question:

Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

$\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)$

Calculate the number of ions obtained from $\mathrm{CaCl}_{2}$ when $222 \mathrm{~g}$ of it is dissolved in water.

Solution:

Molar mass of $\mathrm{CaCl}_{2}=40+(2 \times 35 \cdot 5)=111 \mathrm{~g} \mathrm{~mol}^{-1}$

$111 \mathrm{~g}$ of $\mathrm{CaCl}_{2}$ represent $=1 \mathrm{~mol}$

$222 \mathrm{~g}$ of $\mathrm{CaCl}_{2}$ represent $=\frac{(222 \mathrm{~g})}{(111 \mathrm{~g})} \times 1 \mathrm{~mol}=2 \mathrm{~mol}$

No. of molcules in 2 moles of $\mathrm{CaCl}_{2}=2 \times \mathrm{N}_{\mathrm{A}}=2 \times 6.022 \times 10^{23}$.

1 molecule of $\mathrm{CaCl}_{2}$ in solution form ions $=3$

$2 \times 6.022 \times 10^{23}$ molecules of $\mathrm{CaCl}_{2}$ in solution form ions $=3 \times 2 \times 6.022 \times 10^{23}$

$=36.132 \times 10^{23}$

$=3.6132 \times 10^{24}$