If 500 ml of CaCl 2 solution contains 3 . 01 × 10 22 chloride ions, molarity of the solution will be

${\mathrm{CaCl}}_2$ solution has the two mole of chloride ion. By applying the mole-mole analysis,

The number of molecules of chloride ions = $\frac{3.01\times {10}^{22}}{2}$

Mole of chloride ion = $\frac{\mathrm{Number} \mathrm{of} \mathrm{molecules}}{{\mathrm{N}}_{\mathrm{A}}}=\frac{3.01\times {10}^{22}}{2\times 6.02\times {10}^{23}}=0.025 \mathrm{mol}$

Molarity of solution = $\frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{ml}}\times 1000$

Number of moles of solute = $0.025 \mathrm{mol}$

Volume of solution = $500 \mathrm{ml}$

Molarity of solution = $\frac{0.025}{500}\times 1000=0.05$

Hence, the molarity of solution is $0.05 \mathrm{M}$.