If 500 mL of  $CaC{l}_2$ solution contains  $3.01\times {10}^{22}$ chloride ions, molarity of the solution  will be 

$CaC{l}_2\to C{a}^{2+}+2C{l}^{-}$

$C{l}^{-}$  in solution  $=3.01\times {10}^{22}$  ions
No. of  $CaC{l}_2$  present in solution  $=\frac{3.01\times {10}^{22}}{2}$
 $=1.505\times {10}^{22}CaC{l}_2$
Molarity of solution  $=\frac{No.ofmolesofCaC{l}_2}{Vol.ofsolution\left(inlitre\right)}$
$=\frac{1.51\times {10}^{22}}{6.022\times {10}^{23}}\times \frac{1000}{500}=0.05$