mole of ${\mathrm{CaCl}}_2=$ conC. $($ moldm${}^{-3})\times$ volume $\left({\mathrm{dm}}^2\right)$
$=0.2$ moldm ${}^{-3}\times \left(\frac{50}{1000}\right){\mathrm{dm}}^3$
$=0.01{\mathrm{mol}}^{-3}$
${\mathrm{CaCl}}_2\to {\mathrm{Ca}}^{2+}+2\mathrm{C}$s can be seen from the equation, 1 mole of ${\mathrm{CaCl}}_2$ will produce 2 moles of ${\mathrm{Cl}}^{-}$.
Thus, $0.01\mathrm{mol}$ of ${\mathrm{CaCl}}_2$ will furnish $0.02\mathrm{mol}$ of ${\mathrm{Cl}}^{-}$.