mole of \(\mathrm{CaCl}_2=\) conC. \(\left(\right.\) moldm\(\left.^{-3}\right) \times\) volume \(\left(\mathrm{dm}^2\right)\)
\(=0.2\) moldm \(^{-3} \times\left(\frac{50}{1000}\right) \mathrm{dm}^3\)
\(=0.01 \mathrm{~mol}^{-3}\)
\(\mathrm{CaCl}_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{C}\)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 \(\mathrm{Cl}^{-}\).