A student has 100mL of 0.1 M KCl solution. To make it 0.2 M, he has to

V₁=100 mL  ;   V₂=?

M₁=0.1 M     ;    M₂=0.2 M

V₁M₁=V₂M₂

100 X 0.1=  V₂ X 0.2

V₂=50 mL

By evaporating 50 mL of solution, molarity of the solution becomes 0.2 M.

Moles present in solution of molarity 0.1 M with solution 100 mL is:

$\mathrm{Moles}=\mathrm{Molarity}\times \mathrm{Volume} (\mathrm{in} \mathrm{L})=0.1 \mathrm{M}\times 100 \mathrm{mL}\times \frac{1 \mathrm{L}}{1000 \mathrm{mL}}=0.01 \mathrm{mol}$

Moles present in solution of molarity 0.2 M with solution 100 mL is:

$\mathrm{Moles}=\mathrm{Molarity}\times \mathrm{Volume} (\mathrm{in} \mathrm{L})=0.2 \mathrm{M}\times 100 \mathrm{mL}\times \frac{1 \mathrm{L}}{1000 \mathrm{mL}}=0.02 \mathrm{mol}$

By adding 0.01 moles of KCl, molarity increases to 0.2 M.