The molarity of a solution is M. To decrease the molarity of this solution to $\text{M}/2$ , then

Solution:

Consider initial and final solutions as per dilution law.

Assume concentration (M) and volume (V) for initial and final conditions as follows below.

              Initial                   Final

            ${\text{M}}_1=\text{M}$              ${\text{M}}_2=\text{M}/\text{2}$

               ${\text{V}}_1 =\text{V}$             ${\text{V}}_2=?$

From dilution method,

          ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

   ${\text{V}}_2=\frac{{\text{M}}_1{\text{V}}_1}{{\text{M}}_2}=\frac{\cancel{\text{M}}\times \text{V}}{\frac{\cancel{\text{M}}}{2}}=2\text{V}$

Therefore, to decrease the molarity of the solution to one-half from initial ,

 Then the volume of solution must be doubled.