What volume of a 2.00 m kcl solution is required to prepare 500. ml of a 0.100 m kcl solution?

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

2 M x V1 = 0.1 M x. 5 L

V1 = 0.025 L or 25 mL of the 2 M KCl solution is needed