What volume of 0.25 mol/L solution of lead (II) nitrate would be required to form 500mL of a 0.15 mol/L solution of lead (II) nitrate. Show calculations

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.

0.25 M x V1 = 0.15 M x. 500 L

V1 = 0.3 L or 300 mL