Dilutions of Solutions

Dilution: reducing the concentration of a chemical


  • When a solution is diluted, more solvent is added to it.
        Since M = n ÷ L, and n (the moles of solute) is the same for the original solution and the new diluted solution, it follows that M1L1 = M2L2
        where M1=original concentration of solution
                L1=original volume of solution
                M2=new concentration of solution after dilution
                L2=new volume of solution after dilution



  • To calculate the new concentration (M2) of a solution given its new volume (L2) and its original concentration (M1) and original volume (L1):
        M2 = (M1 x L1) ÷ L2



  • To calculate the new volume (L2) of a solution given its new concentration (M2) and its original concentration (M1) and original volume (L1):
        L2 = (M1 x L1) ÷ M2


    • Examples

    1. M2=(M1L1) ÷ L2

    Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L.
    • M2=(M1L1) ÷ L2
    • M1 = 0.25M
    • L1 = 100mL = 100 ÷ 1000 = 0.100L (volume must be in litres)
    • L2 = 1.5L
    • [NaCl(aq)]new = M2 = (0.25 x 0.100) ÷ 1.5 = 0.017M
          (or 0.0.017 mol/L or 0.0.17mol L-1)

    2. L2=(M1L1) ÷ M2

    Calculate the volume to which 500mL of 0.02M coppper sulfate solution must be diluted to make a new concentration of 0.001M.
    • L2=(M1L1) ÷ M2
    • M1 = 0.02M
    • L1 = 500mL = 500 ÷ 1000 = 500 x 10-3L = 0.500L (since there are 1000mL in 1L)
    • M2 = 0.001M
    • L(CuSO4)new = L2 = (0.02 x 0.500) ÷ 0.001 = 10.00L
    Here is a video found on YouTuBe about dilution: