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Dilution of Solutions Calculations (M1V1=M2V2)

Key Concepts

  • The concentration of a solution is usually given in moles per litre (mol L-1 OR mol/L).
        This is also known as molarity.

  • Concentration, or Molarity, is given the symbol M.
        A short way to write that the concentration of a solution of hydrochloric acid is 0.01 mol/L is to write [HCl]=0.01M
        The square brackets around the substance indicate concentration.

  • The solute is the substance which dissolves.

  • The solvent is the liquid which does the dissolving.

  • A solution is prepared by dissolving a solute in a solvent.

  • When a solution is diluted, more solvent is added to it.
        Since M = n ÷ V, and n (the moles of solute) is the same for the original solution and the new diluted solution, it follows that M1V1 = M2V2
        where M1=original concentration of solution
                V1=original volume of solution
                M2=new concentration of solution after dilution
                V2=new volume of solution after dilution

  • To calculate the new concentration (M2) of a solution given its new volume (V2) and its original concentration (M1) and original volume (V1):
        M2 = (M1 x V1) ÷ V2

  • To calculate the new volume (V2) of a solution given its new concentration (M2) and its original concentration (M1) and original volume (V1):
        V2 = (M1 x V1) ÷ M2

Examples

1. M2=(M1V1) ÷ V2

Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L.

  • M2=(M1V1) ÷ V2

  • M1 = 0.25M

  • V1 = 100mL = 100 ÷ 1000 = 0.100L (volume must be in litres)

  • V2 = 1.5L

  • [NaCl(aq)]new = M2 = (0.25 x 0.100) ÷ 1.5 = 0.017M
        (or 0.017 mol/L or 0.017 mol L-1)

2. V2=(M1V1) ÷ M2

Calculate the volume to which 500mL of 0.02M coppper sulfate solution must be diluted to make a new concentration of 0.001M.

  • V2=(M1V1) ÷ M2

  • M1 = 0.02M

  • V1 = 500mL = 500 ÷ 1000 = 500 x 10-3L = 0.500L (since there are 1000mL in 1L)

  • M2 = 0.001M

  • V(CuSO4)new = V2 = (0.02 x 0.500) ÷ 0.001 = 10.00L
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