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Dilution of Solutions Calculations (c1V1=c2V2)

Key Concepts

  • The concentration of a solution is often given in moles per litre (mol L-1 OR mol/L).
        This is also known as molarity1.

  • Concentration in mol L-1, or Molarity, is given the symbol c (sometimes M).
        For a 0.01 mol L-1 sodium chloride solution, Chemists write either:
          [NaCl(aq)] = 0.01 mol L-1 (square brackets around formula indicate concentration)
        or
          c(NaCl(aq)) = 0.01 mol L-1 (c stands for concentration, formula given in brackets)

  • The solute is the substance which dissolves, for example, sodium chloride, NaCl.

  • The solvent is the liquid which does the dissolving, for example water, (aq).

  • A solution is prepared by dissolving a solute in a solvent, for example, NaCl(aq) tells us that NaCl was dissolved in water.

  • When a solution is diluted, more solvent is added to it.
        Let c1 = n1 ÷ V1
            c1 = initial concentration in mol L-1 before dilution
            n1 = moles of solute
            V1 = initial volume of solution in litres before dilution
        Rearranging: n1 = c1 x V1
        Diluting the solution means adding more solvent:
            volume increases, V2=final volume of solution after dilution (L)
            concentration decreases, c2=final concentration of solution after dilution (mol L-1)
            moles of solvent remains the same, so, n1=moles of solute (mol)
        For the diluted solution n1 = c2 x V2
        Since the moles of solute = n1 both before and after dilution:
    c1V1 = c2V2

  • To calculate the new concentration (c2) of a solution given its new volume (V2) and its original concentration (c1) and original volume (V1):
    c2 = (c1 x V1) ÷ V2

  • To calculate the new volume (V2) of a solution given its new concentration (c2) and its original concentration (c1) and original volume (V1):
    V2 = (c1 x V1) ÷ c2

Examples

Calculate concentration of solution after dilution: c2 = (c1V1) ÷ V2

Calculate the new concentration in mol L-1 (molarity) if enough water is added to 100 mL of 0.25 mol L-1 sodium chloride solution to make up 1.5 L.

  1. Extract the data from the question :
        c1 = 0.25 mol L-1
        V1 = 100 mL = 100 ÷ 1000 = 0.100 L (volume must be in litres)
        V2 = 1.5 L
        c2 = ? mol L-1

  2. Write the equation :
        c2 = (c1V1) ÷ V2

  3. Sustitute in the values and solve :
        c2 = (0.25 x 0.100) ÷ 1.5 = 0.017 mol L-1
        [NaCl(aq)] = 0.017 mol L-1 (or 0.017 mol/L or 0.017 M)

Calculate the volume of solution after dilution (V2 = (c1V1) ÷ c2)

Calculate the volume to which 500 mL of 0.02 mol L-1 coppper sulfate solution must be diluted to make a new concentration of 0.001 mol L-1.

  1. Extract the data from the question :
        c1 = 0.02 mol L-1
        V1 = 500 mL = 500 ÷ 1000 = 500 x 10-3 L = 0.500 L (since there are 1000 mL in 1L)
        c2 = 0.001 mol L-1
        V2 = ? L

  2. Write the equation :
        V2 = (c1V1) ÷ c2

  3. Substitute in the values and solve :
        V2 = V(CuSO4)new = (0.02 x 0.500) ÷ 0.001 = 10.00 L


What would you like to do now?

1There are many other ways to measure concentration in chemistry. Tutorials dealing with other ways to measure the concentration of solutions can be found under the heading Solutions on the AUS-e-TUTE homepage.

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