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Concentration of Solutions (Molarity) Calculations

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 molarity1.

  • Concentration in mol/L or mol L-1, or Molarity, is given the symbol c (sometimes M).
        For a 0.01 mol L-1 HCl solution we can write :
            [HCl] = 0.01 mol L-1 (concentration implied by square brackets around formula)
            c(HCl) = 0.01 mol L-1 (c stands for concentration, formula given in brackets)

  • c = n ÷ V
          c = concentration of solution in mol L-1 (mol/L or M),
          n = moles of substance being dissolved (moles of solute),
          V = volume of solution in litres (L)

  • This formula can be re-arranged to find:
          moles of solute given molarity and volume of solution: n = c x V
          volume of solution given moles of solute and molarity: V = n ÷ c


1. Calculating Concentration (c = n ÷ V)

Calculate the concentration in mol L-1 (molarity) of a sodium chloride solution containing 0.125 moles sodium chloride in 0.5 litres of solution.

  1. Extract the data from the question
        n = 0.125 mol
        V = 0.5 L

  2. Write the equation:
        c = n ÷ V

  3. Substitute in the values and solve:
        [NaCl(aq)] = c(NaCl(aq)) = 0.125 ÷ 0.5 = 0.25 mol L-1 (or 0.25 mol/L or 0.25 M)

2. Calculating Moles of Solute (n = c x V)

Calculate the moles of copper sulfate in 250 mL of 0.02 mol L-1 copper sulfate solution.

  1. Extract the data from the question:
        c = 0.02 mol L-1
        V = 250 mL = 250 ÷ 1000 = 250 x 10-3 L = 0.250 L (since there are 1000 mL in 1 L)

  2. Write the equation:
        n = c x V

  3. Substitute in the values and solve:
       n = 0.02 x 250 x 10-3 = 0.005 mol

3. Calculating Volume of Solution (V = n ÷ M)

Calculate the volume of a 0.80 mol L-1 potassium bromide solution containing 1.6 moles of potassium bromide.

  1. Extract the data from the question:
        n = 1.6 mol
        c = 0.80 mol L-1

  2. Write the equation:
        V = n ÷ c

  3. Substitute in the values and solve:
        V = 1.6 ÷ 0.80 = 2.00 L

What would you like to do now?

1 There are many other ways to measure concentration of solutions in chemistry. You will find tutorials dealing with these under the heading Solutions on the AUS-e-TUTE homepage.

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