Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

11. Introducing Molarity, volumes and the concentration of solutions

How do we define the concentration of a solution? How do we calculate concentration? What units do we use for concentration? What is molarity? How do we use moles to calculate the mass of a substance to make up a specific volume of a solution of specific concentration? All calculation methods are fully explained with fully worked out example questions.

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11. Molarity, volumes and the concentration of solutions

• Revise section 7. moles and mass before proceeding in this section 11 and eventually you may need to be familiar with the use of the apparatus illustrated above, some of which gives great accuracy when dealing with solutions and some do not.

• It is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution. So we need a standard way of comparing the concentrations of solutions. 
• The concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre,
• that is mol dm^-3 and this is called molarity, sometimes denoted in shorthand as M.
  • Note: 1dm³ = 1 litre = 1000ml = 1000 cm³, so dividing cm³/1000 gives dm³, which is handy to know since most volumetric laboratory apparatus is calibrated in cm³ (or ml), but solution concentrations are usually quoted in molarity, that is mol/dm³ (mol/litre).
• Equal volumes of solution of the same molar concentration contain the same number of moles of solute i.e. the same number of particles as given by the chemical formula.
• You need to be able to calculate
  • the number of moles or mass of substance in an aqueous solution of given volume and concentration
  • the concentration of an aqueous solution given the amount of substance and volume of water, for this you use the equation ....
  • (1a) molarity (concentration) of Z = moles of Z / volume in dm³
    • you need to be able to rearrange this equation ... therefore ...
    • (1b) moles = molarity (concentration) x volume in dm³ and ...
    • (1c)  volume in dm³ = moles / molarity (concentration)
  • You may also need to know that ...
    • (2) molarity x formula mass of solute = solute concentration in g/dm³
      • and dividing this by 1000 gives the concentration in g/cm³, and
    • (3) concentration in g/dm³ / formula mass = molarity in mol/dm³
      • both equations (2) and (3) result from equations (1) and (4), work it out for yourself.
  • and don't forget by now you should know:
    • (4) moles Z = mass Z / formula mass of Z
    • and (5) 1 mole = formula mass in grams

• Example 11.1

  • What mass of sodium hydroxide (NaOH) is needed to make up 500 cm³ (0.500 dm³) of a 0.500 mol dm^-3 (0.5M) solution? [A_r's: Na = 23, O = 16, H = 1]

  • 1 mole of NaOH = 23 + 16 + 1 = 40g

  • molarity = moles / volume, so mol needed = molarity x volume in dm³

  • mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH

  • therefore mass = mol x formula mass

  • = 0.25 x 40 = 10g NaOH required

• Example 11.2

  • (a) How many moles of H₂SO₄ are there in 250cm³ of a 0.800 mol dm^-3 (0.8M) sulphuric acid solution?

  • (b) What mass of acid is in this solution? [A_r's: H = 1, S = 32, O = 16]

    • (a) molarity = moles / volume in dm³, rearranging equation for the sulfuric acid

      • mol H₂SO₄ = molarity H₂SO₄ x volume of H₂SO₄ in dm³

      • mol H₂SO₄ = 0.800 x 250/1000 = 0.200 mol H₂SO₄

    • (b) mass = moles x formula mass

      • formula mass of H₂SO₄ = 2 + 32 + (4x16) = 98

      • 0.2 mol H₂SO₄ x 98 = 19.6g of H₂SO₄ 

• Example 11.3

  • 5.95g of potassium bromide was dissolved in 400cm³ of water. Calculate its molarity. [A_r's: K = 39, Br = 80]

  • moles = mass / formula mass, (KBr = 39 + 80 = 119)

  • mol KBr = 5.95/119 = 0.050 mol

  • 400 cm³ = 400/1000 = 0.400 dm³

  • molarity = moles of solute / volume of solution

  • molarity of KBr solution = 0.050/0.400 = 0.125M

• Example 11.4

  • What is the concentration of sodium chloride (NaCl) in g/dm³ and g/cm³ in a 1.50 molar solution?

  • At. masses: Na = 23, Cl = 35.5, formula mass NaCl = 23 + 35.5 = 58.5

  • since mass = mol x formula mass, for 1 dm³

  • concentration = 1.5 x 58.5 = 87.8 g/dm³, and

  • concentration = 87.75 / 1000 = 0.0878 g/cm³

• Example 11.5

  • A solution of calcium sulphate (CaSO₄) contained 0.500g dissolved in 2.00 dm³ of water.

  • Calculate the concentration in (a) g/dm³, (b) g/cm³ and (c) mol/dm³.

    • (a) concentration = 0.500/2.00 = 0.250 g/dm³, then since 1dm³ = 1000 cm³

    • (b) concentration = 0.250/1000 = 0.00025 g/cm³

    • (c) At. masses: Ca = 40, S = 32, O = 64, f. mass CaSO₄ = 40 + 32 + (4 x 16) = 136

      • moles CaSO₄ = 0.5 / 136 = 0.00368 mol

      • concentration CaSO₄ = 0.00368 / 2 = 0.00184 mol/dm³

There are more questions involving molarity in section 12. on titrations

and section 14.3 on dilution calculations