Doc Brown's Chemistry - Chemical Calculations

11. Introducing Molarity, volumes and the concentration of solutions

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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, mol dm^-3, 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 (0.5 dm³) of a 0.5M solution? [A_r's: Na = 23, O = 16, H = 1]

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

  • for 1000 cm³ (1 dm³) of 0.5M you would need 0.5 moles NaOH

  • which is 0.5 x 40 = 20g

  • however only 500 cm³ of solution is needed compared to 1000 cm³

  • so scaling down: mass NaOH required = 20 x 500/1000 = 10g

• Example 11.2

  • How many moles of H₂SO₄ are there in 250cm³ of a 0.8M sulphuric acid solution? What mass of acid is in this solution? [A_r's: H = 1, S = 32, O = 16]

  • formula mass of sulphuric acid = 2 + 32 + (4x16) = 98, so 1 mole = 98g

  • if there was 1000 cm³ of the solution, there would be 0.8 moles H₂SO₄

  • but there is only 250cm3 of solution, so scaling down ...

  • moles H₂SO₄ = 0.8 x (250/1000) = 0.2 mol

  • mass = moles x formula mass, which is 0.2 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 = 0.05 mol

  • 400 cm³ = 400/1000 = 0.4 dm³

  • molarity = moles of solute / volume of solution

  • molarity of KBr solution = 0.05/0.4 = 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

  • Therefore 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 0.5g dissolved in 2dm³ of water. Calculate the concentration in (a) g/dm³, (b) g/cm³ and (c) mol/dm³.

  • (a) concentration = 0.5/2 = 0.25 g/dm³, then since 1dm³ = 1000 cm³

  • (b) concentration = 0.25/1000 = 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