* GCSE/IGCSE & basic GCE-AS Chemical Calculations - Introducing molarity, volumes and solution concentrations *

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study examples carefully11. Introducing Molarity, volumes and the concentration of solutionsstudy examples carefully

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study examples carefully11. Molarity, volumes and the concentration of solutionsstudy examples carefully

volumetric apparatus

  • 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? [Ar'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? [Ar'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. [Ar'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


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    (spanish) Doc Brown Química - Cálculos químicos 11. Introducción a molaridad, los volúmenes y la concentración de soluciones On-line cuantitativos Química
    cálculos *  (thai) เคมีการคำนวณ 11 แนะนำ Molarity, ปริมาณและการแก้ปัญหาความเข้มข้นของ Online เคมีการคำนวณปริมาณ *  (chinese) 督布朗的化学-化学计算 11。引进摩尔浓度,数量和浓度的解决方案 在线定量化学计算 *

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