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Check out what is available? Study the different examples then try the Quizzes!Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

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

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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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. 
  • top sub-indexThe 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
  • Molarity calculation 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? [Ar'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

  • top sub-indexMolarity calculation 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? [Ar'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 

  • Molarity calculation 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 = 0.050 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

  • Molarity calculation 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

  • Molarity calculation 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

top sub-index


  • Self-assessment Quizzes

    [msc] type in answer click me for QUIZ!Honly

    or multiple choice click me for QUIZ!Honly

    See also Advanced level GCE-AS-A2 acid-alkali titration calculation questions


    OTHER CALCULATION PAGES

    1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

    2. Calculating relative formula/molecular mass of a compound or element molecule

    3. Law of Conservation of Mass and simple reacting mass calculations

    4. Composition by percentage mass of elements in a compound

    5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

    6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

    7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

    8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

    9. Moles and the molar volume of a gas, Avogadro's Law

    10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

    11. Molarity, volumes and solution concentrations (and diagrams of apparatus) (this page)

    12. How to do volumetric titration calculations e.g. acid-alkali titrations (and diagrams of apparatus)

    13. Electrolysis products calculations (negative cathode and positive anode products)

    14. Other calculations e.g. % purity, % percentage & theoretical yield, volumetric titration apparatus, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

    15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

    16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

    17. Radioactivity & half-life calculations including dating materials


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