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INTRODUCTION TO MOLARITY and solution concentrations

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

Quantitative chemistry calculations Help for problem solving in doing molarity calculations from given masses, volumes and molecular/formula masses. Practice revision questions on calculating molarity from mass, volume and formula mass data, using experiment data, making predictions. 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. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses

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

volumetric apparatus

Appendix on How to make up a standard solution is on a separate page

See also 14.3 dilution of solutions calculations

  • 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 give 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 in.
    • The more you dissolve in a given volume of solvent, or the smaller the volume you dissolve a given amount of solute in, the more concentrated the solution.
  • The concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre (reminder mole formula triangle on the right),
  • with units  mol dm-3 (or mol/dm3)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).
    • Concentration is also expressed in a 'non-molar' format of mass per volume e.g. g/dm3
  • 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 ....  (reminder molarity formula triangle on the right),
    • (1a) molarity (concentration) of Z = moles of Z / volume in dm3
      • This is sometimes referred to as the mole-concentration,
      • and 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
        • This is sometimes referred to as the mass-concentration,
        • 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 to sum up, by now you should know:
      • (4) moles Z = mass Z / formula mass of Z
      • (5) 1 mole = formula mass in grams
      • (6) molarity = moles/dm3
  • 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

    • 500 cm3 = 500/1000 = 0.50 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 This involves calculating concentration in other ways e.g. mass/volume units

    • 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

        • -

  • Molarity calculation Example 11.6

    • If 5.00g of sodium chloride is dissolved in exactly 250 cm3 of water in a calibrated volumetric flask, what is the molarity of the solution?

    • Ar(Na) = 23, Ar(Cl) = 35.5, so Mr(NaCl) = 23 + 35.5 = 58.5

    • mole NaCl = 5.0/58.5 = 0.08547

    • volume = 250/1000 = 0.25 dm3

    • Molarity = 0.08547/0.25 = 0.342 mol/dm3

    • -

  • Molarity calculation Example 11.7

    • -


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

and section 14.3 on dilution calculations and

Appendix on How to make up a standard solution is on a separate page

top sub-index


How do you find out how soluble a substance is in water?

Reminder: solute + solvent ==> solution

i.e. the solute is what dissolves, the solvent is what dissolves it and the resulting homogeneous mixture is the solution.

The solubility of a substance is the maximum amount of it that will dissolve in a given volume of solvent e.g. water.

The resulting solution is known as a saturated solution, because no more solute will dissolve in the solvent.

Solubility can be measured and expressed in with different concentration units e.g. g/100cm3, g/dm3 and molarity (mol/dm3).

Solubility can also be expressed as mass of solute per mass of water e.g. g/100g of water.

You can determine solubility by titration if the solute reacts with a suitable reagent e.g. acid - alkali titration and it is especially suitable for substances of quite low solubility in water e.g. calcium hydroxide solution (alkaline limewater) can be titrated with standard hydrochloric acid solution.

However, many substances like salts are very soluble in water and a simple evaporation method will do which is described below e.g. for a thermally stable salt like sodium chloride.

(1) A saturated solution is prepared by mixing the salt with 25cm3 of water until no more dissolves at room temperature.

(2) The solution is filtered to make sure no undissolved salt crystals contaminate the saturated solution.

(3) Next, an evaporating dish (basin) is accurately weighed. Then, accurately pipette 10 cm3 of the saturated salt solution into the basin and reweigh the dish and contents.

By using a pipette, its possible to express the solubility in two different units.

(4) The basin and solution are carefully heated to evaporate the water.

(5) When you seem to have dry salt crystals, you let the basin cool and reweigh it.

(6) The basin is then gently heated again and then cooled and weighed again.

This is repeated until the weight of the dish and salt is constant, proving that all the water is evaporated

By subtracting the original weight of the dish from the final weight you get the mass of salt dissolved in the volume or mass of saturated salt solution you started with.

You can repeat the experiment to obtain a more accurate and reliable result.

(7) Calculations

By using a pipette it is possible to calculate the solubility in two ways, expressed as two quite different units.

Suppose the dish weighed 95.6g.

With the 10.0 cm3 of salt solution in weighed 107.7g

After evaporation of the water the dish weighed 96.5g

Mass of 10.0 cm3 salt solution = 107.7 - 95.6 = 12.1g

Mass of salt in 10 cm3 of salt solution = 96.5 - 95.6 = 0.9g

Mass of water evaporated = 107.7 - 96.5 = 11.2g

(a) Expressing the solubility in grams salt per 100 g of water

From the mass data above 0.9g of salt dissolved in 11.2g of water

Therefore Xg of salt dissolves in 100g of water, X = 100 x 0.9 / 11.2 = 8.0

Therefore the solubility of the salt = 8.0g/100g water

You can scale this up to 80.0g/1000g H2O, or calculate how much salt would dissolve in any given mass of water.

You can also express the solubility as g salt/100g of solution.

0.9g salt is dissolved in 12.1g of solution, Xg in 100g of solution

Therefore X = 100 x 0.9 / 12.1 = 7.4, so solubility = 7.4g/100g solution

These calculations do not require the original salt solution to be pipetted. You can just measure out approximately 10cm3 of the salt solution with 10cm3 measuring cylinder, and do the experiment and these calculations in the exactly the same way.

(b) However, if you know the exact volume of salt solution and the mass dissolved in it, then you can calculate the concentration in g/dm3, and if you know the formula mass of the salt, you can calculate the molarity of the solution.

From part (a) we have 0.9g of salt in 10.0 cm3

Therefore Xg will dissolve in 1000cm3 solution, X = 1000 x 0.9 / 10 = 90g/1000 cm3

Solubility of salt = 90g/dm3

Suppose the formula mass of the salt was 200, calculate the molarity of the saturated solution.

moles salt = mass / formula mass = 90/200 = 0.45 moles

Therefore solubility of saturated salt solution in terms of molarity = 0.45 mol/dm3


NOTE Solubility varies with temperature, see Gas and salt solubility in water and solubility curves, and it usually (but not always) increases with increase in temperature. So, in the experiment described above, the temperature of the saturated solution should be noted, or perhaps controlled to be saturated at 20oC or 25oC.

  • Self-assessment Quizzes

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    or multiple choice click me for QUIZ!Honly

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


    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 acid-alkali titration calculations, diagrams of apparatus, details of procedures

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

    14. Other calculations e.g. % purity, % percentage & theoretical yield, 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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