GCSE & A level Chemistry Calculations: Introduction to molarity definition & calculations INTRODUCTION TO MOLARITY and solution concentrations

e.g. g/dm3, g/cm3 and mol/dm3

= g dm-3, g cm-3 and mol dm-3

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations  11. Introducing molarity, volumes and the concentration of solutions in aqueous media - how to make up a standard solution

(a) Explaining the terms solubility, concentration, strength and molarity

(b) Measures of concentration and simple calculations of molarity

(b)(i) Concentration is terms of mass of solute per unit volume of solution

(b)(ii) Concentration in terms of moles of solute per unit volume of solution

Spotted any careless error? or request a type of GCSE calculation not covered? 11. Molarity, volumes and the concentration of solutions  (a) Explaining the terms solubility, concentration, strength and molarity

• Why are the terms 'concentration', 'strength' and 'molarity' important?

• Quite a lot of analytical procedures in chemistry involve the use of solutions of accurately known concentration e.g. titrations.

• If you want to analyse an acid solution you need to titrate it with a standard solution of alkali of accurately known concentration e.g. an accurately known molarity (concentration usually expressed in mol/dm3, lots more on this on the rest of this page!).

• The solubility of a substance is the maximum amount of solute that dissolves in a given volume of solvent.

• It is important to know the solubility of substance in various liquids, quite often quoted the maximum solubility of salts in water, but often quoted, not as molarity, but in g salt /100 g of water and plotted in graphs known as solubility curves.

• This is the maximum concentration possible for a given solute and solvent.

• • For more on solubility see

• Misconceptions

• There are differences in using the words ‘concentration’ and ‘strength’ in science compared to everyday language

• In scientific language concentration is a specifically defined term e.g.

• (i) the mass of solute per unit volume of solvent e.g. g/dm3, g/cm3  (g dm-3, g cm-3), OR,

• (ii) moles per unit volume of solvent e.g. mol/dm3   (mol dm-3),

• neither is a isn't a vague description, they are specific units!

• 'Strength' in terms of solution concentration is not a scientifically defined term and tends to be used in everyday language to 'crudely' indicate a concentration e.g. 'great/high strength' indicating a very concentrated solution, and conversely, 'low/weak strength' to indicate a low concentration solution.

• Unfortunately this every use of the term is widespread, so take care, because it does not apply to the science of chemistry!

• In chemistry, for solutions, the word 'strength' is applied to e.g. an acid to indicate how much it ionises in aqueous solution - is it a weak or strong acid or alkali (soluble base).

• A low strength solution of acid indicates it is a weak acid and only ionises a few % to give hydrogen ions.

• e.g. ethanoic acid: CH3COOH(aq) CH3COO(aq) + H+(aq)

• The equilibrium is about 2% to the right, a very weak acid.

• An acid of high strength indicates that it ionises to a very percent to form hydrogen ions i.e a strong acid.

• hydrochloric acid: HCl(g) + aq ===> H+(aq) + Cl(aq)

• When you dissolve hydrogen chloride in water (aq) you get virtually 100% ionisation into the hydrogen ion and chloride ion.

• In both cases the term concentration applies to the concentration of the original molecules:

•  e.g. ignoring extent of ionisation, the concentration of CH3COOH or HCl in mol/dm3.

• For more details see More on acid-base theory and weak and strong acids and their properties

• 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.
• Note: A standard solution is one whose precise concentration is known
• The concentration may be found by making up the solution from scratch e.g. by measuring out the mass of a solid and dissolving it in a known volume of solvent.
• OR, it may be found by standardising a soltion is some e.g. titrating an acid with an alkali or an alakli with an acid.

• Reminders: The dissolved substance is called the solute and the liquid dissolving it is the solvent.
• The more of the substance you dissolve in a liquid, the more concentrated the solution, the solute particles on average are closer together.
•  • The diagrams represent to substances dissolved in the solvent, the right-hand diagram represents a more concentrated solution e.g. a mixture of two salts in water.
• The pictures do not mean the right hand picture is a stronger solution!
• Unfortunately, in everyday language, it would be described as such, but this is science and the correct use of scientific language is essential!

(b) Measures of concentration and simple calculations of molarity

(b)(i) Concentration in terms of mass of solute per unit volume of solution

• We will look at moles in (b)(ii)
• The simplest measure of concentration is mass of solute per unit volume of solvent e.g.
• concentration = mass of solute / volume of solvent
• Take 5.0 g of salt dissolved in 500 cm3 of water.
• The concentration can be expressed in several ways.
• concentration = 5.0/500 = 0.01 g/cm3
• 1 dm3 = 1000 cm3, so 500 cm3 = 500/1000 = 0.50 dm3
• concentration = 5.0/0.50 = 10.0 g/dm3
• For interconversion: g/cm3 x 1000 = g/dm3  AND  g/dm3/1000 = g/dm3
• -
• Sometimes the general formula c = m/v is used
• c = concentration, m = mass, v = volume
• rearrangements: m = c x v  and  v = m/c
• -
• Example questions (not using moles)
• Q1 What is the concentration in g/dm3 if 6.0 g of salt is dissolved in 150 cm3 of water?
• 150 / 1000 = 0.15 dm3
• concentration =  mass / volume = 6.0 / 0.15 = 40.0 g/dm3
• -
• Q2 Given a salt solution of concentration 16 g/dm3, what mass of salt is in 40 cm3 of the solution?
• 1 dm3 = 1000 cm3
• c = m / v = 16 / 1000 = 0.16 g/cm3  (note this a way of converting g/dm3 to g/cm3)
• Therefore mass of salt: m = c x v = 0.16 x 40 = 6.4 g of salt
• -
• Q3 Given 5.0 g of a salt, what volume of water in cm3, should it be dissolved in to give a solution of concentration of 12.5 g/dm3?
• c = m / v, rearranging gives v = m / c
• v = 5.0 / 12.5 = 0.40 dm3
• volume of water needed = 1000 x 0.40 = 400 cm3
• -
• Its also good to be able to do dilution' calculations in section 14.3 dilution of solutions

(b)(ii) Concentration in terms of moles of solute per unit volume of solution

• For most analytical and calculation purposes the concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre of solution (reminder mole formula triangle on the right). 1 cubic decimetre (dm3) = 1 litre (l) in old money!
• concentration = molarity = moles of solute / volume of solvent
• Make sure you know how to calculate moles, see the triangle on the right!
• Using concentration units of mol dm-3 (or mol/dm3), the concentration is called molarity, sometimes denoted in shorthand as M (old money again, take care!) and the word molar is used too.
• 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
• You need to know all about moles to proceed further on this page and get into 'molarity' ...
• ... so read section 7. on moles and mass - essential pre-reading for section 11 ...
• AND, if you can't understand molarity, you cannot do titration calculations either!
• 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), so, for a substance Z ...
• (1a) molarity (concentration) of Z = moles of Z / volume in dm3
• This is sometimes referred to as the molar concentration (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)
• Use the triangle on the right to help you rearrange the equation for the basic definition of molarity.
• 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

• If 5.00g of sodium chloride is dissolved in exactly 250 cm3 of water in a calibrated volumetric flask,

• (a) what is the concentration in g/dm3?

• volume = 250/1000 = 0.25 dm3

• concentration = mass / volume = 5/0.25 = 20 g/dm3

• -

• (b) 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 = mol of solute / volume of solvent

• Molarity = 0.08547/0.25 = 0.342 mol/dm3

• -

• Molarity calculation Example 11.2

• 5.95g of potassium bromide was dissolved in 400cm3 of water.

• (a) 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 = 0.125 mol/dm3

• -

• (b) What is the concentration in grams per dm3?

• concentration = mass / volume, the volume = 400 / 1000 = 0.4 dm3

• concentration = 5.95 / 0.4 = 14.9 g/dm3

• -

• Molarity calculation Example 11.3

• 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

• -

• Molarity calculation Example 11.4

• (a) How many moles of H2SO4 are there in 250 cm3 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 11.5 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 11.6

• 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   (or from 0.500/2000)

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

• moles CaSO4 = 0.5 / 136 = 0.00368 mol in 2.00 dm3 of water

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

• -

• Molarity calculation Example 11.7

• -

• Its also good to be able to do dilution' calculations in section 14.3 dilution of solutions

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

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(c) APPENDIX 1 on SOLUBILITY and concentration calculations

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 X g 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, X g 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 X g 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 , 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.

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(d) APPENDIX 2 - How to make up a standard solution - a solution of precisely known concentration

The method and procedure of how to make up a standard solution of a soluble solid e.g. a salt, is fully described.

Procedure for making up a standard solution of known molarity

The method and procedure of how to make up a standard solution of a soluble solid e.g. a salt, is fully described Suppose you want to make up 250 cm3 of a salt solution of concentration 20g/dm3 (20g/litre, 20g/1000cm3, 20g/1000ml).

Example 1.

c = m / v, m = c x v, m = 20 x 250 / 1000 = 5g

so 5g of the salt is needed to be made up into an aqueous solution of exactly 250.0 cm3.

The procedure to is described in detail example 2. below.

To prepare a solution of known molarity, you need to work backwards from the volume required and the molarity to see how much solid you need.

Example 2.

Suppose you want to make up 250cm3 of a sodium chloride solution of concentration 0.20 moldm-3  Preliminary calculation:

From molarity formula (on the right): moles = molarity (mol/dm3) x volume (dm3)

and volume in cm3 / 1000 = dm3

moles NaCl needed = 0.20 x 250/1000 = 0.20 x 0.25 = 0.05 mol NaCl

Atomic masses: Na =23 and Cl = 35.5, so molar mass of NaCl = 23 + 35.5 = 58.5

From basic mole formula: mass of NaCl needed = mol NaCl x formula mass NaCl

mass of NaCl needed = 0.05 x 58.5 = 2.925 g (which is ok if you have a 3 decimal place balance!), so

2.295g of pure NaCl salt is needed to made up 250.0 cm3 of solution with a precise concentration of 0.20 mol/dm3.

Procedure to make the standard solution i.e. one of known concentration of solid (in this case)  An accurate one pan electronic balanced is set to zero (preferably with an accuracy of two decimal places). A beaker is placed on the balance and the reading noted (ignore the figures on the diagram). Very carefully, with a spatula (not shown), salt crystals are added to the beaker until it weighs exactly 2.925 grams more than the beaker. This can be a very fiddly procedure if you want exactly 2.925g of salt. Pure water (distilled/deionised) is then added to the beaker to completely dissolve the salt and use of a stirring rod helps to speed up the process. The amount of water you add to the beaker should be much less than 250cm3 to allow for the transfer and rinsing of the solution into the standard volumetric flask using a 'squeezy' wash bottle! Eventually a clear solution of the salt should be seen, there should be no residual salt crystals at the bottom of the beaker or on the sides of the beaker. You can use the wash bottle to rinse down any crystals on the side of the beaker, but watch the volume you use.. An accurately calibrated 250cm3 volumetric flask should be washed out and cleaned several times with pure water. Then, the whole of the solution in the beaker is transferred into the flask with the help of a funnel to avoid the risk of spillage. To make sure every drop of the salt solution ends up in the flask, a wash bottle of pure water is used to rinse out the beaker several times, AND rinse the stirring rod and the funnel too. This is to ensure nothing is lost in the transfer fro beaker to flask. Then, very carefully, the flask is topped up with pure water so the meniscus rests exactly on the 250.0cm3 calibration mark, a teat pipette is useful for the last few drops of water. The stopper is placed on and the flask carefully shaken quite a few times to ensure the salt solution is completely mixed up. Finally, check the meniscus lies on the calibration mark, in case another few drops are needed. Either way, the last drops of water should be added most carefully with a teat pipette.

Job done!

Note on standard solutions of acids and alkalis

You can purchase standard solutions ready for use.

OR, a phial of concentrated acid or alkali, which you dilute into a specified volume to give a specific molarity.

Apart from weighing out a solid, the procedure is the same as and , ensuring every drop from the phial is rinsed down the funnel into the calibrated volumetric flask.

See dilution' calculations in section 14.3 dilution of solutions

(e) Self-assessment Quizzes on molarity calculations:

type in answer QUIZ on molarity  or  multiple choice QUIZ on molarity

multiple choice titration QUIZ

(good revision for A level students) Above is typical periodic table used in GCSE science-chemistry specifications in doing molarity calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

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Keywords: 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 advanced level AS/A2/IB courses. These revision notes and practice questions on how to do molarity calculations in using solutions in chemistry and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.               Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial. 