GCSE & A level Chemistry Calculations: Reacting mass calculations - concentration & volumes

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Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations 6b. Reacting mass chemical calculations - reacting masses, equations, volumes and concentrations - volumetric titrations (BUT method NOT using moles or molarity)

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Keywords: Quantitative Chemistry calculations online Help for problem solving in doing volumetric titration calculations involving masses, volume and concentrations. Practice revision questions on mass, volume, concentration and titrations, using experiment data, making predictions. This page describes and explains, with fully worked out examples, how to do titration calculations just using concentrations measured in mass/volume units. This calculation method also involves reacting masses deduced from the balanced symbol equation. 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. These revision notes and practice questions on how to do chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

See also 12. Volumetric titration methods & calculations using molarity & mole concept

AND see also 11. Molarity calculations and 14.3 dilution of solutions calculations  6b. Reacting masses, concentration of solution and volumetric titration calculations Advanced Level acid-alkali volumetric titration calculations

For quantitative chemistry, it is important to know the concentration of solutions and be able to do calculations based on experimental results for chemical analysis.

Concentration units

The concentration of a solution is often measured in grams per decimetre cubed

so the units are expressed as gdm-3, or g/dm3, which was g/litre.

concentration = mass / volume (c = m / V)

concentration in gdm-3, mass in g, volume in dm3

Its really important remember that 1 dm3 = 1000 cm3 or 1000 ml

and cm3/1000 = dm3

rearrangements of the formula using the triangle to help gives

mass = concentration x volume (m = c x V)

and volume = mass / concentration (V = m / c)

In the example questions I have used the shorthand formulae c = m / V, m = c x V and V = m / c

and RFM as an abbreviation of relative formula mass or molecular mass, and other shorthands:-

sf means numbers rounded to 2/3/4 significant figures and dp means round to 2/3/4 decimal places

A standard solution is one of accurately known concentration. Three simple examples to illustrate using the formula (formula triangle on right)

(a) What is the concentration of a salt solution if you dissolve 10g of sodium chloride in 250 cm3 of water?

250 cm3 is equal to 250/1000 = 0.25dm3

therefore the concentration c = m / V = 10/0.25 = 40 g/dm3

(40g/litre in old units, still in common use!)

(b) What mass of the salt is required to make 200 cm3 of concentration 15g/dm3?

V = 200/1000 = 0.2 dm3

m = c x V = 15 x 0.2 =  3.0 g

(c) If you were given 8.0 g of salt, what volume of water, in dm3 and cm3, should you dissolve it in, to give a salt solution of concentration of 5.0g/dm3?

V = m / c = 8.0 / 5.0 = 1.6 dm3

V = 1.6 x 1000 = 1600 cm3

TOP OF PAGE QUESTIONS involving mass, volume and concentration

• Example Question 6b (Q1)  A potassium sulfate solution A has a concentration of 6.5g/dm3

• (a) What mass of potassium sulfate is present in 25cm3 of solution A?

• (b) What volume of solution A, in cm3, contains 3.0g of potassium sulphate?

• (c) What mass of the salt is required to make 250 cm3 of a solution B so it contains 3.75 g/dm3 of potassium sulfate?

• (d) If 12.5g of potassium sulphate was dissolved in 1500 cm3 of water to make up solution C, what is the concentration of the salt?

• -

• Answers to Question 6b. Q1  A potassium sulfate solution has a concentration of 6.5g/dm3

• (a) What mass of potassium sulfate is present in 25cm3 of the solution?

• 25cm3 = 25/1000 = 0.025 dm3

• m = c x V

• mass = 6.5 x 0.025 =  0.1625 g

• (b) What volume of solution, in cm3, contains 3.0g of potassium sulphate?

• V = m / c = 3.0 / 6.5 =  0.462 dm3 (to 3 sf)

• V = 1000 x 0.462 = 462 cm3

• (c) What mass of the salt is required to make 250 cm3 of a solution B so it contains 3.75 g/dm3 of potassium sulfate?

• volume = 250/1000 = 0.25 dm3

• m = c x V = 3.75 x 0.25 = 0.9375 g

• (d) If 12.5g of potassium sulphate was dissolved in 1500 cm3 of water to make up solution C, what is the concentration of the salt?

• V = 1500 cm3/1000 = 1.5 dm3

• c = m / V = 12.5/1.5 = 8.33 g/dm3

• -

• Example Question 6b (Q2) Nitric acid - sodium hydroxide titration

• The reaction equation for the neutralisation of nitric acid by sodium hydroxide to form sodium nitrate is

• HNO3(aq) + NaOH(aq) ===> NaNO3(aq) + H2O(l)

• (a) Calculate the relative formula masses of nitric acid, sodium hydroxide and sodium nitrate

• Atomic masses: H = 1, N = 14, O = 16, Na = 23

• 25.0 cm3 of a nitric acid solution was pipetted into a conical flask and titrated with a sodium hydroxide solution of concentration 5.0g/dm3. If it took 16.8 cm3 of the alkali to neutralise the acid, calculate the following

• (b) The mass of sodium hydroxide that reacted with the nitric acid.

• (c) Calculate the mass of nitric acid that reacted with the sodium hydroxide.

• (d) What was the concentration of the nitric acid in g/dm3?

• (e) If the resulting neutral solution was carefully evaporated to dryness, what mass of sodium nitrate salt crystals would be left as a residue?

• -

• Answers to Question 6b. Q2 Nitric acid - sodium hydroxide titration

• (a)  nitric acid + sodium hydroxide ===> sodium nitrate + water HNO3(aq) + NaOH(aq) ===> NaNO3(aq) + 2H2O(l) RFM 63 40 85 -

• 25.0 cm3 of a nitric acid solution was pipetted into a conical flask and titrated with a sodium hydroxide solution of concentration 5.0g/dm3. If it took 16.8 cm3 of the alkali to neutralise the acid, calculate the following

• (b) The mass of sodium hydroxide that reacted with the nitric acid.

• V = 16.8 cm3/1000 = 0.0168 dm3

• m = c x V = 5.0 x 0.0168 =  0.084g NaOH

• (c) Calculate the mass of nitric acid that reacted with the sodium hydroxide.

•  HNO3 : NaOH 63 : 40 x g : 0.084 g
• Solving the ratio: x = 0.084 x 63/40 = 0.1323 g HNO3
• (d) What was the concentration of the nitric acid in g/dm3?

• V = 25.0 cm3/1000 = 0.025 dm3

• c = m / V = 0.1323 / 0.025 = 5.29 g/dm3 (3sf, 2dp)

• (e) If the resulting neutral solution was carefully evaporated to dryness, what mass of sodium nitrate salt crystals would be left as a residue?

•  NaOH : NaNO3 40 : 85 0.084 g : x g
• Solving the ratio: x = 0.084 x 85/40 = 0.1785 g NaNO3
• You can also use HNO3 instead of NaOH i.e. 63 : 85, it doesn't matter which you use.
• -

• Example Question 6b (Q3) Limewater analysis

• The symbol equation for the neutralisation reaction between calcium hydroxide solution (limewater) and hydrochloric acid to form calcium chloride and water is

• Ca(OH)2(aq) + 2HCl(aq) ===> CaCl2(aq) + 2H2O(l)

• 50 cm3 of a limewater solution was completely neutralised by 9.7 cm3 of a hydrochloric acid solution. If the concentration of the hydrochloric acid was 3.65 g HCl/dm3 calculate the concentration of calcium hydroxide in the water by the following method.

• (a) calculate the relative formula masses of the reactants.

• Atomic masses: Ca = 40, O = 16, H = 1, Cl = 35.5

• (b) What mass of hydrochloric acid (as HCl) reacted with the limewater?

• (c) What mass of calcium hydroxide reacted with the hydrochloric acid?

• (d) What is the concentration of calcium hydroxide in g/dm3?

• -

• Question 6b (Q3) Limewater analysis

• The symbol equation for the neutralisation reaction between calcium hydroxide solution (limewater) and hydrochloric acid to form calcium chloride and water is

• 50 cm3 of a limewater solution was completely neutralised by 9.7 cm3 of a hydrochloric acid solution. If the concentration of the hydrochloric acid was 3.65 g HCl/dm3 calculate the concentration of calcium hydroxide in the water by the following method.

• (a) Calculate the relative formula masses of the reactants.

•  calcium hydroxide + hydrochloric acid ===> calcium chloride + water Ca(OH)2(aq) + 2HCl(aq) ===> CaCl2(aq) + 2H2O(l) RFM 74 36.5 -
• (b) What mass of hydrochloric acid (as HCl) reacted with the limewater?

• V = 9.7 cm3/1000 = 0.0097 dm3

• m = c x V = 3.65 x 0.0097 =  0.0354 g HCl

• (c) What mass of calcium hydroxide reacted with the hydrochloric acid?

•  Ca(OH)2 : 2HCl 74 : 2 x 36.5 = 73 x g : 0.0354 g
• Solving the ratio: x = 0.0354 x 74/73 = 0.03588 g Ca(OH)2
• (d) What is the concentration of calcium hydroxide in g/dm3?

• V = 50 cm3/1000 = 0.05 dm3

• c = m / V = 0.03588/0.05 = 0.718 g/dm3 Ca(OH)2

• -

• Example Question 6b (Q4) Sulfuric acid - potassium hydroxide reaction to make potassium sulfate

• The neutralisation reaction between sulphuric acid and potassium hydroxide is

• H2SO4(aq) + 2KOH(aq) ===> K2SO4(aq) + 2H2O(l)

• A solution of potassium hydroxide contained 100g/dm3.

• A solution of sulfuric acid contained 120g/dm3.

• What volume in cm3, of the sulfuric acid solution, is required to neutralise 250 cm3 of the potassium hydroxide solution?

• Atomic masses: H = 1, S = 32, O = 16, K = 39

• Answers to Question 6b. Q4 Sulfuric acid - potassium hydroxide reaction to make potassium sulfate

• H2SO4(aq) + 2KOH(aq) ===> K2SO4(aq) + 2H2O(l)

• Potassium hydroxide contained 100g/dm3 and sulfuric acid contained 120g/dm3.

• What volume in cm3, of the sulfuric acid solution, is required to neutralise 250 cm3 of the potassium hydroxide solution?

• To calculate mass of KOH

• V = 250 cm3/1000 = 0.25 dm3

• m = c x V = 100 x 0.25 =  25g of KOH

• To calculate mass of H2SO4 needed

•  H2SO4 : 2KOH 98 : 2 x 56 = 112 x g : 25 g
• Solving the ratio: x = 25 x 98/112 = 21.875 g H2SO4
• To calculate the volume of sulphuric acid needed.

• V = m / c = 21.875/120 =  0.1823 dm3

•  0.1823 x 1000 =  182 cm3 (3sf, nearest cm3)

• - Example Question 6b (Q5)

0.5g of sodium chloride was dissolved in water and you are given a silver nitrate solution of concentration 30g/dm3. When you mix the two solutions you get a white precipitate of silver chloride.

Given the equation and the relative formula masses (RFMs) calculated from the atomic masses

Na = 23, Cl = 35.5, Ag = 107.8, O = 16, N = 14

 sodium chloride + silver nitrate ===> silver chloride + sodium nitrate NaCl(aq) + AgNO3(aq) ===> AgCl(s) + NaNO3(aq) (a) RFM 58.5 169.8 (143.3) (85) (not needed)

Using an accurate burette, precisely what volume (in cm3) of the silver nitrate solution must be added to the sodium chloride solution in order to precipitate the maximum amount of silver chloride without wasting a drop of quite a costly solution! Three steps to the calculation:- (a) calculate the relative formula masses of the reactants, (b) the mass of silver nitrate reacting and (c) the volume of silver nitrate needed.

(b) From the equation calculate the mass of silver nitrate that reacts with 0.5 g of sodium chloride.

According to the symbol equation one RFM or 'molecule' of sodium chloride reacts with one RFM or 'molecule' of silver nitrate.

 NaCl : AgNO3 58.5 : 169.8 0.5g : x g

Solving the ratio gives x = 0.5 x 169.8 / 58.5 = 1.451g of silver nitrate

(b) From your answer to (a) calculate the volume of silver nitrate needed.

V = m / c = 1.451 / 30 = 0.0484 dm3

V = 1000 x 0.0484 = 48.4 cm3

Example Question 6b (Q6)

The reaction between hydrochloric acid and sodium hydroxide is

 hydrochloric acid + sodium hydroxide ===> sodium chloride + water HCl(aq) + NaOH(aq) ===> NaCl(aq) + H2O(l) RFM = 36.5 RFM = 40

According to the symbol equation one RFM or 'molecule' of hydrochloric acid reacts with one RFM or 'molecule' of sodium hydroxide.

(The atomic masses involved are H = 1, Cl = 35.5, Na = 23 and O = 16, check the RFM for yourself)

Therefore from the relative formula masses (RFMs) 36.5g of HCl reacts with 40g of sodium hydroxide.

You are given a standard solution of hydrochloric acid of 7.3 g/dm3 (of HCl). 25.0 cm3 of sodium hydroxide solution was pipetted into a conical flask. On titration with the acid solution using a burette and suitable indicator, it was found that 14.6 cm3 of the acid solution was required to completely neutralise the alkaline sodium hydroxide. Calculate the concentration of the sodium hydroxide solution in g/dm3 via the stages outlined below

(a) The mass of hydrochloric acid reacting.

V = 14.6 cm3/1000 = 0.0146 dm3

m = c x V = 7.3 x 0.0146 =  0.1066 g HCl

(b) Calculate the mass of NaOH that reacts with 0.1066 g of HCl

 HCl : NaOH 36.5 : 40 0.1066g : x g

Solving the ratio x = 0.1066 x 40/36.5 =  0.1168 g

(c) Calculate the concentration of NaOH

V = 25 cm3/1000 = 0.025 dm3

c = 0.1168/0.025 =  4.67 g/dm3 (2dp, 3sf)

Now it is possible to titrate the acid solution with the alkali solution (or vice versa) to obtain an unknown concentration of one of the solutions. One concentration must be known and the two volumes (acid and alkali) which react together exactly i.e. no excess of either reactant solution. How to do this is described in Section 12. acid-alkali titrations

Example 6b (Q7)

Citric acid is the most common acid in citrus fruits and can be estimated by titration with sodium hydroxide.

 citric acid + sodium hydroxide ===> sodium citrate + water (aq) + 3NaOH(aq) ===> (aq) + H2O(l) RFM = 192 for C6H8O7 RFM = 40 but reacting mass is 3 x 40 = 120

You are a given of sodium hydroxide with a concentration of 12.0 g/dm3.

10 cm3 of squeezed lemon juice was pipetted into a flask. Using a burette and suitable indicator the lemon juice required a titration volume of 35.5 cm3 of the sodium hydroxide solution to completely neutralise it.

Assuming all the acidity is due to citric acid, calculate the concentration of citric acid in the lemon juice.

(a) What mass of sodium hydroxide reacted with the lemon juice?

V = 35.5 cm3/1000 = 0.0355 dm3

m = c x V = 12.0 x 0.0355 = 0.426g NaOH

(b) What mass of citric acid reacted with the 0.426 g of sodium hydroxide?

 C6H8O7 : 3NaOH 192 : 3 x 40 = 120 x g : 0.426 g

Solving the ratio x = 0.426 x 192/120 = 0.6816 g C6H8O7

(c) Calculate the concentration of citric acid in the lemon juice?

V = 10 cm3/1000 = 0.01 dm3

c = m / V = 0.6816/0.01 = 68.2 g/dm3 C6H8O7

• Example Question 6b (Q8)

• This is a much more elaborate reacting mass calculation involving solution concentrations and extended ideas from the results.

• In this exemplar Q I've used simple formulae a lot for short-hand.

• A solution of hydrochloric contained 7.3 g HCl/dm3.

• A solution of a metal hydroxide of formula MOH was prepared by dissolving 4.0g of MOH in 250 cm3 of water.

• M is an unknown metal but it is known that the ionic formula of the hydroxide is M+OH-. 25cm3 samples of the MOH solution were pipetted into a conical flask and titrated with the hydrochloric solution using a burette and a few drops of phenolphthalein indicator.

• All the MOH is neutralised as soon as the pink indicator colour disappears (i.e. the indicator becomes colourless).

• On average 19.7 cm3 of the HCl acid solution was required to completely neutralise 25.0 cm3 of the MOH solution.

• [Atomic masses: H = 1, Cl = 35.5, O = 16, M = ?]

• (a) Give the equation for the reaction between the metal hydroxide and the hydrochloric acid.

• MOH(aq) + HCl(aq) ==> MCl(aq) + H2O(l)

• You may or may not be required to give the state symbols in (), or you may be just asked to complete the equation given part of it.

• (b) Calculate the mass of HCl used in each titration.

• 1 dm3 = 1000 cm3, so in 19.7 cm3 of the HCl solution there will be

• 19.7 x 7.3 / 1000 =  0.1438 g HCl

• (c) Calculate the mass of MOH that reacts with the mass of HCl calculated in (b).

• 25cm3 of the 250 cm3 MOH solution was used, so the mass of MOH titrated is

• 25 x 4 / 250 = 0.40 g MOH

• (d) Calculate the formula mass of HCl.

• formula mass = 1 + 35.5 = 36.5

• (e) Calculate the mass in g of MOH that reacts with 36.5g of HCl and hence the formula mass of MOH.

• 0.1438 g HCl reacts with 0.40 g MOH

• therefore 36.5g HCl reacts with z g of MOH

• solving the ratio for z, z = 36.5 x 0.40 / 0.1438 = 101.5 g MOH

• Since the formula mass of HCl is 36.5 and from the equation, 1 MOH reacts with 1 HCl the experimental formula mass of MOH is found to be 101.5

• (f) What is the atomic mass of the metal?

• Since O = 16 and H = 1, total of 17, atomic mass of M = 101.5  - 17 = 84.5

• (g) From the formula information on the metal hydroxide deduce the following giving reasons:

• What group of the periodic table is M likely to be from?

• Group 1 Alkali Metals because they form singly charged positive ions (they readily lose their outer electron to do so).

• Which metal is M likely to be?

• Rubidium. The atomic mass for rubidium is quoted as 85.5 and the experimental value of 84.5 is pretty close - don't forget the experimental procedures will not be perfect, especially at GCSE level!

• -

• Example Question 6b (Q9) Empirical formula calculation - another more elaborate reacting mass calculation involving solution concentrations to arrive at a formula mass.

• I've used HCl and MOH as shorthand in the question and answers.

• The idea is to use the analysis data to work out the atomic mass of metal M, that forms a hydroxide of formula MOH.

• A solution of hydrochloric contained 3.65 g/dm3.

• A solution of a metal hydroxide of formula MOH was prepared by dissolving 5.0g of MOH in 1 dm3 of water (M is an unknown metal).

• 25 cm3 of the MOH solution required 22.3 cm3 of the HCl acid solution to neutralise it in a titration procedure using a pipette (MOH) and burette (HCl).

• The equation for the neutralisation reaction is: MOH + HCl ==> MCl + H2O

• Atomic masses: H = 1, Cl = 35.5, O = 16, M = ?

• (a) Calculate the mass of MOH neutralised in each titration.

• 5 x 25 / 1000 = 0.125g MOH (remember 1 dm3 = 1000 cm3)

• (b) Calculate the mass of HCl reacting in each titration.

• 3.65 x 22.3 / 1000 = 0.0814 g HCl

• (d) Calculate the formula mass of HCl

• Formula mass of HCl = 1 + 35.5 = 36.5

• (c) Calculate the mass of MOH that reacts with 36.5 g HCl and hence the formula mass of MOH.

• If 0.125 g MOH reacts with 0.0814 g HCl

• z g MOH reacts with 36.5 g HCl

• solving the ratio, z = 36.5 x 0.125 / 0.0814 = 56.1 g MOH

• Therefore the experimental formula mass of MOH is 56.1 (~56) because from the equation 1 HCl reacts with 1 MOH.

• (d) If the metal is in the Group 1 of Alkali Metals, what is the atomic mass of M and what metal is M?

• If the formula mass of MOH is 56, atomic mass of M = 56 - 1 -16 = 39

• The atomic mass of potassium is 39, so M is potassium.

• (So 'fictitious' MOH is really KOH, potassium hydroxide. You are likely to very familiar with another in the same group, sodium hydroxide NaOH)

• -

• Example 6b. (Q10) A Purity calculation based on reacting masses and solution concentrations.

• This question is based the titration of aspirin (an acid), with standard sodium hydroxide solution.

• 0.300g of aspirin was titrated with sodium hydroxide solution of concentration 4.00g/dm3.

• If the aspirin required 16.45 cm3 of the NaOH(aq) to neutralise it, calculate the percent purity of the aspirin.

• The simplified equation for the reaction is ...

• C6H4(OCOCH3)COOH + NaOH  ==> C6H4(OCOCH3)COONa + H2O

• Mr(aspirin) = 180,  Mr(NaOH) = 40   (atomic masses: C = 12, H = 1, O = 16, Na = 23)

• Therefore the reacting mass ratio is 180g aspirin reacts with 40g of sodium hydroxide.

• The titration was 16.45 cm3, so, converting the cm3 to dm3,

• the mass of NaOH used in the titration = 4.00 x 16.45/1000 = 0.0658g,

• so we can scale this up to get the mass of aspirin titrated,

• therefore the mass of aspirin titrated = 0.0658 x 180 / 40 = 0.296g

• therefore the % purity = 100 x 0.296 / 0.300 = 98.7%

• -

• Example 6b. (Q11) also based on reacting masses and solution concentrations.

• Deposition from hard water samples

• On analysis, a sample of hard water was found to contain 0.056 mg of calcium hydrogen carbonate per cm3 (0.056 mg/ml).

• If the water is boiled, calcium hydrogencarbonate Ca(HCO3)2, decomposes to give a precipitate of calcium carbonate CaCO3, water and carbon dioxide.

• (a) Give the symbol equation of the decomposition complete with state symbols.

• Ca(HCO3)2(aq) ==> CaCO3(s) + H2O(l) + CO2(g)

• (b) Calculate the mass of calcium carbonate in grammes deposited if 2 litres (2 dm3, 2000 cm3 or ml) is boiled in a kettle.

• [ atomic masses: Ca = 40, H = 1, C = 12, O = 16 ]

• the relevant ratio is based on: Ca(HCO3)2 ==> CaCO3

• The formula masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40 + 12 + 16x3) respectively

• there the reacting mass ratio is 162 units of Ca(HCO3)2 ==> 100 units of CaCO3

• the mass of Ca(HCO3)2 in 2000 cm3 (ml) = 2000 x 0.056 = 112 mg

• therefore solving the ratio 162 : 100 and 112 : z mg CaCO3

• where z = unknown mass of calcium carbonate

• z = 112 x 100/162 = 69.1 mg CaCO3

• since 1g = 1000 mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate

• (c) Comment on the result, its consequences and why is it often referred to as 'limescale'?

• This precipitate of calcium carbonate will cause a white/grey deposit to be formed on the side of the kettle, especially on the heating element.

• Although 0.0691 g doesn't seem much, it will build up appreciably after many cups of tea!

• The precipitate is calcium carbonate, which occurs naturally as the rock limestone, which dissolved in rain water containing carbon dioxide, to give the calcium hydrogen carbonate in the first place.

• Since the deposit of 'limestone' builds up in layers it is called 'limescale'.

See also 12. Volumetric titration methods & calculations using molarity & mole concept

AND see also 11. Molarity calculations and 14.3 dilution of solutions calculations

TOP OF PAGE

APPENDIX - How to make up a standard solution (mass/volume)

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

For making up a standard solution of known molarity see the page Suppose you want to make up 250cm3 of a salt solution of concentration 20g/dm3 (20g/litre, 20g/1000cm3, 20g/1000ml).

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. 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 (92.54g). Very carefully, with a spatula (not shown), salt crystals are added to the beaker until it weighs exactly 5.00 grams more (97.54g). This can be a very fiddly procedure if you want exactly 5.00g of salt. Pure water 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. 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 round the inside if there any crystals on the inside surface of the beaker. 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, the stirring rod and in the end the funnel too. 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 portions 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 is still 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.

For making up a standard solution of known molarity see the page

TOP OF PAGE Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

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