Practise calculating the concentration of a solution from the mass of solute and volume of water using the activities in this lesson plan 

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    Everything you need to teach this lesson: slides, student worksheets, concentration cards and teacher guidance including answers.

Learning objectives

  1. Recall how to convert between volumes in cm3 and dm3.
  2. Define the word ‘concentration’ from a given numerical value.
  3. Calculate the concentration of a solution in both g dm-3 and mol dm-3.
  4. Convert the concentration of a solution from g dm-3 to mol dm-3 and vice versa.

Introduction

In chemistry, concentration describes how much of a substance (the solute) is present in a given volume of solution. Concentration is most commonly expressed in two ways: grams per cubic decimetre (g dm-3) and moles per cubic decimetre (mol dm-3).

Both units relate the amount of solute to the volume of solution, and being confident in using each is essential for quantitative chemistry, including titrations and stoichiometric calculations.

In this activity, learners begin by comparing the concentrations of several solutions and reflecting on their current level of understanding. They then work in pairs using cards to link a mass of solute, volume of water and concentration of a solution, exploring the process of calculating concentration.

The lesson plan includes an extension activity, giving learners additional practise calculating the concentration of solutions.

Equipment

Demonstration

Read our standard health and safety guidance before carying out any practical activity.

  • Beakers, 250 cm3, x3
  • Copper(II) sulfate solution 0.1 mol dm-3, 200 cm3
    WARNING: irritant (skin, eyes)
  • Water, 50 cm3

Other equipment

  • A set of ‘traffic light’ cards for each learner (optional)
  • Scissors to cut out the concentration cards

Sequence of activities

Demonstration

  1. Issue ‘traffic light’ cards to all learners. You could also use thumbs up/down or another appropriate hand signal to indicate responses.

  2. Explain that they are going to:
    • compare the concentration and number of moles of solute in solutions.
    • use the ‘traffic light’ cards to indicate their view: green for ‘the same’, red for ‘different’ and yellow for ‘unsure’.
  3. Pour 100 cm3 of copper(II) sulfate solution into each of two beakers A and B. Pour half of the solution from beaker A into a third beaker C.
  4. Ask the learners to give their view on the following comparisons:
    • The number of moles of copper(II) sulfate in beakers B and C.
    • The concentration of copper(II) sulfate in beakers B and C.
  5. After adding water to beaker C to make the total volume 100 cm3 again, ask learners for a second time to give their view on:
    • The number of moles of copper(II) sulfate in beakers B and C.
    • The concentration of copper(II) sulfate in beakers B and C.
  6. Use their indications as an aid to sharing the learning objective.

Explaining concentrations

Ask learners to work through the first section of the student sheet which assesses the key knowledge needed to calculate and compare concentrations.

Organise learners to:

  1. Work individually to complete the explanations and the ‘can do’/‘can’t do’/‘not sure’ boxes.
  2. Join with another learner.
  3. Compare responses and convert any ‘can’t do’ or ‘not sure’ responses to ‘can do’.
  4. Join with another pair of learners if there are still any ‘can’t do’ or ‘not sure’ responses.

Card matching

Task 1

  1. Move learners back into pairs.
  2. Give a set of ‘Concentration cards’ in the teacher notes to each pair. Ask them to match the correct mass and volume to each concentration.
  3. Circulate and support with prompts while pairs of learners:
    • group cards together showing the mass of sodium hydroxide and volume of water needed to produce the concentration shown on one of the cards
    • record their answers in the table on the student sheet
    • explain the general approach to calculating concentrations.

Task 2 and 3

When pairs have recorded and shown the correct answers, give them a set of ‘Blank Concentration cards’ and a solute chosen from:

  • sodium carbonate.
  • sulfuric acid.
  • potassium hydroxide.
  • calcium bromide.
  • copper(II) sulfate.

Circulate and support with prompts while pairs:

  1. Devise their own set of concentration cards using the solute given to them so that all cards are used up when the mass of solute, volume of water and concentration of solute or ions in solution are matched up.
  2. Join up with another pair.
  3. Exchange the cards they have devised.
  4. Match up and record the cards devised by the other pair on the student sheet.
  5. Help each other pair to select appropriate cards where this is necessary.

Extension activity

As an extension, set the following problem and work through the solution in a plenary.

Problem

Calculate the final concentrations in mol dm-3 of H+, Na+, Cl- and SO42-, when the following three solutions are mixed together to give a total volume of 2 dm3:

  • 1000 cm3 of 0.1 mol dm-3 HCl
  • 500 cm3 of 0.2 mol dm-3 NaCl
  • 500 cm3 of 0.2 mol dm-3 Na2SO4

Commentary

The snapshot of learner confidence, at the start of the session, gives the learners a baseline as well as informing the teacher.

By writing explanations of how to do simple calculations and discussing their competence in a structured way the learners are helped to recognise their own strength and weaknesses. Their learning is embedded when they set a further card matching exercise for their peers.

Answers

Card matching

ConcentrationMass of NaOHVolume of water
4 g dm-3 NaOH 2 g 0.5 dm3
0.4 mol dm-3 NaOH 4 g 250 cm3
0.5 mol dm-3 NaOH 40 g 2.0 dm3
2.5 mol dm-3 Na+ ions 10 g 100 cm3

Extension activity

Total volume = 2 dm3 (ie 2000 cm3)

Assume all species are strong electrolytes and are fully dissociated in aqueous solution.

Final solution contains:

  • 0.05 mol dm-3 HCl – ie 0.05 mol dm-3 H+ and 0.05 mol dm-3 Cl-
  • 0.05 mol dm-3 NaCl – ie 0.05 mol dm-3 Na+ and 0.05 mol dm-3 Cl-
  • 0.05 mol dm-3 Na2SO4 – ie 0.10 mol dm-3 Na+ and 0.05 mol dm-3 SO42-

Therefore:

  • Concentration of H+ = 0.05 mol dm-3
  • Concentration of Cl- = 0.05 + 0.05 = 0.10 mol dm-3
  • Concentration of Na+ = 0.05 + 0.10 = 0.15 mol dm-3
  • Concentration of SO42- = 0.05 mol dm-3