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Concentration, % m/m, % m/v and Solubility Class 9

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ScienceCBSE Class 9 · Exploring Mixtures and their Separation

Concentration of Solutions and Solubility: Percentage Numericals (Class 9)

Updated 17 October 2026 · Fact-checked

The concentration of a solution is the amount of solute in a given amount of solution: % m/m = mass of solute ÷ mass of solution × 100, % m/v = mass of solute (g) ÷ volume of solution (mL) × 100, and % v/v = volume of solute ÷ volume of solution × 100. Solubility is the largest mass of solute that 100 g of solvent holds at a stated temperature; a solution holding that much is saturated, and the mass of crystals formed on cooling is the difference between the solubilities at the two temperatures (per 100 g of water).

Understand Concentration and Solubility of Solutions

Solutions. A solution is a homogeneous mixture of a solute (the substance dissolved) and a solvent (the substance that dissolves it). The mass of solution = mass of solute + mass of solvent. The composition of a solution matters: ORS, medicines and pesticide sprays work only if the amounts are right.

Three ways to give concentration. Mass by mass (% m/m) = mass of solute ÷ mass of solution × 100, used for food labels and alloys. Mass by volume (% m/v) = mass of solute in grams ÷ volume of solution in mL × 100, used for medicines such as a 0.9% saline drip and a 5% glucose drip. Volume by volume (% v/v) = volume of solute ÷ volume of solution × 100, used for liquids dissolved in liquids, such as vinegar. The denominator is always the solution, never just the solvent. A useful rearrangement is: mass of solute = percentage ÷ 100 × mass (or volume) of solution.

Dilution and mixing. When water is added, the mass of solute does not change, only the mass of solution does. When water evaporates the solute stays and the solution becomes more concentrated. When two solutions are mixed, add the solute masses and the solution masses separately and then find the percentage of the final mixture.

Solubility. Solubility is the largest mass of solute that 100 g of solvent can dissolve at a given temperature. When no more solute dissolves at that temperature the solution is saturated. The solubility of most solids in water increases as temperature rises; that of gases decreases (warm soda goes flat faster, and fish get less oxygen in warm water). A solubility curve plots solubility (g per 100 g water) against temperature, and a steep curve means a strong dependence on temperature.

Data (g per 100 g water): potassium nitrate 32 (20 °C), 62 (40 °C), 106 (60 °C), 167 (80 °C); sodium chloride 36, 36.5, 37, 37; potassium chloride 35, 40, 46, 54. Potassium nitrate changes a lot with temperature, but sodium chloride hardly changes, so common salt is obtained from sea water by evaporation, not by cooling.

Crystallisation. A crystal is a solid whose particles are arranged in a regular, repeating pattern. Crystallisation purifies a solid from a hot saturated solution on cooling: soluble impurities stay behind in the liquid. For each 100 g of water, the mass of crystals formed = solubility at the higher temperature − solubility at the lower temperature; for more water, multiply. Slow cooling gives large, regular crystals; rapid cooling gives small ones. Evaporating to dryness is worse because soluble impurities remain in the solid and some solids (such as sugar) char.

In a saturated solution. Per 100 g of water, the solution has mass 100 + S, where S is the solubility, so % m/m of a saturated solution = S ÷ (100 + S) × 100.

Concentration and Solubility of Solutions: formulas and key facts

Mass of solution
Mass of solute + mass of solvent
Mass by mass percentage
Mass by volume percentage
Volume by volume percentage
Crystals on cooling (per 100 g water)
Solubility at the higher temperature − solubility at the lower temperature
Saturated solution

How to solve Concentration and Solubility of Solutions questions

Use this method for concentration and solubility numericals.

  1. 1Identify the solute, the solvent and the solution, and write their masses or volumes with units.
  2. 2Choose the formula by the units: m/m (both masses), m/v (g and mL) or v/v (both volumes).
  3. 3Remember that the denominator is the solution (solute + solvent).
  4. 4For dilution, keep the solute mass fixed and change only the total; for mixing, add solute masses and solution masses.
  5. 5For solubility, scale the data to the actual mass of water, then subtract to find the crystals.
  6. 6Check by recalculating the percentage from your answer.

Concentration and Solubility of Solutions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Saline and glucose drips

Suppose a clinic prepares 300 mL of a 0.9% m/v saline solution and 150 mL of a 5% m/v glucose solution for demonstration, using the concentrations given for common drips.

Find the mass of salt and the mass of glucose needed.

Try it first, then show the solutionSolution
  1. Salt: 0.9% m/v means 0.9 g in 100 mL. For 300 mL: 0.9 × 3 = 2.7 g.
  2. Glucose: 5% m/v means 5 g in 100 mL. For 150 mL: 5 × 1.5 = 7.5 g.
  3. Check: 2.7 ÷ 300 × 100 = 0.9%, and 7.5 ÷ 150 × 100 = 5%.

Answer: 2.7 g of salt and 7.5 g of glucose.

Memory hook: m/v: grams in each 100 mL.

Crystals from a cooling solution

Suppose a student makes a saturated solution of potassium nitrate in 200 g of water at 60 °C and lets it cool to 20 °C overnight.

Find how much potassium nitrate crystallises, using the solubility values 106 g and 32 g per 100 g of water at 60 °C and 20 °C.

Try it first, then show the solutionSolution
  1. In 100 g of water: crystals = 106 − 32 = 74 g.
  2. For 200 g of water: 74 × 2 = 148 g of crystals.

Answer: 148 g of potassium nitrate.

Memory hook: Hot minus cold, times the water.

Tricks to remember Concentration and Solubility of Solutions

Short memory aids. Say each one aloud once, then use it on a question the same day.

  • The solution is the base

    The denominator is always the whole solution. If 20 g sugar is dissolved in 80 g water, divide by 100 g, not 80 g.

    Use it when: Avoiding the commonest percentage error.

  • m/m, m/v, v/v by units

    Look at the units given: grams and grams, m/m; grams and mL, m/v; mL and mL, v/v.

    Use it when: Choosing the formula.

  • Water changes, solute stays

    Dilution adds only water; evaporation removes only water; the solute mass does not change.

    Use it when: Dilution and evaporation questions.

  • Hot minus cold

    Crystals per 100 g water = solubility hot − solubility cold.

    Use it when: Solubility table questions.

Concentration and Solubility of Solutions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Three percentages · 3 marks · Short answerBoard level

Find the concentration of each solution.
(a) 20 g of sugar dissolved in 80 g of water, in % m/m. (1)
(b) 15 g of salt in 250 mL of solution, in % m/v. (1)
(c) 30 mL of alcohol in 150 mL of solution, in % v/v. (1)

Try it first, then show the solutionSolution
  1. (a) Mass of solution = 20 + 80 = 100 g. % m/m = 20 ÷ 100 × 100 = 20%. (1)
  2. (b) % m/v = 15 ÷ 250 × 100 = 6%. (1)
  3. (c) % v/v = 30 ÷ 150 × 100 = 20%. (1)

Answer: (a) 20% m/m; (b) 6% m/v; (c) 20% v/v.

Example 2: How much solute in the solution? · 3 marks · Short answerBoard level

(a) How much salt is in 400 mL of a 0.9% m/v saline solution? (1)
(b) How much glucose is in 200 mL of a 5% m/v glucose solution? (1)
(c) How much acetic acid is in 250 mL of vinegar that is 5% v/v acetic acid? (1)

Try it first, then show the solutionSolution
  1. (a) Salt = 0.9 ÷ 100 × 400 = 3.6 g. (1)
  2. (b) Glucose = 5 ÷ 100 × 200 = 10 g. (1)
  3. (c) Acetic acid = 5 ÷ 100 × 250 = 12.5 mL. (1)

Answer: (a) 3.6 g; (b) 10 g; (c) 12.5 mL.

Example 3: Solubility of potassium nitrate and sodium chloride · 5 marks · Long answerHarder board level

Solubilities in g per 100 g of water: potassium nitrate 106 at 60 °C and 32 at 20 °C; sodium chloride 37 at 60 °C and 36 at 20 °C.
(a) How much potassium nitrate is in a saturated solution made with 200 g of water at 60 °C? (1)
(b) How much crystallises when this is cooled to 20 °C? (1)
(c) Find the % m/m of the saturated potassium nitrate solution at 20 °C. (1)
(d) How much sodium chloride crystallises when a saturated solution made with 200 g of water is cooled from 60 °C to 20 °C? (1)
(e) Use (b) and (d) to say which salt is purified by cooling and which by evaporation. (1)

Try it first, then show the solutionSolution
  1. (a) 2 × 106 = 212 g. (1)
  2. (b) At 20 °C the 200 g of water holds 2 × 32 = 64 g, so crystals = 212 − 64 = 148 g. (1)
  3. (c) In 100 g of water: 32 g of solute in 132 g of solution, so % m/m = 32 ÷ 132 × 100 = 24.2%. (1)
  4. (d) Crystals = 2 × (37 − 36) = 2 g, a very small amount. (1)
  5. (e) Potassium nitrate gives a large yield on cooling (148 g), so it is purified by crystallisation on cooling. Sodium chloride gives only 2 g, so it is obtained by evaporating the water. (1)

Answer: (a) 212 g; (b) 148 g; (c) 24.2%; (d) 2 g; (e) KNO₃ by cooling, NaCl by evaporation.

Example 4: Preparing, diluting, concentrating and mixing · 5 marks · Long answerHarder board level

(a) How much salt and water are needed to make 500 g of an 8% m/m salt solution? (1)
(b) 150 g of a 20% m/m solution is mixed with 50 g of water. Find the new percentage. (1)
(c) How much water must be evaporated from 200 g of a 10% m/m solution to make it 16% m/m? (2)
(d) 100 g of a 10% m/m solution is mixed with 300 g of a 20% m/m solution. Find the percentage of the mixture. (1)

Try it first, then show the solutionSolution
  1. (a) Salt = 8 ÷ 100 × 500 = 40 g; water = 500 − 40 = 460 g. (1)
  2. (b) Solute = 20% of 150 = 30 g. New mass = 150 + 50 = 200 g. New % = 30 ÷ 200 × 100 = 15%. (1)
  3. (c) Solute = 10% of 200 = 20 g (it stays the same). For 16%, final mass = 20 ÷ 0.16 = 125 g. Water to evaporate = 200 − 125 = 75 g. Check: 20 ÷ 125 = 16%. (2)
  4. (d) Solute = 10 + 60 = 70 g; solution = 400 g; % = 70 ÷ 400 × 100 = 17.5%. (1)

Answer: (a) 40 g salt and 460 g water; (b) 15%; (c) 75 g; (d) 17.5%.

Example 5: Reading a solubility table for an unknown salt · 4 marks · Case-basedHOTS / competency

The solubility of salt Z in g per 100 g of water is 20 at 20 °C, 35 at 40 °C, 55 at 60 °C and 80 at 80 °C.
(a) Is Z more likely a solid or a gas dissolved in water? Give a reason. (1)
(b) Estimate its solubility at 50 °C, assuming a straight line between 40 °C and 60 °C. (1)
(c) A saturated solution of Z of mass 270 g at 40 °C is cooled to 20 °C. How much of Z crystallises? (1)
(d) What is the % m/m of the final saturated solution at 20 °C? (1)

Try it first, then show the solutionSolution
  1. (a) A solid, because its solubility increases as temperature rises (gases become less soluble when warmed). (1)
  2. (b) Midway between 35 and 55: (35 + 55) ÷ 2 = 45 g per 100 g water. (1)
  3. (c) At 40 °C, 135 g of saturated solution has 35 g of Z in 100 g of water, so 270 g (twice) has 70 g of Z in 200 g of water. At 20 °C, 200 g of water holds 2 × 20 = 40 g. Crystals = 70 − 40 = 30 g. (1)
  4. (d) Final solution = 200 g water + 40 g Z = 240 g, so % m/m = 40 ÷ 240 × 100 = 16.7%. (1)

Answer: (a) solid; (b) 45 g per 100 g water; (c) 30 g; (d) 16.7%.

Where students lose marks in Concentration and Solubility of Solutions

The slips examiners see most often, and what to write instead.

  • Mistake: Dividing by the mass of the solvent instead of the solution.

    Students use the water mass given in the problem.

    Wrong working

    20 g sugar in 80 g water: 20 ÷ 80 × 100 = 25%

    Right working

    Solution = 100 g, so 20 ÷ 100 × 100 = 20%

    Do this: Solution = solute + solvent. Always add first.

    Usually costs: 1 to 2 marks.

  • Mistake: Using the wrong type of percentage.

    The units are ignored.

    Wrong working

    Treating 15 g in 250 mL as % m/m.

    Right working

    Grams and mL means % m/v: 15 ÷ 250 × 100 = 6%.

    Do this: Match the formula to the units.

    Usually costs: 1 mark.

  • Mistake: Forgetting that solute stays the same in dilution.

    Students recalculate the solute from the new percentage.

    Wrong working

    Adding water to 150 g of 20% solution gives 20% of 200 g = 40 g.

    Right working

    Solute stays 30 g; 30 ÷ 200 × 100 = 15%.

    Do this: Find solute first and keep it fixed.

    Usually costs: 2 marks.

  • Mistake: Not scaling the solubility to the mass of water used.

    The table is per 100 g of water.

    Wrong working

    With 200 g water, crystals = 106 − 32 = 74 g.

    Right working

    74 × 2 = 148 g.

    Do this: Divide the mass of water by 100 g to get the multiple.

    Usually costs: 1 to 2 marks.

  • Mistake: Saying that gas solubility rises with temperature.

    It is assumed that all solubilities behave like those of solids.

    Wrong working

    Warm water holds more dissolved oxygen.

    Right working

    The solubility of gases in water decreases as temperature rises.

    Do this: Solids up, gases down.

    Usually costs: 1 mark.

Exam tips

  • Write the formula first, then substitute with units; the denominator is the whole solution.
  • Check each answer by working the percentage back from your masses.
  • Show the multiplier for solubility questions (for 250 g of water, multiply by 2.5).
  • Say 'at that temperature' when defining saturated solution and solubility.
  • Use exact phrases such as 'g per 100 g of water' for solubility.

Practice questions on Concentration and Solubility of Solutions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

  1. Q1 · 2 marks · Very short answerBoard level

    Define a saturated solution and explain why a warm bottle of soda goes flat faster than a cold one.

    Hint

    Think of the solubility of a gas in water as temperature rises.

    Answer

    A saturated solution is one in which no more solute dissolves at that temperature (1 mark). The solubility of a gas in water decreases as the temperature rises, so the dissolved carbon dioxide escapes faster from the warm soda (1 mark).

  2. Q2 · 3 marks · Short answerHarder board level

    12 g of glucose is dissolved in water to make 150 mL of solution. Find (a) the % m/v, (b) the mass of glucose in 250 mL of this solution, and (c) the volume of the solution that contains 6 g of glucose.

    Hint

    First find the grams in 100 mL.

    Answer

    (a) 12 ÷ 150 × 100 = 8% m/v (1 mark). (b) 8 ÷ 100 × 250 = 20 g (1 mark). (c) 6 ÷ 8 × 100 = 75 mL (1 mark).

  3. Q3 · 2 marks · Assertion-reasonBoard level

    Assertion (A): Common salt is obtained from sea water by evaporation and not by cooling. Reason (R): The solubility of sodium chloride in water hardly changes with temperature. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.

    Hint

    What would cooling give for a salt of nearly constant solubility?

    Answer

    (a). Cooling would give very few crystals because NaCl solubility changes very little (36 g at 20 °C to 37 g at 60 °C per 100 g water), so the water must be evaporated.

  4. Q4 · 3 marks · Short answerHarder board level

    Solubilities in g per 100 g of water: potassium chloride is 54 at 80 °C and 35 at 20 °C; potassium nitrate is 167 at 80 °C and 32 at 20 °C. Starting with saturated solutions in 250 g of water at 80 °C cooled to 20 °C, find the mass of crystals of each salt, and say which gives the better yield.

    Hint

    Scale per 100 g by 2.5.

    Answer

    KCl: (54 − 35) × 2.5 = 19 × 2.5 = 47.5 g (1 mark). KNO₃: (167 − 32) × 2.5 = 135 × 2.5 = 337.5 g (1 mark). Potassium nitrate gives the far better yield because its solubility changes much more with temperature (1 mark).

  5. Q5 · 4 marks · Case-basedHOTS / competency

    A craftsman melts 120 g of copper with 30 g of zinc to make brass. (a) Find the % m/m of each metal. (b) How many grams of each are needed for 1 kg of the same brass? (c) He adds 20 g more zinc to the melt. Find the new % of copper and zinc. (d) Why can brass not be separated by filtration or by a separating funnel?

    Hint

    Brass is an alloy, a homogeneous mixture of metals.

    Answer

    (a) Total = 150 g. Cu = 120 ÷ 150 × 100 = 80%; Zn = 30 ÷ 150 × 100 = 20% (1 mark). (b) 1 kg = 1000 g: Cu = 800 g, Zn = 200 g (1 mark). (c) Zn = 50 g in 170 g: Zn = 50 ÷ 170 × 100 = 29.4%, Cu = 120 ÷ 170 × 100 = 70.6% (1 mark). (d) An alloy is a homogeneous mixture; its metals are mixed throughout on a very small scale and cannot be separated by physical methods such as these (1 mark).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

  1. 1.What is the formula for % m/v?Show answer

    Mass of solute in g ÷ volume of solution in mL × 100.

  2. 2.What is a saturated solution?Show answer

    A solution in which no more solute dissolves at that temperature.

  3. 3.Which dissolved substance becomes less soluble as temperature rises?Show answer

    A gas, such as the carbon dioxide in soda.

  4. 4.What is the concentration of a normal saline drip?Show answer

    0.9% m/v.

  5. 5.Which salt's solubility changes hardly at all with temperature?Show answer

    Sodium chloride (common salt).

  6. 6.What does a steep solubility curve show?Show answer

    That solubility changes a lot with temperature.

Concentration and Solubility of Solutions: frequently asked questions

What is the difference between solubility and concentration?

Solubility is the maximum mass of solute that dissolves in 100 g of solvent at a given temperature. Concentration is the amount of solute in any given solution, which may be less than the maximum.

Why does the mass of the solution include the solute?

The solution is the solute and the solvent together, so its mass is the sum of both.

Why filter a hot saturated solution before cooling to get crystals?

To remove insoluble impurities before the crystals form; otherwise they would be trapped in the crystals.

Why do slowly cooled crystals look better?

Slow cooling gives particles time to arrange in a regular pattern, forming larger, shinier crystals.

Can the percentage concentration exceed 100?

No. The solute is a part of the solution, so the percentage is always less than 100.