This MCQ module is based on: Solutions, Concentration and Solubility
TOPIC 16 OF 50
Solutions, Concentration and Solubility
🎓 Class 9
Science
CBSE
Theory
Ch 5 — Exploring Mixtures and their Separation
⏱ ~20 min
🌐 Language: [gtranslate]
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This assessment will be based on: Solutions, Concentration and Solubility
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Introduction: Why Concentration Matters
When your doctor prescribes a saline drip, the nurse takes great care to use exactly 0.9% sodium chloride solution. Anything stronger or weaker can damage red blood cells. When a farmer sprays pesticide on a crop, the label specifies a precise concentration — too dilute and the pests survive, too strong and the plant dies. From cooking sugar syrup of the right thickness to mixing the correct strength of bleach for cleaning, the idea of how much solute is in a solution rules our daily life. This part takes the idea of a solution introduced in Part 1 and gives it numbers.
5.7 Solute and Solvent — Naming the Players
In every solution there are two roles. The substance that gets dissolved is the solute; the substance in which it dissolves is the solvent. Usually the larger amount is the solvent. When 5 g of salt are dissolved in 95 g of water, water is the solvent.
| Solution | Solute | Solvent | State |
|---|---|---|---|
| Lemonade | Sugar, salt, citric acid | Water | Liquid in liquid |
| Tincture of iodine | Iodine (solid) | Alcohol (liquid) | Solid in liquid |
| Soda water | Carbon dioxide (gas) | Water (liquid) | Gas in liquid |
| Air | Oxygen, argon, CO₂ (gases) | Nitrogen (gas) | Gas in gas |
| Brass | Zinc (~30%) | Copper (~70%) | Solid in solid (alloy) |
| 22-carat gold | Copper / silver | Gold | Solid in solid (alloy) |
5.8 Concentration of a Solution
The concentration of a solution tells us how much solute is dissolved in a given amount of solution. There are two simple ways to express it at the Class 9 level.
(a) Mass by Mass Percent — Mass %
This is used when both solute and solvent are weighed.
\[\text{Mass \%} = \frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\]
Note: Mass of solution = Mass of solute + Mass of solvent.
(b) Mass by Volume Percent — Mass/Volume %
This is used when the solute is weighed but the solution is measured by volume (very common in laboratories).
\[\text{Mass/Volume \%} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}}\times 100\]
Worked Example 1 — Simple Mass Percent
A solution contains 40 g of common salt in 320 g of water. Calculate the mass percent of salt in the solution.
Step 1: Mass of solute = 40 g; Mass of solvent = 320 g.
Step 2: Mass of solution = 40 + 320 = 360 g.
Step 3: \(\text{Mass \%} = \dfrac{40}{360}\times 100 = 11.11\%\) (approx).
Worked Example 2 — Find the Solute Mass
How much sugar must be dissolved in 250 g of water to make a 5% solution by mass?
Step 1: Let mass of sugar = \(x\) g. Mass of solution = \((x+250)\) g.
Step 2: \(\dfrac{x}{x+250}\times 100 = 5\) gives \(100x = 5x + 1250\), so \(95x = 1250\).
Step 3: \(x = \dfrac{1250}{95} = 13.16\) g (approx). About 13.16 g of sugar in 250 g of water.
Worked Example 3 — Mass/Volume Percent
15 g of urea is dissolved in water to make 250 mL of solution. Find its mass/volume percent.
\(\dfrac{15}{250}\times 100 = 6\%\) (mass/volume).
Worked Example 4 — Salt and Water Mixture
A solution contains 25 g of salt in 175 g of water. (i) What is the mass percent? (ii) If you wanted to double the concentration, how much extra salt should be added?
(i) Mass of solution = 200 g; Mass % = \(\dfrac{25}{200}\times 100 = 12.5\%\).
(ii) Target = 25%. Let extra salt = \(x\) g. New mass of solute = \(25 + x\); new mass of solution = \(200 + x\).
\(\dfrac{25+x}{200+x}\times 100 = 25\) gives \(100(25+x) = 25(200+x)\), so \(2500 + 100x = 5000 + 25x\), \(75x = 2500\), \(x = 33.33\) g.
Worked Example 5 — Working Backwards
The label on a 500 mL bottle of glucose solution reads ‘5% (m/v)’. How many grams of glucose are present in the bottle?
\(\text{Mass of glucose} = \dfrac{5}{100}\times 500 = 25\) g.
Worked Example 6 — Mixing Two Solutions
200 g of a 10% (m/m) sugar solution is mixed with 300 g of a 20% (m/m) sugar solution. Find the mass percent of the final solution.
Sugar from first: \(\dfrac{10}{100}\times 200 = 20\) g.
Sugar from second: \(\dfrac{20}{100}\times 300 = 60\) g.
Total sugar: 80 g; Total mass of solution: 200 + 300 = 500 g.
Final mass %: \(\dfrac{80}{500}\times 100 = 16\%\).
Worked Example 7 — NCERT-Style
A solution contains 50 g of sugar in 350 g of water. Calculate the concentration in mass percent.
Mass of solution = 50 + 350 = 400 g. Mass % = \(\dfrac{50}{400}\times 100 = 12.5\%\).
5.9 Saturated, Unsaturated and Supersaturated Solutions
If you keep adding sugar to a glass of water at room temperature, at first every spoon dissolves. Then a point comes when the next spoon just sits at the bottom — the water has taken in all it can. The solution has become saturated.
- Unsaturated solution — can still dissolve more solute at the given temperature.
- Saturated solution — cannot dissolve any more at that temperature; extra solute settles down.
- Supersaturated solution — contains more solute than a saturated one at the same temperature; usually made by heating, dissolving extra solute and cooling carefully. Such solutions are unstable; a tiny seed crystal triggers crystallisation.
5.10 Solubility
The solubility of a solute in a solvent is defined as the amount of the solute (in grams) that dissolves in 100 g of the solvent to form a saturated solution at a particular temperature.
\[\text{Solubility} = \frac{\text{Mass of solute (g) in saturated solution}}{\text{Mass of solvent (g)}}\times 100\]
Worked Example 8 — Calculating Solubility
At 30°C, 36.4 g of potassium chloride (KCl) dissolves in 100 g of water to form a saturated solution. What is the solubility of KCl at this temperature? If a student dissolves 18.2 g of KCl in 50 g of water at 30°C, will the solution be saturated?
Solubility = 36.4 g per 100 g of water at 30°C.
Check the second case: Solubility means 36.4 g per 100 g of water, so per 50 g it would be \(\dfrac{36.4}{2} = 18.2\) g.
The student has dissolved exactly 18.2 g, so the solution is just saturated — one more crystal will not dissolve.
5.11 Factors that Affect Solubility
(i) Nature of solute and solvent
‘Like dissolves like.’ Polar solutes (salt, sugar) dissolve well in polar solvents (water). Non-polar solutes (oils, wax) dissolve well in non-polar solvents (kerosene, petrol).
(ii) Temperature
For most solid solutes in liquid solvents, solubility increases with rising temperature. That is why hot tea dissolves sugar much faster than cold tea. For most gaseous solutes, solubility decreases with rising temperature — warm soda goes flat quickly because the gas escapes.
📈 Solubility Curve — Click each curve to compare L4 Analyse
Two solutes — one solid (KNO₃) and one gas (CO₂) — behave in opposite ways when the water is heated. Click each coloured curve to see why the curves point in different directions.
Click either coloured curve to compare how solid and gas solubility respond to a rise in temperature — and see real examples.
(iii) Pressure (mainly for gases)
The solubility of a gas in a liquid increases as pressure increases. This is summed up by Henry’s Law: at constant temperature, the mass of a gas dissolved in a fixed amount of liquid is directly proportional to the pressure of that gas above the liquid. This is why a sealed soda bottle is full of dissolved CO₂, and why opening it (lowering the pressure) causes a fizz of gas to bubble out.
To remember: Solid in liquid — warm it up. Gas in liquid — cool it down and squeeze the pressure up.
Activity 5.2 — Hot Water Holds More SugarL3 Apply
Predict first: If you keep adding sugar to a fixed amount of cold water and to the same amount of hot water, in which case do you think the solution will become saturated faster?
- Take two beakers, each with 50 mL of water. Heat one to about 60°C; keep the other at room temperature.
- Add sugar one teaspoon at a time and stir until it dissolves.
- Stop when no more sugar will dissolve. Count the spoons added in each beaker.
The hot water dissolves many more spoons of sugar than the cold water before reaching saturation. This shows that solubility of a solid solute increases with temperature.
Competency-Based Questions L4 Analyse
CBQ — Drug Dosage in a Hospital
A hospital pharmacist is preparing 0.9% (m/v) sodium chloride (saline) for a patient. The doctor has ordered a 500 mL drip. The pharmacist has only solid salt and distilled water. Saline must be made carefully because human blood cells are sensitive to the concentration of dissolved salts.
1. (Numerical) Calculate the mass of NaCl required to prepare 500 mL of 0.9% (m/v) saline.
Mass of NaCl = \(\dfrac{0.9}{100}\times 500 = 4.5\) g.
2. (MCQ) The solute in this saline is:
(b) Sodium chloride — it is the substance present in smaller amount and is dissolved in water (the solvent).
3. (Short answer) The pharmacist accidentally adds 9 g of NaCl instead of 4.5 g to the 500 mL of water. What is the new mass/volume percent and why is this dangerous to the patient?
New % = \(\dfrac{9}{500}\times 100 = 1.8\%\). This is twice the body’s normal salt level. Such a hypertonic solution would draw water out of the red blood cells, shrinking and damaging them.
4. (True/False with reason) ‘Heating the saline above 60°C is a good way to dissolve more NaCl into it.’
False. The solubility of NaCl barely changes with temperature (about 36 g per 100 g of water from 0°C to 100°C). Heating offers almost no advantage and may degrade the solution and stress sterility.
5. (Long answer) Define solubility. State two factors that change solubility and one factor that does not have a major effect for solid solutes.
Solubility is the maximum mass of a solute (in g) that dissolves in 100 g of solvent at a given temperature. Two factors that change it: (i) nature of solute and solvent, (ii) temperature. Factor with little effect on solid solutes: pressure — pressure changes affect gas solubility strongly but barely change the solubility of solids in liquids.
Assertion & Reason Questions L5 Evaluate
1. Assertion (A): A bottle of soda fizzes loudly when its cap is opened.
Reason (R): The solubility of a gas in a liquid decreases as the pressure of the gas above the liquid drops.
Answer: A. Both statements are true; the drop in pressure described in R is exactly what causes the dissolved CO₂ to escape and produce the fizz in A. This is Henry’s Law in action.
2. Assertion (A): The solubility of potassium nitrate in water rises sharply as the water is heated.
Reason (R): Solubility of all solutes in water always increases with temperature.
Answer: C. A is true (KNO₃ solubility climbs steeply with temperature). R is false — solubility of gases in water actually decreases with temperature, and even some salts (like sodium chloride) hardly change with it.
3. Assertion (A): A supersaturated solution is more stable than a saturated one.
Reason (R): A supersaturated solution contains more solute than the maximum that can dissolve at that temperature.
Answer: D. A is false — supersaturated solutions are less stable; the slightest disturbance triggers crystal formation. R is the correct definition.
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