New NCERT Solutions for Chapter 5 Exploring Mixtures and their Separation Class 9 Science
- Chapter Name: Exploring Mixtures and their Separation
- Textbook Name: Class 9 Science
Chapter 5 Exploring Mixtures and their Separation NCERT Solutions Class 9 Science
Think it Over
1. Why do suspended particles settle in muddy water over time but not in milk?
Answer
Suspended particles in muddy water are relatively large and heavy. Due to gravity, these particles settle down at the bottom when the mixture is left undisturbed. In milk, the particles are extremely small colloidal particles that remain uniformly dispersed in the liquid and do not settle under normal conditions.
2. How is evaporation different from boiling?
Answer
Evaporation is a slow process that takes place only at the surface of a liquid and can occur at any temperature. Boiling is a rapid process that occurs throughout the liquid when it reaches its boiling point, resulting in the formation of bubbles.
3. Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?
Answer
Bright rays of sunlight are visible because tiny dust particles and water droplets present in the air scatter the sunlight. This scattering makes the path of the light beam visible, a phenomenon known as the Tyndall effect.
Pause and Ponder
1. A common talcum powder contains 4 % m/m zinc oxide, which acts as an antiseptic. How much zinc oxide is present in 300 g of the talcum powder?
Answer
Given: Percentage of zinc oxide = 4% m/m
Mass of talcum powder = 300 g
Calculation:
\(\text{Mass of zinc oxide}=\frac{4}{100}\times300\,\text{g}\)
= 12 g
12 g of zinc oxide is present in 300 g of the talcum powder.
2. Your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. She asks you to mix two tablespoons of the concentrate with water in a glass tumbler. If each tablespoon measures 15 mL and you make 150 mL of juice per person, what is the % v/v of orange juice concentrate in the mixture you prepared?
Answer
Given:
Volume of orange juice concentrate = 2 × 15 mL = 30 mL
Volume of solution = 150 mL
Formula:
\(\%\;v/v=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times100\)
Calculation:
\(\%\;v/v=\frac{30}{150}\times100=20\%\)
The concentration of orange juice concentrate in the mixture is 20% v/v.
3. Vinegar, used as a food preservative and additive, contains 5 % v/v acetic acid. Glacial acetic acid is a liquid, i.e., 100% acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed?
Answer
To prepare vinegar containing 5% v/v acetic acid, take 5 mL of glacial acetic acid and add sufficient water to make the total volume 100 mL. Mix the solution thoroughly.
Mix 5 mL of glacial acetic acid with water and make the total volume up to 100 mL to obtain vinegar containing 5% v/v acetic acid.
4. Refer to the solubility curves given in Activity 5.2. If equal masses of hot, saturated solutions of compounds ‘A’ and ‘B’ are cooled from 80 °C to 60 °C, which solution is likely to deposit more solid?
Answer
Compound ‘B’ is likely to deposit more solid on cooling from 80 °C to 60 °C.
This is because the solubility curve of compound ‘B’ shows a much larger decrease in solubility between 80 °C and 60 °C than that of compound ‘A’. As the temperature decreases, a greater amount of dissolved compound ‘B’ comes out of the solution in the form of crystals.
Answer: The saturated solution of compound ‘B’ will deposit more solid.
5. Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.
Answer
Yes, the size of common salt crystals depends on the rate of evaporation.
If evaporation occurs slowly, salt particles get sufficient time to arrange themselves in a regular pattern, forming larger and well-shaped crystals.
If evaporation occurs rapidly, crystals form quickly and do not get enough time to grow, resulting in smaller crystals.
Answer: Slow evaporation produces larger crystals, whereas rapid evaporation produces smaller crystals.
6. State whether the following statements are True or False. Also, correct the False statements.
(i) Salt can be separated from a salt solution by evaporation or distillation.
(ii) Distillation can be used for separation of two liquids even when these have the same boiling point.
(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.
(iv) Evaporation and crystallization are the same processes.
Answer
(i) True.
Salt can be obtained from a salt solution by evaporation. Distillation can also be used, where water is collected as the distillate and salt remains behind.
(ii) False.
Correct Statement: Distillation can be used to separate two liquids only when they have different boiling points.
(iii) False.
Correct Statement: In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment.
(iv) False.
Correct Statement: Evaporation is the process of converting a liquid into vapour, whereas crystallization is the process of obtaining pure solid crystals from a saturated solution.
7. Why do immiscible liquids form two separate layers in a separating funnel?
Answer
Immiscible liquids do not mix with each other because their molecules do not attract one another sufficiently. Due to the difference in their densities, they form separate layers when left undisturbed. The denser liquid settles at the bottom, while the less dense liquid remains on the top.
Immiscible liquids form two separate layers in a separating funnel because they do not mix and have different densities.
8. Is sublimation different from evaporation? Justify.
Answer
Yes, sublimation is different from evaporation.
In sublimation, a solid changes directly into vapour without passing through the liquid state. Examples include camphor, naphthalene and dry ice. In evaporation, a liquid changes into vapour from its surface at temperatures below its boiling point.
Sublimation involves the direct conversion of a solid into vapour, whereas evaporation involves the conversion of a liquid into vapour. Therefore, the two processes are different.
9. Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions and colloids, what type of mixture do you think clouds are and why?
Answer
Clouds are colloids. They consist of tiny water droplets or ice crystals dispersed in air. These particles are very small and remain suspended in the air for a long time without settling down. They are intermediate in size between the particles of a solution and a suspension.
Clouds are colloidal mixtures because tiny water droplets or ice crystals remain uniformly dispersed in air without settling.
10. Why do cities with a lot of smoke and dust in the air often look hazy?
Answer
Cities with a lot of smoke and dust appear hazy because these fine particles remain suspended in the air and scatter sunlight in different directions. Due to this scattering of light, visibility decreases and the atmosphere looks cloudy or hazy.
Revise, Reflect, Refine
1. Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option.
(i) Air — Hm, Milk — Ht, Sugar solution — Hm, Smoke — Hm
(ii) Brass — Ht, Fog — Ht, Vinegar — Ht, Muddy water — Hm
(iii) Copper sulfate solution — Hm, Salt solution — Hm, Milk — Hm, Bronze — Hm
(iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm
Answer
(ii) Brass — Ht, Fog — Ht, Vinegar — Ht, Muddy water — Hm
Brass is a homogeneous mixture (alloy), fog is a heterogeneous mixture (colloid), vinegar is a homogeneous mixture (solution), and muddy water is a heterogeneous mixture (suspension).
2. Choose the correct options, and explain the reason for the correct and incorrect options.
Which among the following mixtures show the Tyndall Effect?
A mixture of:
(a) air and dust particles
(b) copper sulfate and water
(c) starch and water
(d) acetone and water
(i) a and b
(ii) b and d
(iii) a and c
(iv) c and d
Answer
(iii) a and c
Air and dust particles form a colloidal mixture, and starch in water also forms a colloid. Colloidal particles scatter light and show the Tyndall effect.
Copper sulfate in water and acetone in water form true solutions, which do not show the Tyndall effect.
3. A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in Table 5.2. Words and phrases may be used more than once.
Answer
| Solution | Suspension | Colloid |
|---|---|---|
| Properties | Properties | Properties |
| Small-sized particles (less than 1 nm diameter) | Large-sized particles (more than 1000 nm diameter) | Moderate-sized particles (1–1000 nm) |
| Homogeneous mixture | Heterogeneous mixture | Heterogeneous mixture |
| Transparent | Settles down when left undisturbed | Does not settle down |
| Cannot be separated by filtration | Separates by filtration | Scatters light |
| Examples | Examples | Examples |
| Salt solution | Mud | Milk |
| Brass | Sand in water | Smoke |
| Butter |
4. Solve the following problems:
(i) A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogen carbonate. Express the concentration of each component in the mixture using an appropriate method.
(ii) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.
Answer
(i) Sugar = 75 g
All-purpose flour = 420 g
Sodium hydrogen carbonate = 5 g
Total mass of mixture = 75 + 420 + 5 = 500 g
Mass by mass percentage of sugar
= (75/500) × 100
= 15%
Mass by mass percentage of flour
\(=\frac{75}{500}\times100\)
\(=15\%\)
Mass by mass percentage of sodium hydrogen carbonate
\(=\frac{5}{500}\times100\)
\(=1\%\)
Sugar = 15% m/m
All-purpose flour = 84% m/m
Sodium hydrogen carbonate = 1% m/m
(ii) Mass of brass = 120 g
Copper = 70%
Mass of copper
\(=\frac{70}{100}\times120\)
= 84 g
Mass of zinc
= 120 − 84
= 36 g
Copper present = 84 g
Zinc present = 36 g
5. The label on a cooking oil pack says one litre (910 g). If this oil is mixed with water, will it form a separate layer? If so, which substance will be on top? How will you separate the two layers? Also, draw the diagram of the apparatus used.
Answer
Yes, cooking oil and water are immiscible liquids and will form separate layers.
One litre of oil has a mass of 910 g, so its density is 0.91 g/mL, which is less than the density of water (1 g/mL).
Therefore, oil will form the upper layer and water will form the lower layer.
The two liquids can be separated using a separating funnel. The lower layer of water is allowed to flow out first through the stopcock, and the oil is collected separately.
6. Assertion (A): Solutions do not exhibit the Tyndall effect.
Reason (R): The particles in solutions are larger than 100 nm, so they cannot scatter light.
Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer
(iii) A is true, but R is false.
Solutions do not show the Tyndall effect because their particles are extremely small (less than 1 nm) and cannot scatter light. The reason is false because it incorrectly states that the particles are larger than 100 nm.
7. How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.
Answer
| Mixture | Method of separation | Reason for selection |
|---|---|---|
| Mud from muddy water | Filtration | Mud particles are insoluble and larger in size, so they can be retained by a filter. |
| Plasma from other components in the blood sample | Centrifugation | Blood components have different densities and can be separated by rapid spinning. |
| Naphthalene and sand | Sublimation | Naphthalene sublimes on heating whereas sand does not. |
| Chalk powder and common salt | Dissolve in water, filtration and evaporation/crystallization | Salt dissolves in water while chalk powder does not. |
| Common salt and water | Evaporation or distillation | Water can be evaporated or distilled, leaving salt behind. |
| Oil from water | Separating funnel | Oil and water are immiscible liquids and form separate layers. |
| Pigments of the flower | Paper chromatography | Different pigments move at different rates in a solvent and get separated. |
8. Two miscible liquids, A and B, are present in a mixture. The boiling point of A is 60 °C and the boiling point of B is 90 °C. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.
Answer
The difference between the boiling points of liquids A and B is 30 °C (90 °C − 60 °C).
Since the difference in boiling points is greater than 25 °C, the mixture can be separated by distillation.
On heating the mixture, liquid A (boiling point 60 °C) vaporises first. The vapours are cooled in a condenser and collected separately. Liquid B remains in the distillation flask.
The mixture should be separated by distillation.
9. Compare evaporation, crystallization and distillation. In which situation would you prefer each of these over the others?
Answer
| Process | Description | Preferred Situation |
|---|---|---|
| Evaporation | The solvent is allowed to evaporate, leaving the solute behind. | Used when only the solute is required, such as obtaining salt from seawater. |
| Crystallization | Pure crystals are obtained from a saturated solution by cooling or evaporation. | Used for obtaining pure solids and removing impurities from them. |
| Distillation | A liquid is vaporised and then condensed to recover it. | Used when the solvent is also required or when separating miscible liquids with different boiling points. |
I would prefer evaporation when only the solute is needed, crystallization when pure crystals are required, and distillation when the solvent must also be recovered or when separating miscible liquids.
10. Blood is an example of a colloidal mixture.
(i) What would happen if blood behaved like a true suspension inside the body?
(ii) In a blood sample, identify the dispersed phase and the dispersion medium.
Answer
(i) If blood behaved like a true suspension, its particles would settle down when the blood remained stationary. This would prevent the uniform transport of oxygen, nutrients and other substances throughout the body and could be harmful.
(ii) In a blood sample, identify the dispersed phase and the dispersion medium.
Dispersed phase: Blood cells (RBCs, WBCs and platelets)
Dispersion medium: Plasma
11. You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.
Answer
The mixture contains sand, common salt and naphthalene.
Step 1: Sublimation (Apparatus 1)
Naphthalene sublimes on heating and gets separated from the mixture.
Step 2: Dissolve the remaining mixture in water and perform filtration (Apparatus 3)
Common salt dissolves in water while sand remains insoluble. Sand is separated by filtration.
Step 3: Evaporation/Crystallization (Apparatus 2)
The salt solution is heated to obtain common salt.
Correct sequence:
Sublimation → Filtration → Evaporation (or Crystallization)
12. Why is distillation an effective method for separating a mixture of water and acetone?
Answer
Distillation is effective because water and acetone are miscible liquids with significantly different boiling points.
Acetone boils at about 56°C, whereas water boils at 100°C. On heating the mixture, acetone vaporises first. The vapours are cooled and condensed to obtain pure acetone, while water remains behind.
13. Answer the following questions with the help of the data given in Table 5.4.
(i) What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C?
(ii) A student makes a saturated solution of potassium chloride in water at 80 °C and leaves the solution to cool at room temperature (25 °C). What would she observe as the solution cools? Explain.
(iii) What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts by increasing temperature from 10 °C to 80 °C.
Answer
(i) From Table 5.4, the solubility of potassium nitrate at 40°C is 62 g per 100 g of water.
Calculation:
Mass of potassium nitrate required in 50 g water
\(=\frac{62\times50}{100}\)
= 31 g
31 g of potassium nitrate is required.
(ii) At 80°C, potassium chloride has a solubility of 54 g per 100 g of water.
At about 25°C, its solubility is much lower (between 35 g and 37.4 g per 100 g of water).
As the solution cools, excess potassium chloride can no longer remain dissolved and starts separating out as crystals.
(iii) Generally, the solubility of salts increases with an increase in temperature.
| Salt | Solubility at 10°C (g/100 g water) | Solubility at 80°C (g/100 g water) | Increase |
|---|---|---|---|
| Potassium nitrate | 21 | 167 | 146 |
| Sodium chloride | 36 | 37 | 1 |
| Potassium chloride | 35 | 54 | 19 |
| Ammonium chloride | 24 | 66 | 42 |
Comparison:
Potassium nitrate shows the maximum increase in solubility.
Ammonium chloride shows a moderate increase.
Potassium chloride shows a smaller increase.
Sodium chloride shows very little change in solubility.
The solubility of all four salts increases with temperature, but the increase is greatest for potassium nitrate and least for sodium chloride.
14. Three students, A, B and C, are preparing sugar solutions for an experiment.
- Student A dissolves 20 g of sugar in 80 g of water.
- Student B dissolves 20 g of sugar in 100 g of water.
- Student C dissolves 30 g of sugar in 80 g of water.
(i) Calculate the mass percentage (% m/m) concentration of sugar in each student’s solution.
(ii) Whose solution is the most concentrated? Explain why.
Answer
(i) For Student A
Mass of sugar = 20 g
Mass of water = 80 g
Mass of solution = 20 + 80 = 100 g
\(\%\;m/m=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times100\)
\(\%\;m/m=\frac{20}{100}\times100\)
= 20%
Mass percentage of Student A's solution = 20% m/m
For Student B
Mass of sugar = 20 g
Mass of water = 100 g
Mass of solution = 20 + 100 = 120 g
\(\%\;m/m=\frac{20}{120}\times100\)
= 16.67%
Mass percentage of Student B's solution = 16.67% m/m
For Student C
Mass of sugar = 30 g
Mass of water = 80 g
Mass of solution = 30 + 80 = 110 g
\(\%\;m/m=\frac{30}{110}\times100\)
= 27.27%
Mass percentage of Student C's solution = 27.27% m/m
(ii) The concentration of the three solutions is:
Student A = 20% m/m
Student B = 16.67% m/m
Student C = 27.27% m/m
Since Student C's solution has the highest mass percentage of sugar, it contains the greatest amount of sugar per 100 g of solution.
Student C's solution is the most concentrated because it has the highest sugar concentration, 27.27% m/m.
15. Examine Fig. 5.26.
(i) Identify the separation technique marked as ‘S’.
(ii) Label the apparatus A, B and C.
(iii) Which of the following mixtures can be separated by the technique identified above? Use the data given in Table 5.5. Mixtures:
(a) water — acetone
(b) water — salt
(c) acetone — alcohol
(d) sand — salt
(e) alcohol — chloroform
(f) alcohol — benzene
Answer
(i) is used for distillation. The separation technique marked as ‘S’ is Distillation.
(ii) A – Distillation flask
B – Water condenser (Liebig condenser)
C – Conical flask (Receiver)
(iii) Distillation is used to separate:
- A liquid from a dissolved solid.
- Two miscible liquids having a difference in boiling points of about 25°C or more.
(a) Water — Acetone
Boiling point difference = 100°C − 56°C = 44°C
Can be separated by distillation.
(b) Water — Salt
Salt is a non-volatile solid dissolved in water.
Can be separated by distillation.
(c) Acetone — Alcohol
Boiling point difference = 78°C − 56°C = 22°C
Cannot be separated by simple distillation.
(d) Sand — Salt
Both are solids.
Cannot be separated by distillation.
(e) Alcohol — Chloroform
Boiling point difference = 78°C − 61°C = 17°C
Cannot be separated by simple distillation.
(f) Alcohol — Benzene
Boiling point difference = 80°C − 78°C = 2°C
Cannot be separated by simple distillation.
The mixtures that can be separated by the technique identified above are:
(a) Water — Acetone
(b) Water — Salt
