Explore Exploring Mixtures and Their Separation Class 9 NCERT Solutions with detailed and easy-to-understand answers based on the latest CBSE syllabus. This chapter explains the concepts of mixtures, types of mixtures, solutions, suspensions, colloids, and various separation techniques such as filtration, evaporation, distillation, and chromatography.
Exploring Mixtures and their Separation Class 9 NCERT Solutions
1. Why do suspended particles settle in muddy water over time but not in milk?
Answer: In muddy water particles settle because they are large, but in milk particles are very small and remain mixed, and they do not settle down.
2. How is evaporation different from boiling?
Answer: Evaporation happens at the surface of the liquid at any temperature. In evaporation no bubbles are formed, and it is a slow process. Boiling happens in the whole liquid, not only on the surface. It happens only at a fixed temperature, which is known as the boiling point. In boiling, bubbles are formed, and it is a fast process.
3. Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?
Answer: Sunlight is made of tiny particles called photons. When the sunlight passes through small gaps between leaves, then the gaps act like pinholes. These pinholes allow only straight beams of light to pass. These beams look like bright rays because of dust and water particles in the air.
4. 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: The talcum powder contains 4% m/m zinc oxide, and there is a total of 300 g of talcum powder.
Formula,
\[ \text{Mass of zinc oxide} = \frac{\text{Percentage}}{100} \times \text{Total mass} \]\[ \text{Mass of zinc oxide} = \frac{4}{100} \times 300 \]= 12 g
The talcum powder contains 12 g of zinc oxide.
5. 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
- Concentrate used = 2 tablespoons
- Each tablespoon = 15 mL
- So, concentrate volume = 2 × 15 = 30 mL
- Total juice prepared = 150 mL
Formula,
\[ \%\text{ v/v} = \frac{\text{Volume of solute (concentrate)}}{\text{Total volume of solution}} \times 100 \]\[ \%\text{ v/v} = \frac{30}{150} \times 100 \]= 20%
The orange juice mixture contains 20% v/v concentrate.
6. 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:
Given:
- Vinegar contains 5% v/v acetic acid.
- Glacial acetic acid = 100% acetic acid.
To prepare 100 ml of Vinegar
- Step 1: Use 5 ml of glacial acetic acid.
- Step 2: Add 95 ml of water (now the total valuem will be 100 ml)
- Step 3: Now, in the solution 5% v/v acetic acid, which is vinegar.
7. 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? Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.
Answer: Compound A deposits more solid than B when cooled from 80 °C to 60 °C. Slow evaporation gives large salt crystals. Fast evaporation gives small salt crystals.
8. 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.
Answer: True
Explanation:
Salt is left behind after evaporation or distallation.
(ii) Distillation can be used for separation of two liquids even when these have the same boiling point.
Answer: False
Explanation:
Distillation works only if liquids have different boiling points.
(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.
Answer: False
Explanation:
The solvent level must be below the sample spot other wise the sample will dissolve directly into the solvent.
(iv) Evaporation and crystallization are the same processes.
Answer: False
Explanation:
Evaporation means liquid change to vapour and Crystallization means pure solid crystals form from solution.
9. Two immiscible liquids of the same density are mixed in a separating funnel, how will the layers form?
Answer: Immiscible liquids do not mix like oil and water. Normally heavier liquid goes down, and lighter liquid stays up. Here both liquids have the same density. So, there will be no upper or lower layer form.
10. Why do immiscible liquids form two separate layers in a separating funnel?
Answer: Immiscible liquids do not mix, like oil and water. In a separating funnel the liquids arrange themselves by density. The denser liquid goes to the bottom layer, and the lighter liquid stays on the top layer. That’s why you see two separate layers.
11. Is sublimation different from evaporation? Justify.
Answer: The liquid changes into vapor, and it is done at the surface of the liquid. For example, water drying from clothes. Sublimation is a method where a solid changes directly into vapor without becoming liquid. For example, camphor or naphthalene balls.
12. 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: The clouds contain tiny water or ice crystals. These particles are very small and do not settle down quickly like in a suspension. They scatter light; because of this, clouds look white or gray. This behavior is the same as a colloid.
13. Why do cities with a lot of smoke and dust in the air often look hazy?
Answer: The smoke and dust particles are very tiny and float in the air. When the sunlight passes through air, then these particles scatter the light; because of this scattering, light cannot go straight, and air looks unclear or hazy. This scattering of light by small particles is called the Tyndall effect.
14. 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: (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm
Explanation:
Homogeneous mixtures (Hm) look the same everywhere; they are uniform. Heterogeneous mixture (HT): mixture not uniform; you can see different parts.
- Muddy water: heterogeneous.
- Milk: heterogeneous (colloid).
- Blood: heterogeneous (cells + plasma).
- Brass: homogeneous (alloy, uniform mixture).
15. 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.
Explanation:
- (a) Air and dust particles show the Tyndall effect because dust particles scatter light.
- (b) Copper sulfate and water do not show the Tyndall effect because they are true solutions and their particles are too small to scatter light.
- (c) Starch and water show the Tyndall effect due to colloids and particles scattering light.
- (d) Acetone and water do not show the Tyndall effect because they are true solutions and have no scattering.
16. 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 the Table 5.2. Words and phrases may be used more than once.
Answer:
| Mixture Type | Properties | Examples |
|---|---|---|
| Solution | The particles are very tiny (less than 1 nm). They are not visible to the eye. They do not settle down on standing. They do not scatter light (no Tyndall effect). | Salt dissolved in water, sugar dissolved in water |
| Suspension | The particles are large (more than 100 nm). They can be seen with the naked eye. They settle down if left undisturbed. They scatter light strongly (show Tyndall effect). | Mud mixed in water, sand in water |
| Colloid | The particles are small but not as tiny as in a solution (1–100 nm). They are not visible to the eye. They do not settle down on standing. They scatter light (show Tyndall effect). | Milk, smoke, fog, blood |
17. 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 hydrogencarbonate. 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) Cake mixture concentrations
Given
- Sugar = 75 g
- Flour = 420 g
- Sodium hydrogencarbonate = 5 g
- Total = 75 + 420 + 5 = 500 g
Formula,
\[ \%\ \text{by mass} = \frac{\text{Mass of component}}{\text{Total mass}} \times 100 \]Calculations
\[ \text{Sugar: } \frac{75}{500} \times 100 = 15\% \]\[ \text{Flour: } \frac{420}{500} \times 100 = 84\% \]\[ \text{Sodium hydrogencarbonate: } \frac{5}{500} \times 100 = 1\% \]- Sugar = 15% by mass
- Flour = 84% by mass
- Sodium hydrogencarbonate = 1% by mass
(ii) Brass alloy composition
Given
- Brass = 120 g
- Copper = 70%
- Zinc = 30%
Formula,
\[ \text{Mass of component} = \frac{\%\ \text{of component}}{100} \times \text{Total mass} \]\[ \text{Copper: } \frac{70}{100} \times 120 = 84 \, \text{g} \]\[ \text{Zinc: } \frac{30}{100} \times 120 = 36 \, \text{g} \]- Copper = 84 g
- Zinc = 36 g
18. 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:
- Oil pack: 1 litre = 910 g.
- Water: 1 litre = 1000 g.
Oil is lighter than water. So, that Oil float on top and water stay at the bottom
Separation
- Use a separating funnel.
- Open the tap and water will come out first because of bottom layer.
- Close the tap the oil reamin in the funnel which is top layer.
19. 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.
Explanation:
Solutions do not show the Tyndall effect because their particles are very tiny (<1 nm). The reason given is wrong because it says particles are larger than 100 nm, which is not true.
20. 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 is solid and water is liquid, so a filter paper can be used to trap mud. |
| Plasma from blood | Centrifugation | Blood is spun at high speed; heavier cells settle down and plasma remains at the top. |
| Naphthalene and sand | Sublimation | Naphthalene changes directly from solid to vapour on heating, while sand does not. |
| Chalk powder and common salt | Dissolve + Filtration + Evaporation | Salt dissolves in water but chalk does not. Filter out chalk, then evaporate water to get salt. |
| Common salt and water | Evaporation | Water evaporates, leaving salt behind. |
| Oil from water | Separating funnel | Oil and water are immiscible (do not mix) and form two layers. |
| Pigments of flower | Chromatography | Different pigments travel at different speeds on paper, so they separate. |
21. 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: Two liquids A and B are mixed.
- A boils at 60 °C
- B boils at 90 °C
- They are miscible (mix completely).
Method
Use simple distillation.
- Step 1: Heat the mixture.
- Step 2: Liquid A has a lower boiling point, 60° C which will boils first and come outfirst.
- Step 3: Then liquid B has higher boiling point 90° C which will remains and can be collected later.
22. Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?
Answer:
| Method | Meaning | When to use |
|---|---|---|
| Evaporation | Liquid slowly changes to vapour, solid left behind | To get salt from sea water |
| Crystallization | Pure solid crystals form from solution | To make pure copper sulfate crystals |
| Distillation | Liquid boils → vapour → cooled back to liquid | To separate alcohol from water |
23. 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:
In a suspension the particles are big and heavy. They settle down at the bottom. If the blood also behaved like a true suspension, then blood cells would sink and not stay mixed. As a result, the blood could not flow properly, and life would not be possible.
(ii) Dispersed phase and dispersion medium in blood:
- Dispersed phase: Blood cells (RBSs, WBCs, platelets)
- Dispersion medium: Plasma this is liquid part of blood.
24. 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 correct sequence is: Sublimation -> Dissolving in water -> Filtration -> Evaporation.
- Sublimation: First, remove naphthalene because it changes directly from solid to vapour when heated, while sand and salt do not.
- Dissolving in water: Add water to the remaining mixture. Salt dissolves in water, but sand does not.
- Filtration: Filter the mixture. The sand stays on the filter paper, and the salt solution passes through.
- Evaporation: Heat the salt solution. The water evaporates, leaving behind pure common salt.
25. Why is distillation an effective method for separating a mixture of water and acetone?
Answer: If the water and acetone mix fully, it means they are miscible liquids. They cannot be separated by filtration or evaporation. Both have a boiling points different:
- Acetone boils at 56 °C
- Water boils at 100 °C
When we heat:
- Acetone boils first and turns into vapour and collected.
- Water boils later and collected separately.
26. 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 with increasing temperature from 10 °C to 80 °C.
Answer:
- 31 g potassium nitrate needed in 50 g water at 40 °C.
- On cooling potassium chloride solution, crystals appear because less salt can dissolve at lower temperature.
- Solubility increases with temperature. Potassium nitrate shows the greatest rise, sodium chloride almost no change, potassium chloride and ammonium chloride moderate rise.
27. 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) Mass percentage of sugar
Formula,
\[ \%\ m/m = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \]Student A
- Sugar = 20 g
- Water = 80 g
- Total = 100 g
Student B
- Sugar = 20 g
- Water = 100 g
- Total = 120 g
Student C
- Sugar = 30 g
- Water = 80 g
- Total = 110 g
(ii) Most concentrated solution
A = 20%
B = 16.7%
C = 27.3%
Student C’s solution is most concentrated because it has the highest mass percentage of sugar.
28. 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) Identify the technique marked ‘S’.
The technique shown in the figure is distillation. Distillation is a method where we can separate liquids that boil at different temperatures.
(ii) Label the apparatus A, B and C.
- Apparatus A is the flask where we can heat the mixture.
- Apparatus B is the condenser where the vapour cools down and changes back into liquid.
- Apparatus C is the receiver where the separated liquid is collected.
(iii) Mixtures that can be separated by distillation (using Table 5.5)
- Water and Acetone: Yes, we can separate the water and acetone because water boils at 100°C and acetone boils at 56°C.
- Water and Salt: No, we cannot separate the water and salt using distillation because salt is a solid and does not boil.
- Acetone and Alcohol: Yes, we can separate the acetone and alcohol because acetone boils at 56°C and alcohol boils at 78°C.
- Sand and Salt: No, they cannot be separated by distillation because both are solids.
- Alcohol and Chloroform: Yes, we can separate them because alcohol boils at 78°C and chloroform boils at 61°C.
- Alcohol and Benzene: Yes, but the biggest problem is their boiling points. Both have very close boiling points: alcohol at 78°C and benzene at 80°C. In this case fractional distillation is needed.
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