In this article, you will find solutions to the NECRT exercise solutions of class 9 Science chapter 5 i.e. Exploring Mixtures and their Separation.
Solved Examples
Example 5.1: If 10 g of salt is dissolved in 90 g of water, calculate the mass by mass percentage of the solution formed.
Answer: Mass of salt (solute) = 10 g
Mass of water (solvent) = 90 g
Total mass of solution = Mass of solute + Mass of solvent
= 10 g + 90 g = 100 g
Example 5.2: If 5 g of glucose is dissolved in water to make 100 mL of solution, calculate its concentration in mass by volume percentage.
Answer: Mass of glucose (solute) = 5 g
Volume of solution = 100 mL
Example 5.3: If 1 mL of a liquid pesticide is mixed with a sufficient amount of water to form 100 mL of a pesticide spray for rice crop, calculate its volume by volume percentage.
Answer: Volume of pesticide (solute) = 1 mL
Total volume of solution = 100 mL
Volume by volume percentage =
In-text Questions (Pause & Ponder)
Q1. 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?
Ans.
The term 4% m/m means that 4 g of zinc oxide is present in every 100 g of talcum powder. Therefore, in 300 g of talcum powder:
(4 × 300) / 100 = 12 g
Hence, 12 g of zinc oxide is present in 300 g of the talcum powder.
Q2. 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?
Ans.
Each tablespoon contains 15 mL of concentrate. Therefore, two tablespoons contain:
2 × 15 = 30 mL
The total volume of juice prepared is 150 mL.
% v/v = (30 / 150) × 100 = 20%
Thus, the orange juice concentrate is 20% v/v in the prepared mixture.
Q3. 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?
Ans.
To prepare vinegar containing 5% v/v acetic acid, 5 mL of glacial acetic acid should be mixed with water and the total volume should be made up to 100 mL.
In this way, a dilute solution containing 5% acetic acid is obtained, which is called vinegar.
Q4. 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?
Ans.
On cooling a saturated solution, the excess dissolved substance separates out as solid crystals. The amount of solid deposited depends on the decrease in solubility during cooling.
Therefore, the compound whose solubility decreases more sharply from 80 °C to 60 °C will deposit more solid.
Q5. Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.
Ans.
Yes, the size of salt crystals changes with the rate of evaporation. When evaporation takes place slowly, the particles get enough time to arrange themselves properly and form large crystals.
If evaporation occurs rapidly, smaller crystals are formed because the particles do not get enough time for proper crystal growth.
Q6. 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.
Ans. True.
Evaporation removes water and leaves behind salt, while distillation can also separate water from the salt solution.
(ii) Distillation can be used for separation of two liquids even when these have the same boiling point.
Ans. False.
Distillation cannot separate liquids having the same boiling point.
(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.
Ans. False.
The solvent level should be below the sample spot at the beginning of the experiment.
(iv) Evaporation and crystallization are the same processes.
Ans. False.
Evaporation removes the solvent, whereas crystallization is used to obtain pure crystals of a substance.
Q7. Why do immiscible liquids form two separate layers in a separating funnel?
Ans.
Immiscible liquids do not mix with each other because their particles have different properties. Therefore, they remain separated.
The liquid with greater density settles at the bottom, while the lighter liquid forms the upper layer in the separating funnel.
Q8. Is sublimation different from evaporation? Justify.
Ans.
Yes, sublimation is different from evaporation. Sublimation is the direct change of a solid into vapour without passing through the liquid state.
Evaporation is the process in which a liquid changes into vapour from its surface. Thus, sublimation occurs in solids, while evaporation occurs in liquids.
Q9. 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?
Ans.
Clouds are colloids because tiny water droplets or ice crystals remain suspended in air without settling down quickly.
These particles are small enough to stay dispersed in the atmosphere and scatter light, which is a property of colloidal mixtures.
Q10. Why do cities with a lot of smoke and dust in the air often look hazy?
Ans.
Smoke and dust particles remain suspended in the air and scatter sunlight in different directions.
This scattering of light reduces visibility and gives the atmosphere a hazy appearance in polluted cities.
Q10. Why do polluted cities often look hazy?
Ans.
Cities with a large amount of smoke, dust, and pollutants in the air often appear hazy because these tiny particles remain suspended in the atmosphere.
These suspended particles scatter sunlight in different directions, reducing visibility and giving the air a foggy or smoky appearance. This phenomenon is common in highly polluted urban areas.
NECRT Exercise Solutions
Q1. 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 sulphate solution—Hm, Salt solution—Hm, Milk—Hm, Bronze—Hm
(iv) Muddy water—Hm, Milk—Ht, Blood—Ht, Brass—Hm
Answer: The correct option is (iii).
Copper sulphate solution — Hm
Salt solution — Hm
Milk — Hm
Bronze — Hm
Q2. Choose the correct options, and explain the reason for the correct and incorrect options.
Which among the following mixtures show the Tyndall Effect?
(a) air and dust particles
(b) copper sulphate 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: The correct option is (iii) a and c. Air and dust particles and starch and water show Tyndall effect because they are colloids.
Q3. 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.
Answer:
 |
| Table showing difference between solutions, colloids and suspensions |
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 hydrogencarbonate. Express the concentration of each component in the mixture using
an appropriate method.
Answer:
First, find the total weight of the mix:
Total mass = 75g (sugar) + 420g (flour) + 5g (soda) = 500g
Now, the percentage for each:
Sugar: (75 / 500) x 100 = 15%
Flour: (420 / 500) x 100 = 84%
Soda: (5 / 500) x 100 = 1%
(ii) A brass alloy contains 70% copper by mass. Calculate the quantities
of copper and zinc present in 120 g of brass.
Answer:
Copper: 70% of 120g = (70 / 100) x 120 = 84g
Zinc: 120g - 84g = 36g
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, they will form two layers because they do not mix. Oil will be on top because it is lighter (less dense) than water. For separating them, we can use a separation funnel.
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.
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:
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 in boiling points is more than 25 degree celsius. So, they can be separated by distillation (Simple as well as fractional).
9. Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?
Answer: These processes include separation of a solid and liquid. But, purpose of the processes is different.
- Evaporation: Used to get a solid back by drying up the liquid.
- Crystallization: Better than evaporation because it gives very pure crystals and doesn't damage the substance with high heat.
- Distillation: Used when you want to keep the liquid (by cooling the steam) instead of letting it disappear into the air.
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 acted like a suspension, the cells would settle at the bottom of our legs or arms whenever we stopped moving. This would block our blood flow. If the heavy particles inside blood separate out inside veins, it might cause a heart attack.
(ii) Dispersed phase: The blood cells. Dispersion medium: The liquid 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:
- Sublimation (Step 1): Heat the mixture to turn naphthalene into gas and collect it.
- Dissolution and Filtration (Step 3): Mix the remaining sand and salt in water. Filter it to get the sand out.
- Evaporation (Step 2): Heat the salt water to get the solid salt back
12. Why is distillation an effective method for separating a mixture of water and acetone?
Answer: Distillation works well because water and acetone have different boiling points. Acetone boils at a much lower temperature (56°C) than water (100°C). When heated, acetone turns to vapor first, which is then cooled and collected separately.
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?
Answer: Potassium Nitrate at 40°C
Table says: 62g is needed for 100g of water.
For 50g of water (half the amount), you need half the salt: 31g.
(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.
Answer: Cooling Potassium Chloride
Observation: Crystals of potassium chloride will start to appear.
Explanation: Solubility decreases as temperature drops. Since the water cannot hold as much salt at 25°C as it did at 80°C, the extra salt "comes out" as crystals.
(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: Effect: Generally, solubility increases when temperature increases.
Comparison: Potassium nitrate shows the biggest increase (from 21g to 167g). Sodium chloride shows almost no change (it stays around 36-37g).
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.
Answer:
Calculate Mass Percentage
Formula: (Mass of sugar / Total mass of solution) x 100
Student A: (20 / 100) x 100 = 20%
Student B: (20 / 120) x 100 = 16.6%
Student C: (30 / 110) x 100 = 27.2%
(ii) Whose solution is the most concentrated? Explain why
Answer:
Most Concentrated Solution
Student C's solution is the most concentrated. This is because it has the highest mass of sugar (30g) dissolved in the same amount of water (80g) as Student A, giving it the highest percentage.
15. Examine Fig. 5.26.
(i) Identify the separation technique marked as ‘S’.
Answer: The technique shown in the image is Simple Distillation.
(ii) Label the apparatus A, B and C.
Answer:
A: Distillation Flask (or Round Bottom Flask)
B: Water Condenser (or Liebig Condenser)
C: Conical Flask (or Receiver Flask)
(iii) Which of the following mixtures can be separated by the technique identified above? Use the data given in Table 5.5. Mixtures:
Answer: Simple distillation is used for mixtures where the boiling point difference is more than 25°C. Based on Table 5.5:
(a) water — acetone: Water (100°C) and Acetone (56°C): Yes (Difference = 44°C).
(b) water — salt: Water and Salt: Yes (Solid-liquid mixtures can be separated this way).
(c) acetone — alcohol: Acetone (56°C) and Alcohol (78°C): No (Difference is only 22°C).
(d) sand — salt: No (Both are solids; this method is for liquids).
(e) alcohol — chloroform: Alcohol (78°C) and Chloroform (61°C): No (Difference is only 17°C).
(f) alcohol — benzene: Alcohol (78°C) and Benzene (80°C): No (Difference is only 2°C).