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.
Exploring Mixtures and their Separation - NECRT 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).