Activity 5.6 – Science Form 2 Chapter 5


Activity 5.6:
Aim: To distinguish between a solution and a suspension.

Materials: Copper(II) sulphate crystal, water, filter paper and chalk powder

Apparatus: Beaker, spatula, conical flask, filter funnel, measuring cylinder, glass rod, torchlight and white screen

Instruction
1. Measure and pour 100 ml of water into a beaker and add one spatula of copper(II) sulphate crystals.

2. Stir the mixture until even and observe the appearance of the mixture formed.

3. Use a torchlight to direct a beam of light towards the beaker and observe whether the light can pass through the mixture as shown in Figure 5.15(a).

4. Let the mixture stand for 10 minutes and filter its content as shown in Figure 5.15(b).

5. Observe if there is any residue left on the filter paper.

6. Repeat steps 1 to 5 by replacing copper(II) sulphate crystals with chalk powder and record your observations.



Questions
1. What is the appearance of the mixture of water and chalk powder?

2. How can the solute in the mixtures be related to the ability of light to pass through them?

3. Give your inference for the filtration test done on both mixtures.


Answer:
1. The mixture of water and chalk powder looks cloudy.

2. When a mixture of solute and solvent produces a clear solution, this allows light to pass through the solution. A bright spot of light will be observed on the white screen. When a mixture of solute and solvent does not produce a clear solution, a suspension is formed. The light cannot pass through. A dim spot of light will be observed on the white screen.

3. Copper(II) sulphate solution that is filtered does not leave any residue because the particles of copper(II) sulphate crystals (solute) dissolve evenly throughout the water (solvent). The mixture of water and chalk powder leaves residue on the filter paper because the mixture produces a suspension that contains the particles of insoluble chalk powder.

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