NCERT Solutions for Class 8 Science Chapter 9 - The Amazing World of Solutes, Solvents, and Solutions

NCERT Solutions for Class 8 Science Chapter 9: Download PDF

Students can download the Science Class 8 Chapter 9 solutions PDF from the icon below and access detailed solutions to all the questions present in the textbook.

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NCERT Solutions for Class 8 Science Chapter 9 Exercise Questions

The detailed solutions to exercise questions given below. The NCERT solutions for class 8 science chapter 9 will help you score well in exams.

Keep The Curiosity Alive

Question 1. State whether the statements given below are True [T] or False [F]. Correct the false statement(s).

(i) Oxygen gas is more soluble in hot water rather than in cold water.

(ii) A mixture of sand and water is a solution.

(iii) The amount of space occupied by any object is called its mass.

(iv) An unsaturated solution has more solute dissolved than a saturated solution.

(v) The mixture of different gases in the atmosphere is also a solution.

Answer:

(i) False - Oxygen gas is less soluble in hot water. It dissolves better in cold water; that is why aquatic life survives better in cooler water.

(ii) False - A mixture of sand and water is not a solution because sand does not dissolve and settles at the bottom. It is a non-uniform mixture.

(iii) False - The amount of space an object occupies is its volume, not mass. Mass is the amount of matter in the object.

(iv) False - An unsaturated solution contains less solute than a saturated solution, so more solute can still dissolve in it.

(v) True - Air is a gaseous solution. Since nitrogen is present in the largest amount in the air, it is considered the solvent, while oxygen, argon, carbon dioxide, and other gases are considered solutes.

Question 2. Fill in the blanks.

(i) The volume of a solid can be measured by the method of displacement, where the solid is __________ in water and the ____________ in water level is measured.

(ii) The maximum amount of _______________ dissolved in _______________ at a particular temperature is called solubility at that temperature.

(iii) Generally, the density ____________ with increase in temperature.

(iv) The solution in which glucose has completely dissolved in water, and no more glucose can dissolve at a given temperature, is called a __________ solution of glucose.

Answer:

(i) The volume of a solid can be measured by the displacement method, where the solid is immersed in water and the rise in water level is measured.

(ii) The maximum amount of solute dissolved in a solvent at a specific temperature is called solubility.

(iii) Generally, the density decreases with an increase in temperature.

(iv) The solution in which glucose has completely dissolved in water, and no more glucose can dissolve at a given temperature, is called a saturated solution of glucose.

Question 3. You pour oil into a glass containing some water. The oil floats on top. What does this tell you?

(i) Oil is denser than water

(ii) Water is denser than oil

(iii) Oil and water have the same density

(iv) Oil dissolves in water.

Answer:

The correct answer is option (ii).

Oil floats on top of water because it is less dense, which means for the same volume, oil has less mass than water. Since denser liquids sink and lighter ones float, water stays below and oil forms a layer on top. This also shows that oil and water do not mix which means they are immiscible liquids.

Question 4. A stone sculpture weighs 225 g and has a volume of 90 cm³. Calculate its density and predict whether it will float or sink in water.

Answer:

Given,
Mass = 225 g, Volume = 90 cm³
Density = Mass ÷ Volume

= 225 ÷ 90 = 2.5 g/cm³

Since the water has a density of 1 g/cm³, therefore, the sculpture is denser than water.
Hence, the stone sculpture will sink in water.

Question 5. Which one of the following is the most appropriate statement, and why are the other statements not appropriate?

(i) A saturated solution can still dissolve more solute at a given temperature.

(ii) An unsaturated solution has dissolved the maximum amount of solute possible at a given temperature.

(iii) No more solute can be dissolved into the saturated solution at that temperature.

(iv) A saturated solution forms only at high temperatures

Answer:

The correct answer is option (iii).

This is because a saturated solution already has the maximum amount of solute it can hold at that temperature. The other options are incorrect because,

(i) A saturated solution cannot dissolve more solute.
(ii) An unsaturated solution is the one that can still dissolve more.
(iv) A saturated solution can form at any temperature, not only at high temperatures

Question 6. You have a bottle with a volume of 2 litres. You pour 500 mL of water into it. How much more water can the bottle hold?

Answer:

The bottle can hold 2 L = 2000 mL.
The amount of water poured = 500 mL
Therefore, the remaining volume is = 2000 – 500 = 1500 mL

So, the bottle can still hold 1500 mL (or 1.5 L) of more water.

Question 7. An object has a mass of 400 g and a volume of 40 cm³. What is its density?

Answer:

Given,
Mass = 400 g, Volume = 40 cm³
As we know,

Density = Mass ÷ Volume

= 400 ÷ 40 = 10 g/cm³

Therefore, the density of this object is 10 g/cm³, which means the object has 10 grams of mass in every cubic centimeter of space.

Question 8. Analyse Fig. 9.25a and 9.25b. Why does the unpeeled orange float, while the peeled one sinks? Explain.

Answer:

The unpeeled orange has a spongy outer peel filled with air pockets that make the whole fruit less dense than water, and so it floats. But when we peel the orange, the air layer is removed, and the orange becomes denser than water, so it sinks. This shows that the presence of air can affect whether an object floats or sinks.

Question 9. Object A has a mass of 200 g and a volume of 40 cm³. Object B has a mass of 240 g and a volume of 60 cm³. Which object is denser?

Answer:

As we know, Density = Mass ÷ Volume

Therefore, for object A, density (d) = 200 g ÷ 40 cm³ = 5 g/cm³
And for object B, density (d`) = 240 g ÷ 60 cm³ = 4 g/cm³
Since d > d`, therefore, object A is denser than object B.

Question 10. Reema has a piece of modeling clay that weighs 120 g. She first molds it into a compact cube that has a volume of 60 cm³. Later, she flattens it into a thin sheet. Predict what will happen to its density.

Answer:

Reema has a clay weighing 120 g. She changed the shape of the modeling clay (cube to and thin sheet), but the mass and volume will remain the same. Even though they acquire different shapes, they still have the same amount of matter and will take up the same space.

Since density = mass/volume, and the mass and volume both remained the same, the density will also remain the same. The change in shape does not affect density if the material and volume are unchanged.

Question 11. A block of iron has a mass of 600 g and a density of 7.9 g/cm³. What is its volume?

Answer:

From the density formula we have,

Volume = Mass ÷ Density

= 600 g ÷ 7.9 g/cm³

= 75.95 cm³

So, the volume of the iron block is about 76 cm³.

This means that 600 g of iron takes up 76 cubic centimeters of space.

Question 12. You are provided with an experimental setup as shown in Fig. 9.26a and 9.26b. On keeping the test tube (Fig. 9.26b) in a beaker containing hot water (~70°C), the water level in the glass tube rises. How does it affect the density?

Answer:

When the test tube is placed in hot water, the water inside the glass tube gets heated and expands. This causes the water level to rise. The mass remains the same, but the volume increases.

As we know, density = mass/volume.

That is, d ∝ 1/V. So, as the volume increases, the density will decrease. This shows that heating reduces density, which is why hot air or water rises.

Approach to Solve Chapter 9 Questions Effectively

To solve questions from Class 8 Science Chapter 9 The Amazing World of Solutes, Solvents, and Solutions effectively, you need to work on the following points given below.

1. Understand the key concepts

The understanding of the topics is crucial to attempt the questions. Some important topics are

• The types of solution- saturated or unsaturated.
• The type of mixtures- uniform or non-uniform.

Learn these concepts with the help of examples, as they can be asked in exams directly. You can also refer to The Amazing World of Solutes, Solvents, and Solutions NCERT notes for conceptual clarity and quick revision.

2. Learn about density and solubility

Density is defined as the mass present in a unit volume of that substance. Learn to apply the formula

Density = Mass ÷ Volume

The solubility of compounds is often asked in exams. The effect of temperature on solubility is one of the most important topics of this chapter. You can learn these concepts by solving the NCERT solutions for class 8 science chapter 9.

3. Make notes

You can make use of tables and charts to revise the concepts. This will help you in answering the questions effectively. These short notes will help you revise the concepts anytime.

4. Understand the activities

The activities discussed in the chapter are so important and are often asked in exams. Do practice them attentively.

5. Practice more

Solve all the questions given in the Class 8 NCERT textbook. Practice the numericals on the density formula and focus on the units. For better understanding refer NCERT solutions for class 8 science chapter 9.

NCERT Solutions for Class 8 Science Chapter 9: Topics

The topics and subtopics covered in the textbook are listed below. Students can use the The Amazing World of Solutes, Solvents, and Solutions NCERT notes to understand these topics in detail.

9.1 What Are Solute, Solvent, and Solution?

9.2 How Much Solute Can a Fixed Amount of Solvent Dissolve?

9.2.1 How does temperature affect the solubility of a solute?

9.3 Solubility of Gases

9.4 Why Do Objects Float or Sink in Water?

9.5 What Is Density

9.5.1 Determination of density

9.5.2 Effect of temperature on density

9.5.3 Effect of pressure on density

NCERT Chapter-Wise Solutions For Class 8 Science

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