Students need to have a clear idea of how substances dissolve, mix and interact with each other; NCERT Solutions for Class 8 Science chapter 9 on "The Amazing World of Solutes, Solvents & Solutions" aims to achieve this goal by introducing the basic concept of solutions and the role they play in our daily lives.
In Chapter 9, the process by which solutes disperse through a solvent is defined, with examples of real-life situations such as saltwater, sugar solutions, and medicinal mixtures. The discussion includes what factors influence the solubility of solutes, including but not limited to: temperature, amount of solute, and the properties of the solvent.
The answers are structured in a manner that will enable students to build a conceptual foundation and confidence in their ability to answer questions related to these topics in an examination setting.
For convenient revision and offline study, students have access to the complete NCERT Solutions for Class 8 Science Chapter 9 as a downloadable PDF.
These solutions, which guarantee accuracy, clarity, and comprehensive syllabus coverage, have been meticulously prepared by ALLEN's subject experts.
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.
Sol.
(i) False
Oxygen gas is more soluble in cold water rather than in hot water. The solubility of most of the gases decreases as the temperature increases.
(ii) False
A mixture of sand and water is a non-uniform mixture (a suspension), not a solution. In a solution, the components must be evenly distributed.
(iii) False
The amount of space occupied by any object is called its volume. Mass is the amount of matter in an object.
(iv) False
A saturated solution has more (or equal) solute than an unsaturated solution. An unsaturated solution can still dissolve more solute.
(v) True
Air is an example of a gaseous solution, with nitrogen as the solvent and other gases like oxygen as the solute.
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.
Sol.
(i) The volume of a solid can be measured by the method of displacement, where the solid is immersed in water and the rise in water level is measured.
(ii) The maximum amount of solute dissolved in a fixed quantity of a solvent (or solution) at a particular temperature is called solubility at that temperature.
(iii) Generally, the density decreases 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 saturated solution of glucose.
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
Sol. (ii) Water is denser than oil. Therefore, oil floats on water.
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.
Sol.
Mass of stone sculpture is 225 g
Volume of stone sculpture is 90 cm³
Density = Mass / Volume
Density = 225 g / 90 cm³ = 2.5 g/cm³
The density of water is approximately 1 g/cm³.
Since the sculpture’s density (2.5 g/cm³) is greater than the density of water, it will sink.
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.
Sol. The most appropriate statement is (iii) No more solute can be dissolved into the saturated solution at that temperature.
(i) This statement describes an unsaturated solution, not a saturated one.
(ii) This statement describes a saturated solution, not an unsaturated one.
(iv) Saturated solutions can be formed at any temperature; the amount of solute needed to reach saturation depends on temperature.
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?
Sol. 2 litres = 2 × 1000 mL = 2000 mL
Amount of water the bottle can still hold
= Total Volume − Poured Volume
= 2000 mL − 500 mL
= 1500 mL
The bottle can hold 1500 mL more water (or 1.5 litres).
7. An object has a mass of 400 g and a volume of 40 cm³. What is its density?
Sol. Density = Mass / Volume
Density = 400 g / 40 cm³ = 10 g/cm³
The density of the object is 10 g/cm³.
8. Analyse Fig. (a) and (b). Why does the unpeeled orange float while the peeled one sinks?
Sol. The unpeeled orange floats because its peel is full of tiny air pockets. This trapped air increases the orange’s total volume significantly without adding much mass. This makes the overall density of the unpeeled orange less than the density of water, causing it to float.
When we peel the orange, we remove the peel and all the trapped air. The remaining orange flesh is denser than water, so it sinks.
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?
Sol. Density of Object A = 200 g / 40 cm³ = 5 g/cm³
Density of Object B = 240 g / 60 cm³ = 4 g/cm³
Object A is denser.
10. Reema has a piece of modelling clay that weighs 120 g. She moulds it into a cube of volume 60 cm³ and later flattens it into a thin sheet.
Predict what will happen to its density.
Sol. The density of the modelling clay will not change because density is an intrinsic property of a substance. Changing shape does not change mass or volume.
11. A block of iron has a mass of 600 g and a density of 7.9 g/cm³. What is its volume?
Sol. Volume = Mass / Density
Volume = 600 g / 7.9 g/cm³ ≈ 75.95 cm³
The volume of the iron block is approximately 75.95 cm³.
12. You are provided with an experimental setup as shown in Fig. (a) and (b). On heating the test tube in hot water (~70°C), the water level rises. How does it affect density?
Sol. When water is heated, its particles gain energy, move faster, and the interparticle space increases. This causes water to expand, increasing volume. Since mass remains the same and volume increases, density decreases. Heating water decreases its density.
(Session 2026 - 27)