The Amazing World of Solutes Solvents and Solutions Class 8 Question Answer Science Chapter 9

Start by reviewing these Class 8 Science Curiosity Solutions Chapter 9 The Amazing World of Solutes Solvents and Solutions Question Answer to strengthen your knowledge.

Class 8 Science Curiosity Chapter 9 Question Answer

Class 8 Science Ch 9 The Amazing World of Solutes Solvents and Solutions Question Answer

Class 8 Science Chapter 9 The Amazing World of Solutes Solvents and Solutions Question Answer (InText)

Question 1.
What happens when you add too much sugar to your tea and it stops dissolving? How can you solve this problem? (Page 134)
Answer:
When too much sugar is added, the solution becomes saturated and to solve this problem heat the tea to increase solubility or add more solvent (water).

Question 2.
Why do sugar and salt dissolve in water but not in oil? Why is water considered a good solvent? (Page 134)
Answer:
Water is a polar solvent that can dissolve polar substances like sugar and salt, while oil is non-polar and cannot dissolve them. Water is a good solvent because it can dissolve a lot of substances compared to any other solvent.

Question 3.
Why are water bottles usually tall and cylindrical instead of spherical? (Page 134)
Answer:
Cylindrical shapes are easier to grip, stack and manufacture compared to spherical ones.

Question 4.
Why does every sip of ORS taste the same? (Page 135)
Answer:
ORS is a uniform mixture (solution) where sugar and salt are evenly distributed.

Question 5.
Is the mixture of chalk powder and water uniform? (Page 135)
Answer:
No, it is a non-uniform mixture as chalk particles settle and are visible.

Question 6.
What will happen if we keep on adding more salt in a given amount of water? (Page 136)
Answer:
The solution will become saturated and any further salt will settle down on the button of the container.

Question 7.
Which solution is more concentrated: 2 spoons of salt in 100 mL water or 4 spoons in 50 mL water? (Page 137)
Answer:
The second solution (4 spoons/50 mL) is more concentrated as it has more solute per unit volume.

Question 8.
Does temperature affect solubility of a solute? (Page 137)
Answer:
Yes, solubility of a solute increases with an increases in temperature, e.g. water at 70°C dissolves more baking soda than water at 50°C.

Question 9.
What happens to undissolved baking soda when heated? (Page 138)
Answer:
It dissolves because solubility increases with temperature.

Question 10.
Do gases also dissolve in water? (Page 139)
Answer:
Yes, gases also dissolve in water.

Question 11.
Is the mixture of gases in water a uniform or non-uniform mixture? (Page 139)
Answer:
A mixture of gases forms a uniform mixture in water.

Question 12.
Does temperature affect gas solubility in water? (Page 139)
Answer:
Yes, gas solubility decreases as temperature rises (e.g., cold water holds more oxygen).

Question 13.
Why does sawdust float on water while sand sinks? (Page 139)
Answer:
Sawdust is less dense than water, while sand is denser.

Question 14.
Why do some oil packets list a volume of 1 litre but a weight of only 910 grams? (Page 141)
Answer:
The oil’s density (0.91 g/cm³) is less than water’s (1 g/cm³), so it weighs less for the same volume.

Question 15.
How much volume does one small division indicate in a measuring cylinder? (Page 144)
Answer:
One small division indicates 1 mL in a measuring cylinder.

Question 16.
Why are measuring cylinders always designed narrow and tall instead of wider and short like a beaker? (Page 144)
Answer:
Measuring cylinders are narrow and tall to increase measurement precision—their shape makes small volume changes (like 1 mL) easier to read compared to wide beakers. The narrow design also reduces errors when reading the meniscus.

Question 17.
How is the level of a coloured liquid measured? (Page 145)
Answer:
For colored liquids, the top of the meniscus (the upper curve of the liquid surface) where the liquid touches the measuring cylinder’s markings is considered.

Question 18.
What change can you make to make a raw egg float in water instead of sinking? (Page 148)
Answer:
Add salt to increase the water’s density until it exceeds the egg’s density.

The Amazing World of Solutes Solvents and Solutions Class 8 Questions and Answers (Exercise)

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:
Here are the answers to the true/false statements
(i) F – Oxygen gas is more soluble in cold water than in hot Water.
(ii) F – A mixture of sand and water is a suspension, not a solution.
(iii) F – The amount of space occupied by an object is called its volume and not mass.
(iv) F – A saturated solution has more solute dissolved than an unsaturated solution at a given temperature.
(v) T – The mixture of gases in the atmosphere (like nitrogen, oxygen, etc.) is a gaseous solution.

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) immersed, rise
(ii) solute, solvent
(iii) decreases
(iv) saturated

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:
(ii) Oil floats on water because it is less dense than water.

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:
(i) Density calculation :
Density = \(\frac{\text { Mass }}{\text { Volume }}\) = \(\frac{225 \mathrm{~g}}{90 \mathrm{~cm}^3}\) = 2.5 g/cm³

(ii) Compare to water.

• Density of water = 1g/ cm³
• Since the stone’s density (2.5 g/ cm³) > water’s density, it will sink.
  The stone’s density is 2.5 g/cm³ and it 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:
(iii) No more solute can be dissolved into the saturated solution at that temperature. While the other statements are incorrect.
(i) False – A saturated solution cannot dissolve more solute at the given temperature (it has already reached maximum solubility).
(ii) False – An unsaturated solution has not dissolved the maximum solute and more can still be added.
(iv) False – Saturated solutions can form at any temperature (not just high temperatures), but solubility limits vary with temperature.

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:
(i) Convert units

• Bottle’s total volume = 2 litres = 2000 mL
• Water already added = 500 mL.

(ii) Calculate remaining capacity
Remaining space = 2000 mL – 500 mL = 1500 mL (or 1.5 litres).
The bottle can hold 1500 mL (1.5 litres) more water.

Question 7.
An object has a mass of 400 g and a volume of 40 cm3. What is its density?
Answer:
Density calculation = \(\frac{\text { Mass }}{\text { Volume }}\) = \(\frac{400g \mathrm{~g}}{90 \mathrm{~cm}^3}\) = 10 g/ cm³
The object’s density is 10 g/cm³.

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

Answer:
Floating vs sinking orange:

1. Unpeeled Orange floats The peel contains air pockets, reducing its average density below water’s density (1 g/cm³).
2. Peeled Orange sinks Without air pockets, its density becomes higher than water.
3. The peel’s air makes the unpeeled orange less dense, allowing it to float.

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:
Calculation of density of object A
Density = \(\frac{200 \mathrm{~g}}{40 \mathrm{~cm}^3}\) = 5 g/cm³
Calculation of density of object B
Density = \(\frac{240 \mathrm{~g}}{60 \mathrm{~cm}^3}\) = 4 g/ cm³
Hence, object A (5 g/cm³) is denser than objects (4 g/cm³).

Question 10.
Reema has a piece of modeling clay that weighs 120 g. She first moulds 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:
The density of the clay remains unchanged (2 g/cm³)
when flattened into a sheet because density depends only on mass and total volume and not shape.

Question 11.
A block of iron has a mass of 600 g and a density of 7.9 g/cm³. What is it’s volume?
Answer:
We are given
Mass of iron block = 600 g
Density of iron = 7.9 g/cm³
Formula of density
density = \(\frac{\text { Mass }}{\text { Volume }}\) ⇒ volume = \(\frac{\text { Mass }}{\text { Density }}\)
Calculation of density
Volume = \(\)\frac{600}{7.9 \mathrm{~g} / \mathrm{cm}^3}\(\) = 75.95 cm³
The volume of the iron block is approximately 75.95 cm³.

Question 12.
You are provided with an experimental setup as shown in Fig. (a) and (b). On keeping the test tube Fig (b) 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 Fig. (b) is placed in hot water (-70 °C), the air inside it heats up and expands, causing the water level in the glass tube to rise. Since the mass of the air remains unchanged while its volume increases, the density of the air inside the tube decreases. This happens because density is calculated as mass divided by volume (Density = Mass/Volume).

The expansion of air due to heating demonstrates how temperature affects gas volume and density—a key principle behind phenomena like hot air balloons, where heated, less dense air rises. Thus, the rising water level directly indicates a reduction in the air’s density within the tube.

Class 8 Science Chapter 9 Question Answer (Activities)

Activity 1 (Page 136)

Aim
To observe what happens when salt is added to water repeatedly and to understand the concepts of saturated and unsaturated solutions.

Materials Required
A clean glass tumbler, water, common salt, a spoon and a stirrer.

Procedure

1. Take a clean glass tumbler and fill it halfway with water.
2. Add one teaspoon of salt to the water and stir it well until it dissolves completely.
3. Gradually add more teaspoons of salt, one at a time and stirring after each addition.
4. Continue this process until you observe that the added salt no longer dissolves and starts settling at the bottom.
5. Record your observations after each addition in the observation table.

Observation Table

S.No.	Amount of Salt Added (teaspoon)	Observation
1.	One	Salt dissolves completely
2.	Two	Salt dissolves completely
3.	Three	Salt dissolves completely
4.	Four	Salt begins to settle at the bottom

Conclusion
We conclude that initially the salt dissolves in water, forming an unsaturated solution. As more salt is added, a point is reached where no more salt dissolves and it starts to settle at the bottom. This indicates the solution has become saturated. The maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature is known as its solubility.

Viva Questions

1. What is a saturated solution?
2. Why does salt stop dissolving after a certain point in water?
3. What is meant by the solubility of a substance?
4. What kind of change is the dissolving of salt in water—physical or chemical? Why?
5. How does the amount of solute affect whether a solution is dilute or concentrated?

Activity 2 (Page 137)

Aim
To understand how temperature affects the solubility of a solute (baking soda) in water.

Materials Required
Glass beaker, water (50 mL), baking soda (sodium hydrogen carbonate), laboratory thermometer, spirit lamp, tripod stand, wire gauze and glass rod.

Procedure

1. Take about 50 mL of water in a clean glass beaker.
2. Measure its temperature using a laboratory thermometer; suppose it is 20 °C.
   
3. Add one spoonful of baking soda to the water and stir it well.
4. Keep adding small amounts of baking soda while stirring until some remains undissolved at the bottom (saturated solution at 20 °C).
5. Heat the contents gently using a spirit lamp to raise the temperature to about 50 °C while stirring.
6. Observe whether the undissolved baking soda now dissolves.
7. Add more baking soda at 50 °C and stir. When it again remains undissolved, heat the mixture further to 70 °C and continue stirring.
8. Observe if more baking soda dissolves at the higher temperature.

Observation Table

S.No.	Temperature (°C)	Observation
1.	20	Baking soda dissolves to a limit rest remains undissolved
2.	50	Previously undissolved baking soda dissolves completely
3.	70	More baking soda dissolves at higher temperature

Conclusion
We conclude that the solubility of baking soda increases with temperature. A solution saturated at a lower temperature becomes unsaturated when the temperature is increased. This shows that most substances become more soluble as the temperature increases.

Viva Questions

1. What happens to the solubility of baking soda when the temperature is increased?
2. At which temperature does the maximum amount of baking soda dissolve—20 °C, 50 °C or 70 °C?
3. What type of solution is formed when no more solute dissolves in the solvent?
4. Is this change physical or chemical? Why?

Activity 3 (Page 142)

Aim
To measure the mass of a solid object accurately using a digital weighing balance.

Materials Required
Digital weighing balance, watch glass (or butter paper), a dry solid object such as a stone.

Procedure

1. Switch ON the digital weighing balance and observe the initial reading.
2. If the display does not show zero, press the tare or reset button to bring the reading to zero.
3. Place a clean and dry watch glass on the weighing pan.
4. Again, press the tare button to reset the reading to zero with the watch glass in place.
5. Now, carefully place the solid object (like a stone) on the watch glass.
6. Observe and note the mass displayed on the digital balance, for example, 16.400 g.

Observation Table

S.No.	Step Performed	Observation (Reading on Display)
1.	Tare digital balance	Display shows 0.000 g
2.	Watch glass placed	Display reset to 0.000 g
3.	Stone placed on watch glass	Display shows 16.400 g

Conclusion
We conclude that the digital weighing balance helps us measure the mass of a solid accurately. Taring ensures that only the mass of the object (not the container) is measured. This method can also be used to weigh liquids using a beaker instead of a watch glass.

Viva Questions

1. Why do we press the tare button on a digital balance?
2. What is the importance of using a watch glass while weighing solids?
3. What does the reading on the digital balance indicate after placing the object?
4. Can this method be used to weigh liquids? How?
5. Is the mass measured using this method affected by the shape of the object?

Activity 4 (Page 143)

Aim
To observe a measuring cylinder and understand how to calculate the smallest volume it can measure.

Materials Required
Measuring cylinder (preferably 100 mL) and water (optional for demonstration).

Procedure

1. Take a 100 mL measuring cylinder and observe it carefully.
2. Note the maximum volume marked on the cylinder.
3. Observe the two larger markings (for example, between 10 mL and 20 mL).
4. Count the number of smallest divisions between any two large marks.
5. Divide the volume difference between the two large marks by the number of smaller divisions to find the value of one small division.
6. Determine the smallest volume this measuring cylinder can read.

Observation Table

Observation Parameter	Reading/Value Noted
Maximum volume on the cylinder	100 mL
Volume difference between two large markings	10 mL
Number of smaller divisions between large marks	10
Volume value of one small division	1 mL
Smallest measurable volume	1 mL

Conclusion
A 100 mL measuring cylinder has 10 small divisions between two large markings, each representing 1 mL. Therefore, the smallest volume it can measure is 1 mL. Such narrow and tall cylinders allow for more accurate volume readings than wider containers like beakers.

Viva Questions

1. What is the total volume that a 100 mL measuring cylinder can measure?
2. How do you calculate the value of one small division on a measuring cylinder?
3. Why are measuring cylinders designed tall and narrow instead of short and wide?
4. Which measuring cylinder would give the most accurate reading for 70 mL – 50 mL or 100 mL? Why?
5. What is the smallest volume that a 250 mL measuring cylinder usually reads?

Activity 5 (Page 144)

Aim
To measure 50 mL of water accurately using a measuring cylinder and understand how to read the meniscus.

Materials Required
Clean and dry 100 mL measuring cylinder, dropper and water.

Procedure

1. Place the measuring cylinder on a flat and horizontal surface.
   
2. Slowly pour water into the measuring cylinder until it reaches close to the 50 mL mark.
3. Use a dropper to add or remove water drop by drop until the bottom of the curved surface (meniscus) just touches the 50 mL mark.
4. Ensure that your eye is at the same level as the bottom of the meniscus while reading the measurement.
   
5. After confirming the correct level, carefully transfer the measured water to the required container.

Note
For colourless liquids like water, the reading is taken at the bottom of the meniscus and for coloured liquids, the reading is taken at the top of the meniscus.

Observation Table

S.No.	Step Performed	Observation
1.	Water poured into measuring cylinder.	Curved surface (meniscus) formed.
2.	Level adjusted using dropper.	Water level reaches exactly 50 mL at bottom meniscus.
3.	Eye level adjusted for accurate reading.	Measurement is correct and free of parallax error.

Conclusion
The volume of water should be read at the bottom of the meniscus for colourless liquids.
Accurate reading depends on keeping the eye level with the meniscus. This technique helps measure precise liquid volumes using a measuring cylinder.

Viva Questions

1. At which point of the meniscus should you take the reading for water?
2. Why should the measuring cylinder be kept on a flat surface?
3. How is the reading different for coloured liquids?
4. What happens if the eye is not at the level of the meniscus while reading?

Activity 6 (Page 145)

Aim
To calculate the volume of objects having cuboid shapes using the formula for volume.

Materials Required
A notebook, shoe box or dice (cuboid-shaped objects), scale (ruler), notebook and pencil.

Procedure

1. Collect any object that has a cuboid shape, such as a notebook, a shoebox, or a dice.
2. Using a scale, measure the length (l), width (w), and height (h) of the object.
3. Record the measurements. For example, for a notebook:
   • Length = 25 cm
   • Width = 18 cm
   • Height = 2 cm
4. Use the formula for volume, Volume = l × w × h
5. Calculate the volume of the object and note it down in your notebook.

Observation table

Object	Length (l)	Width (w)	Height (h)	Volume (l × w × h)
Notebook	25 cm	18 cm	2 cm	900 cm3
Shoe box	……………	…………….	…………….	…………….
Dice	……………	…………….	…………….	…………….

Conclusion
The volume of cuboid-shaped objects can be calculated by multiplying their length, width, and height. This helps us understand how much space an object occupies.

Viva Questions

1. What is the formula to calculate the volume of a cuboid?
2. What does the unit cm³ represent?
3. Can volume be measured for solid objects? How?
4. If the dimensions of a box are doubled, how will it affect the volume?
5. What is the importance of measuring volume accurately?

Activity 7 (Page 146)

Aim
To measure the volume of irregular-shaped solid objects using the water displacement method.

Materials Required
Measuring cylinder (100 mL), water, stone, metal key (or any other small irregular object), thread, notebook and pencil.

Procedure

1. Fill the measuring cylinder with water up to a certain level (e.g., 50 mL) and record this as the initial volume.
   
2. Tie the object (e.g., a stone) with a thread.
3. Slowly lower the object into the water in the measuring cylinder.
4. Record the new level of water as the final volume.
5. Subtract the initial volume from the final volume to get the volume of the object.
6. Repeat with other irregular objects.
7. Note all values in the observation table.

Observation Table

S.No.	Object	Initial Volume (mL)	Final Volume (mL)	Volume of the object (mL or cm3)
1.	Stone	50	55	5
2.	Metal key	50	58	8
3.	Rubber plug	50	53	3

Conclusion
We conclude that when an irregular solid is immersed in water, it displaces an amount of water equal to its own volume. This method is useful for measuring the volume of irregular solids.

Viva Questions

1. What is the principle used to measure volume in this activity?
2. Why do we use a measuring cylinder in this method?
3. What happens to the water level when an object is immersed?
4. What is the SI unit of volume for solids?