Chapter 7 of the Class 9 Chemistry syllabus for Sindh and Karachi boards covers the topic of “Chemical Equilibrium”. This chapter is important for students to understand the concept of equilibrium, its types, factors affecting equilibrium, and the equilibrium constant.
To help students prepare for their exams, there are various notes and study materials available online. These notes cover all the important topics related to Chapter 7 of the Class 9 Chemistry syllabus and include fill in the blanks, multiple-choice questions (MCQs), and question-answer exercises.
The fill in the blanks exercises are designed to help students remember key concepts related to equilibrium, such as the law of mass action, Le Chatelier’s principle, and the equilibrium constant. The MCQs and question-answer exercises are designed to test the students’ understanding of the concepts and help them prepare for their exams.
Here are some tips for students to effectively study Chapter 7 of the Class 9 Chemistry syllabus:
- Read the textbook carefully and take notes.
- Understand the concept of equilibrium and its different types.
- Learn the factors that affect equilibrium and how to apply Le Chatelier’s principle.
- Memorize the formula for the equilibrium constant.
- Practice solving MCQs and question-answer exercises to improve your understanding of the topics.
By following these tips and studying the notes and practice questions available online, students can effectively prepare for Chapter 7 of the Class 9 Chemistry syllabus and perform well in their exams.
Fill in the Blank Solution and Suspension
- The Component of sloutio present in smaller amount is called_____. Solute
- The component present in greater quantity is called ________. Solvent
- Soft drinks are the examples of gas in _______ Liquid
- When water is solvent solution is called __________ solution. aqueous
- The states of matter solid, liquid and gas produce ________ types of solutions on mixing. Nine
- Cloud and steam arce examples of liquid in ________ gas
- Smoke is solid in ________ gas
- The solubility of gases in a liquid decreases with the inverse in temperature.
- The Solubility of solids of solids and liquids arc ot affected by__________ Pressure.
- The Solubility of a gas in a liquid is directly proportional to the pressure gas. This is cllled _______ Henry’s Law.
- for crytallization supter ______________ is prepared. Saturated solution
- Solution is a _____________ of two or more substances. homogenous mixture
- The most common solvent in nature is _____________ . water
- An _ ___is the solution when the liquid solvent is water. aqueous solution
- 10% M/M solution contains 10 gram solute, dissolved in __ g solvent. 90
- _______________ is the symbol for the concentration unit of molarity. M
Read more: Sindh Class 9 Chemistry Notes Cha 6 States of Matter
Multiple Choice Question Chemistry Chapter 7 Class 9th Notes – MCQs
i) The suspended particles in suspensions are generally of the size.
A. 10nmCorrectly unselected
B. 100nmIncorrectly selected
C. 1200nmIncorrectly unselected
D. 1nm
ii) The sum of the mole fractions of solute and solvent is equal to:
A. 5
B. 2
C. 0
D. 1
iii) Solubility is defined as the amount of solute in gram at a given temperature, dissolved in _______of the solvent.
A. 20g
B. 100g
C. 10g
D. 2000g
iv) The process in which a solid directly changes to vapors is known as:
A. Sublimation
B. Evaporation
C. Diffusion
D. Fusion
v) The solubility of a gas ________with the rise in temperature.
A. Increase
B. Decrease
Short question Solution and Suspension chemistry Class 9
Q.1) Define the following terms:
a) Solute
b) Solvent
c) Solubility
d) Crystallization
Answer:
a) Solute
In a mixture solute is a substance which is dissolved in lesser amount in the solvent. For example, in a 5% glucose solution in water, solute is glucose and solvent is water.
b) Solvent
A solvent is a substance that dissolves a solute resulting in a solution. A solvent is usually a liquid but it can be gas or solid. For example, 10 ml alcohol is dissolved in water to make a solution. Here alcohol is solute in small amount and water is a solvent in greater quantity.
c) Solubility
The solubility of a substance is the amount of substance that dissolves in a given quantity of a solvent at a given temperature to produce a saturated solution. Solubility is usually expressed in grams of solute per 100 g of solvent.
d) Crystallization
The process in which dissolved solute comes out of solution and forms crystals is called crystallization.
Q.2) Name the solute and solvent in the following solutions:
a) Syrup
b) Haze (Dust in air)
c) Butter (Water in fat)
d) Fog
e) Jellies (Water in fruit pulp)
f) Smoke
g) Sodium amalgam
h) Cheese (Water in fat)
i) Foam (Water in air)
j) Mist
Answer:
| Name of solution | Solute | Solvent |
| Syrup | Sucrose | Water |
| Haze | Dust | Air |
| Butter | Water | Fat |
| Fog | Water vapour | Air |
| Jellies | Water | Fruit pulp |
| Smoke | CO2 | Air |
| Sodium amalgam | Mercury | Sodium |
| Cheese | Fat | Water |
| Foam | Water | Air |
| Mist | Water | Air |
Q.3) Discuss the factors affecting the solubility.
Answer:
There are some factors that affect the solubility of the solute in a solvent. These are:
i) Temperature:
Solubility of a solid in liquid increases with increase in temperature. For example solubility of sugar in water at 0°C is 179 g /100ml of water whereas at 100°C the solubility is 487g / 100 ml. This means the temperature has a profound effect on the solubility of a substance.
However, solubility of gases in a liquid decreases with the increase in temperature. For example when water is heated the small bubbles come out at the side of beaker. This shows that bubbles are composed of air. Since, air is less soluble in hot water than in cold water, air comes out in the form of bubbles.
ii) Pressure:
The solubilities of solids and liquids are not affected by pressure. But solubility of gas in liquid is affected by the pressure. Solubility of a gas in liquid is directly proportional to the pressure of a gas.
For example, soft drinks like coca cola and 7-up etc are bottled under increased pressure usually greater than 1 atm. When bottles are opened, pressure decreases, solubility of CO2 decreases, so bubbles of CO2 come out of solution. This is called Henry’s law,
m = KP
where m is the amount of gas dissolved.
iii) Nature of Solute and Solvent:
Solute and solvents obey the “like dissolve like” rule. Solute and solvent can be polar and non-polar. Polar and ionic solutes dissolve in polar solvents while non-polar solutes can easily dissolve in non-polar solvents.
For example, NaCl is polar solute so can easily be dissolved in polar water.
Q.4) Explain Why?
a) Common salt dissolves in water but not in petrol.
b) Cold drinks are bottled under a CO2 pressure greater than 1 atmosphere.
c) 100 ml solution of KNO3 can not hold more than 37gm of KNO3 in a dissolved state.
Answer:
a) Solute and solvents obey the like dissolve like a rule. Polar and ionic solutes are dissolved in polar solvents while non-polar solutes can easily dissolve in non-polar solvents. As NaCl is a polar solute, so can easily be dissolved in polar water. But it cannot dissolve in petrol because petrol is non-polar. When we mix them they do not show attractions.
b) Cold drinks like coca-cola and 7-up etc are bottled under increased pressure of CO2 usually greater than 1 atm. At this pressure sufficient amount of CO2 is dissolved in the solution. The solubility of a gas in a liquid is directly proportional to the pressure of a gas.
c) 100 ml solution of KNO3 cannot hold more than 37 g of KNO3 in the dissolved state because there is a limit to how much solute will dissolve in a given volume of solvent. If we add more than 37 g of KNO3 in 100 ml of water, the added salt will not dissolve, no matter how vigorously or how long we dissolve or stir it.
Q.5) Calculate molarity of solution containing 16 gm glucose per 300 ml solution.
Answer:
Data:
Mass of glucose = 16 g
Volume of solution = 300 ml
= 300 / 1000 = 0.3 L
Molarity = ?
Solution:
Molar mass of glucose = 6 × 12 + 12 × 1 + 6 × 16 = 180 g
Molarity = Mass of glucose / (Molar mass of glucose x Volume of solution)
= 16 / (180 x 0.3) = 0.296 M
Q.6) Find the mass of sucrose (Molecular Mass=342) required to be dissolved per 600cm3 solution to prepare a semi-molar solution.
Answer:
Data:
Molecular mass of sucrose = 342 g
Volume of solution = 600 cm3
Mass of sucrose = ?
Solution:
1000 ml of 1 molar solution contain = 342 g
1000 ml of .5 molar solution contain = 342 / 0.5 = 171 g
Thus,
1000 ml of solution contain = 171 g
1 ml of solution contain = 171 / 1000
600 ml of solution contain = 171 / 1000 × 600
= 102.6 g
Mass of sucrose = 102.6g
Thus,
102.6 g of sucrose is required to prepare the semi molar solution.
Q.7) 5.3 gm Na2CO3 was dissolved in 800gm water, calculate the molality of solution.
Answer:
Data:
Mass of Na2CO3 = 5.3 g
Volume of Solvent = 800 g
= 800 / 1000 = 0.8 kg
Molarity = ?
Solution:
Molecular mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 ×16 = 106
Molality = Mass / (Gram formula mass x mass of solvent)
= 5.3 / (106 x 0.8)
= 0.0625 m
Q.8) It is desired to prepare 3 molal solution of NaOH. How much mass of it must be dissolved in 1500gm water.
Answer:
Data:
Molality of solution = 3 m
Volume of solution = 1500 g
= 1500 / 1000 = 1.5 kg
Mass of NaOH = ?
Solution:
Molecular mass of NaOH = 23 + 16 + 1 = 40
Molality = Mass of NaOH / (Molar mass of NaOH x Volume of solution)
Mass of NaOH = 3 x 40 x 1.5 = 18o g
Q.9) Differentiate between
a) Saturated and unsaturated solution,
b) Solution and suspension.
Answer:
a) Saturated and unsaturated solution
| Saturated solution | Unsaturated solution |
| i) The solution contains the amount of solute as the capacity of the solvent. | ii) The solution contains less amount of solute than the capacity of the solvent. |
| ii) Heating is required to dissolve further amount of solute. | ii) Solute can be dissolved without heating. |
| iii) It is dynamic condition. | iii) It is static condition. |
| iv) It is an example of equilibrium. | iv) It is not the example of equilibrium. |
