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Important Questions for CBSE Class 11 Chemistry Chapter 5 - States of Matter

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Last updated date: 13th Jul 2024
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CBSE Class 11 Chemistry Chapter-5 Important Questions - Free PDF Download

Important Questions Of Chapter 5 Chemistry Class 11 States of Matter Gases & Liquids are created by subject matter experts at Vedantu. The main concept behind creating the important questions for the chapter is to enable students to prioritize the concepts of the chapter introduced in Chemistry class 11. Ch 5 Chemistry Class 11 Important Questions are prepared to give a better conceptual understanding. With the help of Class 11 Chemistry Chapter 5 Important Questions, it would be easy for students to predict the type of questions to be asked in the examination. The language used while preparing the important questions is proficient and understandable by a Class 11 student.


Download CBSE Class 11 Chemistry Important Questions 2024-25 PDF

Also, check CBSE Class 11 Chemistry Important Questions for other chapters:

CBSE Class 11 Chemistry Important Questions

Sl.No

Chapter No

Chapter Name

1

Chapter 1

Some Basic Concepts of Chemistry

2

Chapter 2

Structure of Atom

3

Chapter 3

Classification of Elements and Periodicity in Properties

4

Chapter 4

Chemical Bonding and Molecular Structure

5

Chapter 5

States of Matter

6

Chapter 6

Thermodynamics

7

Chapter 7

Equilibrium

8

Chapter 8

Redox Reactions

9

Chapter 9

Hydrogen

10

Chapter 10

The s-Block Elements

11

Chapter 11

The p-Block Elements

12

Chapter 12

Organic Chemistry - Some Basic Principles and Techniques

13

Chapter 13

Hydrocarbons

14

Chapter 14

Environmental Chemistry

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Study Important Questions for Class 11 Chemistry Chapter 5- States of Matter

Very Short Answer Questions                                                                                           1 Mark

1. Define Van der waals forces.

Ans: Van der Waals forces are very small attractive forces that exist between the molecules of a substance.


2. Give an example to show dipole-dipole forces.

Ans: Water molecules possess dipole-dipole forces. The partial positive end of hydrogen attracts the partially negative end of the oxygen molecule.


3.What type of bond exists between ${{H}_{2}}O,HF,N{{H}_{3}},{{C}_{2}}{{H}_{5}}OH$ molecule?

Ans: Since all of the molecules contain hydrogen, as well as an electronegative atom attached, hence the molecules will show hydrogen bonding.


4. Define Boyle’s law.

Ans: According to Boyle’s Law, the pressure of a gas has an inverse relationship with the volume, provided the temperature and number of moles of the gas are fixed.


5. Why are helium and hydrogen gases not liquefied at room temperature by applying very high pressure?

Ans: They have a critical temperature that is lower than the room temperature. For the liquefaction of gases, the temperature must be below the critical temperature.


6. How is the pressure of a given sample of a gas related to the temperature at volume?

Ans: \[Pressure\:\alpha \,\:Temperature\]


7. Define absolute zero temperature.

Ans: It is the temperature, where, theoretically, gases are said to occupy zero volume.


8. Define an ideal gas.

Ans: An ideal gas is a gas that follows the ideal gas equation($PV=nRT$), on any value of temperature and pressure.

9. What is aqueous tension?

Ans: Pressure exerted on the walls of a container, by the saturated water vapour is called aqueous tension.


10. What is the value of R at STP?

Ans:  $R=8.20578\times {{10}^{-2}}L\,atm\,{{K}^{-1}}\,mo{{l}^{-1}}$


11. Molecule A is twice as heavy as molecule B. Which of these has higher kinetic energy at any temperature?

Ans: As kinetic energy is dependent on mass, so Molecule A will have more kinetic energy.

($K.E=\dfrac{1}{2}m{{v}^{2}}$). 


12. Write the Van der Waals equation for n moles of a gas.

Ans:  $(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$


13. Out of $N{{H}_{3}}$ and ${{N}_{2}}$, which will have (i) larger value of ‘a’ and (ii) larger value of ‘b’?

Ans: (i) $N{{H}_{3}}$, due to hydrogen bonding

(ii) ${{N}_{2}}$, due to large molecule size.


14. What property of molecules of real gases is indicated by van der Waals constant ‘a’?

Ans: Attraction between the molecules, or intermolecular attraction.


15. Under what conditions do real gases tend to show ideal gas behaviour?

Ans: High temperature, Low Pressure.


16. How are Van der waals constants ‘a’ and ‘b’ related to the tendency to liquefy?

Ans: “a” measures intermolecular attractions, more value of “a” means more attractive forces, which means it is easier to liquefy. “b” measures volume, less the volume or the value of “b”, the easier it is to liquefy.


17. When does a gas show ideal behaviour in terms of volume?

Ans: When the volume becomes so large that we can neglect the molecules that are associated in space.


18. Define Boyle point.

Ans: Boyle point is the temperature at which a real gas obeys ideal gas law over an appreciable range of pressure.


19. Define standard boiling point.

Ans: It is the boiling point of a liquid, when the atmospheric pressure is 1 bar.


20. What is surface energy?

Ans: Surface energy is the energy required to expand the surface area of the liquid by one unit.


21. What is surface tension? What is its S.I unit?

Ans: It is a phenomenon that occurs when a liquid's surface comes into contact with another phase (it can be a liquid as well). Liquids prefer to have as little surface area as feasible. The liquid's surface behaves like an elastic sheet. Its SI unit is $N{{m}^{-1}}$.


22. How does surface tension change when the temperature is raised?

Ans: On the increasing temperature, surface tension decreases.


23. Why is glycerol highly viscous?

Ans: It is because of the presence of hydrogen bonding, which is a strong attractive force(${{C}_{3}}{{H}_{8}}{{O}_{3}}$).


24. Some tiny light hollow spheres are placed in a flask. What would happen to these 4 spheres, if the temperature is raised?

Ans: As the temperature is raised, the kinetic energy will increase and the spheres will follow zigzag Brownian motion, and collide with themselves and the walls of the flask.


25. The boiling points of a liquid rises on increasing pressure. Give a reason.

Ans: A liquid boils when $V.P=atm\,pressure$. So, on increasing pressure, more energy is needed for the molecules to escape, hence, the boiling point increases.


Short Answer Questions                                                                                                    2 Marks

1. Ice has a lower density than water. Give a reason.

Ans: It is because of the presence of different types of hydrogen bonding in both of the compounds. Ice consists of intermolecular hydrogen bonding and a cage-like 3D structure. When ice melts, some hydrogen bonds break, allowing water molecules to fill the empty gaps. The molecules in liquid water are more densely packed than those in ice. As a result, ice is denser than water.


2. Water has maximum density at ${{4}^{0}}C$. Give a reason.

Ans: The density of the liquid increases as the water molecules become closer together. The water molecules slow down and the density increases when the temperature of warm water drops. As a result, at 4 °C, the density of water reaches its maximum.


3. Define thermal energy.

Ans: We know that atoms and molecules are in constant motion. The energy which is liberated by virtue of this motion is called thermal energy.

As the temperature of a substance increases, the molecules gain more energy and their speed increases, which means, the thermal energy also increases.


4. What are the factors responsible for the strength of hydrogen bonds?

Ans: The strength of the hydrogen bond depends on the distance between the lone pair of negatively charged electrons of the electronegative atom, and the partially positively charged hydrogen atom of the other molecule.


5. At what temperature will the volume of a gas at ${{0}^{0}}C$ double itself, pressure remaining constant?

Ans: Let us take the volume at absolute zero to be $V\,ml$. So, we have:

$  {{V}_{1}}=V\,ml $

 $ {{V}_{2}}=2V\,ml $

 $ {{T}_{1}}=0+273=273K $

 $ {{T}_{2}}=? $

Now, in order to find the temperature, we will use Charles's Law:

$\dfrac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}$ 

 $ \Rightarrow {{T}_{2}}=\dfrac{{{V}_{2}}\times {{T}_{1}}}{{{V}_{1}}}$ 

 $\Rightarrow {{T}_{2}}=\dfrac{2V\times 273}{V} $

 $\Rightarrow {{T}_{2}}=546K $ 

 Now the temperature in Kelvin is  ${{(546-273)}^{0}}C={{273}^{0}}C$.


6. 50 $c{{m}^{3}}$of hydrogen gas enclosed in a vessel maintained under a pressure of 1400 Tor, is allowed to expand to 125 $c{{m}^{3}}$ under constant temperature conditions. What would be its pressure?

Ans: The parameters are:

$  {{P}_{1}}=1400$

 $ {{P}_{2}}=? $

 ${{V}_{1}}=50c{{m}^{3}}$

 ${{V}_{2}}=125\,c{{m}^{3}} $

 Now, we will use ideal gas equation to find the required pressure:

$ {{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$ 

 $ \Rightarrow {{P}_{2}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{V}_{2}}}$ 

 $ \Rightarrow {{P}_{2}}=\dfrac{1400\times 50}{125}$

 $\Rightarrow {{P}_{2}}=560\,Torr $

So, the value of the required pressure is 560 Torr.


7. State the law depicting the volume-temperature relationship.

Ans: The law depicting the volume-temperature relationship is called the Charles Law. It states that when pressure remains constant, then the volume of a given mass of a gas changes by $\dfrac{1}{273}$ of its volume at absolute zero for every one degree centigrade or fall in temperature. Mathematically:

$   {{V}_{t}}={{V}_{0}}+\dfrac{{{V}_{0}}}{273}$

$ \Rightarrow {{V}_{t}}={{V}_{0}}(\dfrac{273+t}{273})$

${{V}_{t}}$ is volume of gas at ${{t}^{0}}C$, ${{V}_{0}}$ is volume at absolute zero.


8. State Avogadro’s Law. Is the converse of Avogadro’s law true?

Ans: According to Avogadro’s law, all gases under the same conditions of pressure and temperature contain the same number of molecules, provided their volumes are equal. For example, molar mass of oxygen molecule is $32g\,mo{{l}^{-1}}$, and that of nitrogen molecule is $28g\,mo{{l}^{-1}}$. So, when the different masses are taken respectively, then we find that both of the molecules have the same number of particles, which is $6.022\times {{10}^{23}}$.


9. Deduce the relation $PV=nRT$where R is a constant called universal gas constant.

Ans: Let us recall the 3 gas laws:

  $ V\alpha \dfrac{1}{p}\,(T,n\,\,const,Boyle's\,Law) $

 $V\alpha T(P,n\,\,const,\,Charle's\,Law) $ 

 $ V\alpha \,n(T,P\,const,\,Avogadro's\,Law) $

On combing the three laws, we have:

$   V\alpha \,n\times \dfrac{1}{P}\times T $

 $ \Rightarrow V\alpha \dfrac{nT}{P} $

 $\Rightarrow PV=nRT $


10. At ${{25}^{0}}C$ and 760 mm of Hg pressure a gas occupies 600ml volume. What will be its pressure at a height where the temperature is ${{10}^{0}}C$ and volume of the gas is 640mL. Calculate the volume occupied by 5.0 g of acetylene gas at ${{50}^{0}}C$ and 740mm pressure.

Ans: The parameters are:

$   {{P}_{1}}=760mm $

$ {{V}_{1}}=600mL $

 $ {{T}_{1}}=25+273=298K $

 ${{V}_{2}}=640mL $ 

 ${{T}_{2}}=10+273=283K$ 

As $PV$ is constant, so, from the combined as equation, we have:

 $  \dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$

 $ \Rightarrow{{P}_{2}}=\dfrac{{{P}_{1}}{{T}_{2}}{{V}_{1}}}{{{T}_{1}}{{V}_{2}}}$

 $\Rightarrow {{P}_{2}}=\dfrac{760\times 283\times 600}{640\times 298} $

 $\Rightarrow {{P}_{2}}=676.6\,\,mm $


11. Explain how the function $PV/RT$ can be used to show gases behave non-ideally at high pressure.

Ans: According to the ideal gas equation, $PV=nRT$, so the ratio given above shows us the number of moles of an ideal gas.

Now, on changing the pressure, there should be no effect on the number of moles, as it is fixed for every ideal gas. However, if it changes, it means that the gas is not behaving like an ideal gas.


12. Mention the two assumptions of the kinetic theory of gases that do not hold good.

Ans: The two assumptions of the kinetic theory that do not hold good are:

  • Molecules of a gas do not have any force of attraction between them.

  • The volume of the molecules of a gas is negligibly small when compared to the space which is occupied by the gas.


13. Calculate the pressure exerted by one mole of $C{{O}_{2}}$ at 273K if the Van der waals constant a =$3.592\,d{{m}^{6}}\,atm\,mo{{l}^{-1}}$. Assume that the volume occupied by $C{{O}_{2}}$ molecules is negligible.

Ans: We will use Van der Waals equation to solve this. As volume occupied by the molecules is negligible we can take $b=0$. The parameters are:

$ a=3.592\,d{{m}^{6}}\,atm\,\,mo{{l}^{-1}} $

$ V=22.4\,d{{m}^{3}} $

 $ R=0.082L\,atm\,{{K}^{-1}}\,mo{{l}^{-1}} $

$ T=273K $

Now, as b is 0, so the modified equation is:

$  P=\dfrac{RT}{V}-\dfrac{a}{{{V}^{2}}}$

 $\Rightarrow P=\dfrac{0.082\times273}{22.4}-\dfrac{3.592}{{{(22.4)}^{2}}} $

 $ \Rightarrow P=0.0003-0.0071$

 $\Rightarrow P=0.9922\,atm$ 


14. Why does viscosity of liquids decrease as the temperature is raised?

Ans: As the temperature of liquids is raised, the molecules gain more energy and tend to break the intermolecular bonds. The volume increases and bond weakens, and the molecules move away from each other, which reduces the density, as well as, decreases the viscosity.


15. What is the effect of temperature on (i) density (ii) vapor pressure of a liquid?

Ans: 

  • $Density=\dfrac{Mass}{Volume}$, so as volume is directly proportional to temperature, and inversely proportional to density, it means that density is inversely proportional to temperature.

  • As the temperature of a liquid is increased, the molecules gain more kinetic energy and their speed increases. This means, with a higher speed, they can exert more force per unit area on the wall of the container. This means that vapour pressure is directly proportional to temperature.


Important Questions of Chapter 5 Chemistry Class 11

Class 11 Chemistry Chapter 5 Important Questions - Important Topics

Topics covered under Ch 5 Chemistry Class 11 Important Questions are Intermolecular Forces, Thermal Energy, Intermolecular Forces vs Thermal Interactions, The Gaseous State, The Gas Laws,  Ideal Gas Equation, Kinetic Molecular Theory of Gases, Behaviour of real gases: Deviation from Ideal Gas Behaviour, Liquefaction of Gases,  Liquid State In this chapter, we also get to know about the main difference between a real gas equation and Ideal gas equation.

From the mentioned topics many different questions can be framed like Questions on critical temperature and pressure, numerical problems based on Boyle’s law, Charles’s law, Gay-Lusscac’s law, and Avogadro’s law, numerical problems on calculating partial pressure, etc can be asked in Class 11 Chemistry Chapter 5 Important Questions.


Benefits of Chemistry Class 11 Chapter 5 Important Questions

Ch 5 Chemistry Class 11 Important Questions are made available in PDF format. This chapter includes several fundamental concepts related to intermolecular forces and how they affect the physical state of a substance. This topic also deals with several other important concepts associated with the liquid and gaseous states of matter. This is the reason why “States of Matter” is regarded as one of the most important chapters in the Class 11 Chemistry textbook.


Conclusion

Students can also find the Class 11 Chemistry Chapter 5 Extra Questions on the Vedantu website, which has been prepared by experts to help students in securing a good score in their annual exam preparation. The topic includes 1 mark, 2 marks, 3 marks, 4 marks, 5 marks, and other questions as per the latest CBSE curriculum for the current session and various competitive examinations like IIT/JEE, NEET, and other competitive exams. Chapter 5 has a good weightage in the final exam. By revising the concepts and solving numericals of chapter 5, you can achieve almost full marks from this chapter.


Important Related Links for CBSE Class 11 Chemistry

FAQs on Important Questions for CBSE Class 11 Chemistry Chapter 5 - States of Matter

1. How to study Chapter 5 efficiently?

Class 11 Chemistry Chapter 5 "States of Matter" has some important definitions, properties, formulas, and numerical questions.

To study this chapter efficiently,

  • Read the NCERT textbook multiple times.

  • Make concise notes of all the important points and formulas.

  • Practice the NCERT numerical questions thoroughly.

  • Refer to Vedantu's important questions and revision notes for this chapter.

  • Practice questions from sample papers and previous years’ question papers. 

2. What are the important topics of Class 11  Chemistry Chapter 5?

Chapter 5-"The States of Matter"  has the following important topics:

  • Intermolecular forces

  • Van der Waals forces

  • London forces

  • Hydrogen bond

  • Thermal energy

  • Intermolecular forces vs Thermal interactions

  • Boyle's laws

  • Charles law

  • Gay Lussac’s Law

  • Avogadro Law 

  • Ideal gas  equation

  • Density and Molar Mass of a Gaseous Substance 

  • Liquefaction of gases

  • Kinetic molecular theory of gases

  • The behaviour of real gases

All the numerical questions are also very important. And to read about all these topics students can download the vedantu app.

3. Is Chapter 5 of Chemistry Class 11 important?

"States of Matter" is a significant chapter in Class 11 Chemistry.  Along with "Chemical Bonding and Molecular Structure,"  "Chemical Thermodynamics", and "Equilibrium”, this chapter carries a total of 21 marks in exams. Chapter 5 has many important topics, properties, definitions, and explanations. Added to that, several numerical questions can be asked from this chapter. To refer to some important questions from this chapter, visit the page  Important Questions for Class 11 Chemistry and download the PDf free of cost.

4. Define Boyle’s point.

The pressure of a fixed amount of gas, with constant temperature, varies inversely with its volume. This is known as Boyle’s law as discovered by Robert Boyle. Boyle point is the exact temperature at which a real gas obeys ideal gas law over a reasonable range of pressure. More such important questions from "States of Matter" can be accessed through the page  Important Questions for Class 11 Chemistry. These can also be downloaded at no cost. 

5. What factors determine the strength of Hydrogen bonds?

The coulombic interaction between the lone pair of electrons of the electronegative atom of one molecule and the hydrogen atom of the other molecule determines the strength of the hydrogen bonds. Many such important questions are listed in Vedantu's Important Questions for Class 11 Chemistry Chapter 5. These questions are handpicked by our experts who have categorized the questions as per the marks in the examinations. Hence, you can find questions of 1,2, and 5 marks answered suitably. All the study material can be accessed through the vedantu app.