Careers360 Logo
NCERT Exemplar Class 11 Chemistry Solutions Chapter 5 States of Matter

NCERT Exemplar Class 11 Chemistry Solutions Chapter 5 States of Matter

Edited By Shivani Poonia | Updated on Mar 29, 2025 08:03 PM IST

The chapter States of Matter tells you the fundamental principles that govern the different physical forms in which matter exists—solids, liquids, and gases. The topics like intermolecular forces, kinetic molecular theory, gas laws (Boyle’s law, Charles’ law, Avogadro’s law, and Gay-Lussac’s law), and the ideal gas equation are all covered in this chapter. The chapter will help students understand the behavior of substances under various conditions of temperature and pressure that will provide a strong foundation for advanced concepts in chemistry and physics.

New: Get up to 90% Scholarship on NEET/JEE Coaching from top Coaching Institutes

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

NEET Scholarship Test Kit (Class 11): Narayana | Physics Wallah Aakash | ALLEN

This Story also Contains
  1. NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: MCQ (Type 1)
  2. NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: MCQ (Type 2)
  3. NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Short Answer Type
  4. NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Matching Type
  5. NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Assertion and Reason Type
  6. NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Long Answer Type
  7. NCERT Exemplar Chapter 5 Include The Following Topics:
  8. Main Subtopics in NCERT Exemplar: States of Matter
  9. What will Students Learn From NCERT Exemplar Chapter 5?
  10. NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise
  11. Important Topics To Cover For Exams From Chapter 5: States of Matter
  12. Check NCERT Textbook Chapter Solutions

This article "NCERT Exemplar Class 11 Chemistry solutions Chapter 5 States of Matter" contains detailed and step-by-step solutions to every question that is listed in the exemplar exercise. Students also learn about the deviations from ideal gas behavior, the concept of real gases and the significance of the van der Waals equation. The NCERT Exemplar solutions also explain several important concepts associated with the liquid and the gaseous states of matter.

The exemplar solution further aids learning by offering detailed explanations, numerical problems, and conceptual questions that will help students to apply theoretical knowledge to practical scenarios. Also, the students can access PDF solutions from the official NCERT website for last-minute revisions and effective study sessions.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: MCQ (Type 1)

Question:1

A person living in Shimla observed that cooking food without using pressure cooker takes more time. The reason for this observation is that at high altitude:
(i) pressure increases
(ii) temperature decreases
(iii) pressure decreases
(iv) temperature increases
Answer:

The answer is the option (iii) pressure decreases.
Explanation: Due to low pressure at high altitudes, boiling takes place at low temperature.

Question:2

Which of the following property of water can be used to explain the spherical the shape of rain droplets?
(i) viscosity
(ii) surface tension
(iii) critical phenomena
(iv) pressure
Answer:

The answer is the option (ii) surface tension.
Explanation: We know that for a given volume, a sphere has a lower state of energy and minimum surface area. Hence, the raindrops are spherical.

Question:3

A plot of volume (V ) versus temperature (T ) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in Fig. 5.1. Which of the following order of pressure is correct for this gas?
q3-image
(i) p1 > p2 > p3 > p4
(ii) p1 = p2 = p3 = p4
(iii) p1 < p2 < p3 < p4
(iv) p1 < p2 = p3 < p4
Answer:

The answer is the option (iii) p1<p2<p3<p4
Explanation: ……… (PV = constant, at constant temperature)
Hence, if V1=V2=V3=V4,
Then, P1=P2=P3=P4,

Question:4

The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends upon
(i) charge of interacting particles
(ii) mass of interacting particles
(iii) polarizability of interacting particles
(iv) strength of permanent dipoles in the particles.
Answer:

The answer is the option (iii) polarizability of interacting particles
Explanation: Because, if the magnitude of interaction energy will be greater, then the polarizability of the interacting particles will also be greater.

Question:5

Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘partial charges’. The partial charge is
(i) more than unit electronic charge
(ii) equal to unit electronic charge
(iii) less than unit electronic charge
(iv) double the unit electronic charge
Answer:

The answer is the option (iii) less than unit electronic charge
Explanation: The unit charge is always greater than partial charges.

Question:6

The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?
(i) 0.8×105atm
(ii) 0.008Nm2
(iii) 8×104Nm2
(iv) 0.25atm
Answer:

We know that,
The partial pressure of O2 = Mole fraction of O2 × total pressure of the mixture
=45×1atm=0.8atm
=0.8×105Nm2=8×104Nm2
Hence opiton (iii) is correct

Question:7

As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
(i) increases
(ii) decreases
(iii) remains same
(iv) becomes half
Answer:

The answer is the option (i) increases
Explanation: Kinetic energy of molecules α temperature
Therefore, according to Gay Lussac's law, at constant volume, temperature as well as pressure both increases.

Question:9

What is SI unit of viscosity coefficient (η)?
(i) Pascal
(ii) Nsm–2
(iii) km–2 s
(iv) N m–2
Answer:

The answer is the option (ii) Nsm-2

Question:10

Atmospheric pressures recorded in different cities are as follows:

Cities

Shimla

Bengaluru

Delhi

Mumbai

p in N/m2

1.01 x 105

1.2 x 105

1.02 x 105

1.21 x 105

Consider the above data and mark the place at which liquid will boil first.
(i) Shimla
(ii) Bangalore
(iii) Delhi
(iv) Mumbai
Answer:

The answer is the option (i) Shimla
Explanation: Boiling point of liquid α atmospheric pressure. Shimla has the lowest atmospheric pressure; hence, the liquid will boil first in Shimla.

Question:11

Which curve in Fig. 5.2 represents the curve of ideal gas?
q11-image

(i) B only
(ii) C and D only
(iii) E and F only
(iv) A and B only
Answer:

The answer is the option (i) B only
Explanation: Since we know that PV is constant at all temperatures for an ideal gas, hence the curve B.

Question:12

Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?
(i) Increase
(ii) No effect
(iii) Decrease
(iv) No regular pattern will be followed
Answer:

The answer is the option (iii) Decrease
Explanation: Since kinetic energy increases with temperature, it can overcome intermolecular forces, and liquid starts flowing; hence the viscosity decreases.

Question:13

How does the surface tension of a liquid vary with increase in temperature?
(i) Remains same
(ii) Decreases
(iii) Increases
(iv) No regular pattern is followed
Answer:

The answer is the option (ii) Decreases
Explanation: We know that, Surface tension α 1/temperature & kinetic energy α temperature, hence the intermolecular attraction decreases.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: MCQ (Type 2)

Question:14

With regard to the gaseous state of matter which of the following statements are correct?
(i) Complete order of molecules
(ii) Complete disorder of molecules
(iii) Random motion of molecules
(iv) Fixed position of molecules
Answer:

The answer is the option (ii) Complete disorder of molecules, (iii) Random motion of molecules
Explanation: Entropy of gases is very high; hence, molecules are in random motion and disorderly arranged.

Question:15

Which of the following figures does not represent 1 mole of dioxygen gas at STP?
(i) 16 grams of gas
(ii) 22.7 litres of gas
(iii) 6.022 × 1023 dioxygen molecules
(iv) 11.2 litres of gas
Answer:

The answer is the option (i) 16 grams of gas, (iv) 11.2 litres of gas
Explanation: 1 mole of O2 has 6.023 × 1023 molecules of O2, & at STP it occupies 22.4-litre volume and has a molar mass of 32g.

Question:16

Under which of the following two conditions applied together, a gas deviates most from the ideal behaviour?
(i) Low pressure
(ii) High pressure
(iii) Low temperature
(iv) High temperature
Answer:

The answer is the option (ii) High pressure, (iii) Low temperature
Explanation: =Gases deviate from PV=nRT under high pressure and low temperature, these are real gases, and it deviates from ideal behaviour.

Question:17

Which of the following changes decrease the vapour pressure of water kept in a sealed vessel?
(i) Decreasing the quantity of water
(ii) Adding salt to water
(iii) Decreasing the volume of the vessel to one-half
(iv) Decreasing the temperature of water
Answer:

The answer is the option (ii) Adding salt to the water, (iv) Decreasing the temperature of the water
Explanation: Adding salt to water decreases the surface area for a water molecule to vaporize, and hence vapour pressure of water also decreases.
Now, vapour pressure α temperature, hence, option (iv).

NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Short Answer Type

Question:18

If 1 gram of each of the following gases are taken at STP, which of the gases will occupy (a) greatest volume and (b) smallest volume? CO,H2O,CH4,NO.
Answer:

According to Avogadro's law, we know that at STP, Volume of
1 mole gas = 22.4 L
1g CO = 22.4/28g L
1g H2O = 22.4/18g L
1g CH4 = 22.4/16g L
1g NO = 22.4/30g L
Therefore,
(a) 1g CH4 will occupy maximum volume &
(b) 1g NO will occupy the minimum volume

Question:19

Physical properties of ice, water and steam are very different. What is the chemical composition of water in all the three states.
Answer:

The molecular arrangement of water is different in the solid, liquid and gaseous state; however, the chemical composition is the same in all the states.

Question:20

The behaviour of matter in different states is governed by various physical laws. According to you what are the factors that determine the state of matter?
Answer:

The factors that determine the different states of matter are volume, mass, temperature and pressure.
q-20-image

Question:21

Use the information and data given below to answer the questions (a) to (c):

  • Stronger intermolecular forces result in higher boiling point.

  • Strength of London forces increases with the number of electrons in the molecule.

  • Boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively.

(a) Which type of intermolecular forces are present in the molecules HF, HCl, HBr and HI?
(b) Looking at the trend of boiling points of HCl, HBr and HI, explain out of dipole-dipole interaction and London interaction, which one is predominant here.
(c) Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?

Answer:

(a) Dipole-dipole are present in HCl, HBr, HI and HF; but, in HF along with these Intermolecular hydrogen bonding is also present.
(b)Dipole moment decreases from HCl to HI due to decrease in electronegativity, whereas, from HCl to HI its boiling point increases, hence we can say that London forces are predominant here.
(c) HF has hydrogen bonding as well as the highest electronegativity; hence it has the highest dipole moment as well as the highest boiling point.

Question:22

What will be the molar volume of nitrogen and argon at 273.15K and 1 atm?
Answer:

The conditions given above are of STP where the volume occupies by q mole of gas is 22.4 L. Hence, the volume of N2 and argon, here, will be 22.4 L.

Question:24

Two different gases ‘A’ and ‘B’ are filled in separate containers of equal capacity under the same conditions of temperature and pressure. On increasing the pressure slightly the gas ‘A’ liquefies but gas B does not liquefy even on applying high pressure until it is cooled. Explain this phenomenon.
Answer:

This is based on the phenomenon of critical temperature; here, Gas 'A' liquefies easily, hence, it is below the critical temperature, whereas Gas' B' doesn't liquefy even on applying high pressure; hence it is above the critical temperature.

Question:25

Value of universal gas constant (R) is the same for all gases. What is its physical significance?
Answer:

The eq. of universal gas constant (R) is R = PV/nRT. Therefore, the unit of R will depend on P, V & T viz, Pa m3 K-1 mol-1 or Jmol-1K-1. Since Joule is the unit of work; therefore, R is the work done by gas mole per kelvin.

Question:26

One of the assumptions of the kinetic theory of gases states that “there is no force of attraction between the molecules of a gas.” How far is this statement correct? Is it possible to liquefy an ideal gas? Explain.
Answer:

Since there are no intermolecular forces of attraction between gas molecules of an ideal gas. Hence, the above statement is correct for ideal gases.

Question:27

The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension : water, alcohol (C2H5OH) and hexane [CH3(CH2)4CH3]
Answer:

The molecules above are of water, alcohol, and hexane. Here water and alcohol are polar and have dipole-dipole interaction and intermolecular hydrogen bonding, whereas hexane is non-polar and has weak London dispersion forces.
H-bonding is stronger in water; hence, surface tension is also more in water than alcohol. Therefore, Hexane < Alcohol < Water.

Question:28

Pressure exerted by saturated water vapour is called aqueous tension. What correction term will you apply to the total pressure to obtain pressure of dry gas?
Answer:

Let us consider that at temperature 'T' pressure of moist gas is 'P'; then, the pressure of dry gas can be obtained by the equation:
PDrygas=PTotal - aqueous tension

Question:29

Name the energy which arises due to motion of atoms or molecules in a body. How is this energy affected when the temperature is increased?
Answer:

Thermal energy arises due to the motion of atoms or molecules in a body. It is a measure of kinetic energy. Thermal energy α temperature.

Question:30

Name two intermolecular forces that exist between HF molecules in liquid state.
Answer:

Hydrogen bond and dipole-dipole interactions are the two forces existing in HF molecules when they are in a liquid state.

Question:31

One of the assumptions of kinetic theory of gases is that there is no force of attraction between the molecules of a gas.State and explain the evidence that shows that the assumption is not applicable for real gases.
Answer:

The assumption is not applicable to real gases because they can be liquified by cooling and applying high pressure. Hence, the force of attraction exists between the molecules of real gases.

Question:32

Compressibility factor, Z, of a gas is given as Z = (pV/nRT)
(i) What is the value of Z for an ideal gas?
(ii) For real gas what will be the effect on value of Z above Boyle’s temperature?

Answer:

(i) Z is a compressibility factor. Its value for an ideal gas is Z = 1.
(ii) Real gases show positive deviation above Boyle's temperature, i.e., Z>1.

Question:33

The critical temperature (Tc) and critical pressure ( pc) of CO2 are 30.98°C and 73 atm respectively. Can CO2 (g) be liquefied at 32°C and 80 atm pressure?
Answer:

It does not matter that how high temperature or high-pressure is CO2 at; it cannot be liquified above 30.98o and 73 atm. Hence, it cannot be liquified at 32o and 80 atm pressure.

Question:34

For real gases the relation between p, V and T is given by van der Waals equation:
(p+an2v2)(vnb)=nRT
where ‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas. ‘a’ is the measure of magnitude of intermolecular attraction.
(i) Arrange the following gases in the increasing order of ‘b’. Give reason.O2,CO2,H2,He
(ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason.CH4,O2,H2


Answer:

(i) We know that volume α size of the molecules. Hence, the increasing order of the value of 'b' will be:
H2<He<O2<CO2
(ii) 'a' is the Van der Waal's constant which represents the magnitude of intermolecular attraction. An α size of the electron cloud. Hence, the greater the size of the electron cloud, the greater will be the dispersion forces and polarizability of the molecule. Hence, the gases in decreasing order of the magnitude of 'a' will be: CH4>O2>H2

Question:35

The relation between pressure exerted by an ideal gas (pideal) and observed pressure (preal) is given by the equation Pideal =Preal+(an2/V2)
If pressure is taken in Nm–2, number of moles in mol and volume in m3, Calculate the unit of ‘a’.
What will be the unit of ‘a’ when pressure is in atmosphere and volume in dm3?

Answer:

It is given that Pideal=Preal+an2/V2
We know that the unit of p = Nm-2, unit of V = m3, number of moles(n) = mol
Unit of ‘a’ = Nm-2(m3)2/(mol)2 = Nm4mol-4
If we take the unit of p = atm, V = dm3 and n = mol
Then unit of ‘a’ = pV2/n2 = atm(dm3)2/(mol)2 = atm dm6mol-2.

Question:36

Name two phenomena that can be explained on the basis of surface tension.
Answer:

The two phenomena that can be explained on the basis of surface tension are-
(i) The spherical shape of liquid drops
(ii) Capillary action i.e. rise and dip of a liquid in the column of a capillary.

Question:37

Viscosity of a liquid arises due to strong intermolecular forces existing between the molecules. Stronger the intermolecular forces, greater is the viscosity. Name the intermolecular forces existing in the following liquids and arrange them in the increasing order of their viscosities. Also give reason for the assigned order in one line. Water, hexane (CH3CH2CH2CH2CH2CH3), glycerine (CH2OHCH(OH)CH2OH).
Answer:

(a) Hexane has Vander Waal force of attraction present as the intermolecular force.
(b) Water has hydrogen bonding present in the form of intermolecular forces
(c) Glycerin has hydrogen bonding present as the intermolecular force
Hexane < water < glycerine
The reason behind this order is that in hexane, weak bonding forces exist; therefore, it has the least viscosity, whereas, in water and glycerine dipole-dipole interaction and extensive H-bonding are present. Glycerine has the strongest intermolecular forces. Therefore, the order of viscosity is-

Question:38

Explain the effect of increasing the temperature of a liquid, on intermolecular forces operating between its particles, what will happen to the viscosity of a liquid if its temperature is increased?
Answer:

When there is an increase in the temperature, the intermolecular force which operates between the particle reduces, subsequently the strength of the bond increases along with the kinetic energy. Thus, when there is an increase in the temperature, there is a reduction in the viscosity. This is a result of the fact that viscosity decreases when there is reduction in the intermolecular forces.

Question:39

The variation of pressure with the volume of the gas at different temperatures can be graphically represented as shown in Fig. 5.3. Based on this graph answer the following questions.

1660110821933Answer:

(i) According to Boyle's law, The pressure of gas α 1/volume of gas, at a constant temperature.
Therefore, if volume decreases at a constant temperature, then the pressure increases and vice versa. For example, at 200K if pressure increases from p1 to p2, volume decreases from v­1 to v2(v2 < v1)
(ii) According to Charles's law, The volume of gas α temperature, at constant pressure.Therefore, if temperature decreases from 200K to 400K, at a constant temperature, the volume of gas increases.

Question:40

Pressure versus volume graph for a real gas and an ideal gas is shown in Fig. 5.4. Answer the following questions based on this graph.
q-40-image
(i) Interpret the behaviour of real gas with respect to an ideal gas at low pressure.
(ii) Interpret the behaviour of real gas with respect to an ideal gas at high pressure.
(iii)Mark the pressure and volume by drawing a line at the point where real gas behaves as an ideal gas.

Answer:

(i) Real gases show a very small deviation from ideal behaviour at low pressure because both the curves coincide with each other.
(ii) Real gases show large deviation at high pressure as the curves fall apart.
(iii)The point at which the curves intersect each other shows that the real gas behaves like an ideal gas.
q-40-image

NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Matching Type

Question:41

Match the graph between the following variables with their names.

Column I (Graphs)

Column II (Names)

(i) Pressure vs temperature graph at constant molar volume.

(a) Isotherms

(ii) Pressure vs volume graph at constant temperature.

(b) Constant temperature curve

(iii) Volume vs temperature graph at constant pressure.

(c) Isochores


(d) Isobars

NEET/JEE Offline Coaching
Get up to 90% Scholarship on your NEET/JEE preparation from India’s Leading Coaching Institutes like Aakash, ALLEN, Sri Chaitanya & Others.
Apply Now

Answer:

(i)→(c); (ii) →(a); (iii) →(d)
Explanation:
(i) Constant molar volume graphs are called isochore.
(ii) Constant temperature graphs are called isotherm.
(iii) Constant pressure graphs are called isobar.

Question:42

Match the following gas laws with the equation representing them.

(i) Boyle’s law
(ii) Charle’s law
(iii) Dalton’s law
(iv) Avogadro law


(a) V ∝ n at constant T and p
(b) pTotal = p1 + p2 + p3 +…… at constant T, V
(c) PV/T = Constant
(d) V ∝ T at constant n and p
(e) p ∝ 1/V at constant n and T

Answer:

(i) →(e); (ii) →(d); (iii) →(b); (iv) →(a)
Explanation:
(i) pressure of gas α 1/volume of gas (T constant) ….. (Boyle's law)
(ii) The volume of gas α temperature of the gas, at constant pressure …… (Charles's law)
(iii) At constant V and T, the total pressure of the gaseous mixture of two or more non-reacting gases is equal to the algebraic sum of their partial pressures ……. (Dalton's law)
(iv) Under same temperature and pressure conditions, equal volumes of all gases contain equal no. of moles. V α n. ……. (Avogadro's law)

Question:43

Match the following graphs of ideal gas with their co-ordinates :

Graphical representation x and y co-ordinates
capture-43

Answer:

(i) →(b); (ii) →(c); (iii) →(a)
Explanation:
(i) pressure of gas α 1/volume of gas (T constant)
(ii) The pressure of gas α 1/V
(iii) Product of pressure and volume is constant. PV = constant

NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Assertion and Reason Type

Question:44

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Three states of matter are the result of balance between intermolecular forces and thermal energy of the molecules.
Reason (R): Intermolecular forces tend to keep the molecules together but thermal energy of molecules tends to keep them apart.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (i) Both A and R are right, and R is the correct explanation of A.
Explanation: Balance is required in both intermolecular forces and thermal energy to decide the state of matter.

Question:45

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): At constant temperature, pV vs V plot for real gases is not a straight line.
Reason (R) : At high pressure all gases have Z > 1 but at intermediate pressure most gases have Z < 1.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (ii) Both A and R are true but R is not the correct explanation of A.

Question:46

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): The temperature at which vapour pressure of a liquid is equal to the external pressure is called boiling temperature.
Reason (R) : At high altitude atmospheric pressure is high.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (iii) A is true, but R is false.
Explanation: Boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the external pressure. As the altitude is high, the pressure is low.

Question:47

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Gases do not liquefy above their critical temperature, even on applying high pressure.
Reason (R) : Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (i)Both A and R are true, and R is the correct explanation of A.
Explanation: Gases do not liquefy even on applying high pressure if they are above the critical temperature as the Intermolecular forces of attraction cannot hold the molecules together, and the molecular speed is also high.

Question:48

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): At critical temperature liquid passes into gaseous state imperceptibly and continuously.
Reason (R) : The density of liquid and gaseous phase is equal to critical temperature.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the optio (i) Both A and R are true, and R is the correct explanation of A.
Explanation: Density of liquid becomes equal to its vapour phase at a critical temperature which causes the liquid to change into gaseous state imperceptibly and continuously.

Question:49

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Liquids tend to have maximum number of molecules at their surface.
Reason (R) : Small liquid drops have spherical shape.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (iv) A is false, but R is true.
Explanation: Liquid drops have a spherical shape as they try to reduce the no. of molecules and tend to reduce surface tension at their surface.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 5: Long Answer Type

Question:50

Isotherms of carbon dioxide at various temperatures are represented in Fig. 5.5. Answer the following questions based on this figure.
capture-50
(i) In which state will CO2 exist between the points a and b at temperature T1?
(ii) At what point will CO2 start liquefying when temperature is T1?
(iii) At what point will CO2 be completely liquefied when temperature is T2.
(iv) Will condensation take place when the temperature is T3.
(v) What portion of the isotherm at T1 represent liquid and gaseous CO2 at equilibrium?

Answer:

(i) Between the points, a and b, CO2 will exist in a gaseous state at temperature T1.
(ii) At point b there is phase transition, and the plot becomes linear.
(iii) At Temperature T2, CO2 will be completely liquefied at point g.
(iv) T3 > Tc. At T3 temperature, condensation will not take place.
(v) Liquid and gaseous CO2 are in equilibrium between b and c.

Question:52

Why does the boundary between liquid phase and gaseous phase disappear on heating a liquid upto critical temperature in a closed vessel? In this situation what will be the state of the substance?
Answer:

The density of the liquid and that of vapour becomes equal at a critical point due to which there is no boundary of separation left. These are called superficial fluids, and they dissolve in many organic substances.

Question:53

Why does sharp glass edge become smooth on heating it to its melting point in a flame? Explain which property of liquids is responsible for this phenomenon.
Answer:

It is possible due to the surface tension. On applying heat, the glass starts to melt and takes a rounded shape at edges having minimum surface area. This is known as the fire polishing of glass.

Question:54

Explain the term ‘laminar flow’. Is the velocity of molecules the same in all the layers in laminar flow? Explain your answer.
Answer:

Laminar flow can also be defined as the flow in which there is a regular gradation of velocity in passing from one layer to the next.
When a liquid flows over a fixed surface, it comes in contact with it. The layer of molecules in the immediate contact of the surface is stationary. As the distance of layers increases from the fixed layer, the velocity of the upper layer also increases.
capture-54

Question:55

Isotherms of carbon dioxide gas are shown in figure. Mark a path for changing gas into liquid such that only one phase (i.e. either a gas or liquid) exists at any time during the change. Explain how the temperature, volume and pressure should be changed to carry out the change.
capture-55

Answer:

In the figure above, we can increase the temperature to move from A to F vertically and can reach G by compressing the gas at constant temperature along isotherm, thus, increasing the pressure. Then by lowering the temperature, we can move vertically downwards to D. We get liquid as soon as we cross the point H on isotherm. The substances will remain in one phase if we carry out this process at a critical temperature. This is called continuity of state between the gaseous and liquid states.

The topics covered in Class 11 Chemistry NCERT exemplar solutions chapter 5 include gas laws, the ideal gas equation, intermolecular forces the kinetic theory of gases. Candidates can avail the NCERT Exemplar Class 11 Chemistry solutions chapter 5 PDF download feature to get the solutions offline. It is convenient for learning and understanding the subject better.

NCERT Exemplar Chapter 5 Include The Following Topics:

  • Numerical problems based on Boyle’s law, Charles’s law, Gay-Lusscac’s law and Avogadro’s law.

  • Numerical problems on calculating partial pressure.

  • Questions on critical temperature and pressure.

  • Questions on Van der Waals forces and other types of intermolecular forces.

Main Subtopics in NCERT Exemplar: States of Matter

  • Intermolecular Forces
  • Dispersion Forces or London Forces
  • Dipole-dipole Forces
  • Dipole-induced Dipole Forces
  • Hydrogen Bond
  • Thermal Energy
  • Intermolecular Forces Vs Thermal Interactions
  • The Gaseous State Ex
  • The Gas Laws Ex
  • Boyle’s Law (Pressure-volume Relationship)
  • Charles’ Law (Temperature-volume Relationship)
  • Gay Lussac’s Law (Pressure-temperature Relationship)
  • Avogadro Law (Volume – Amount Relationship)
  • Ideal Gas Equation
  • Density and Molar Mass of a Gaseous Substance
  • Dalton’s Law of Partial Pressures
  • Kinetic Molecular Theory of Gases
  • Behaviour Of Real Gases: Deviation from Ideal Gas Behaviour
  • Liquefaction of Gases
  • Liquid State
  • Vapour Pressure
  • Surface Tension and Viscosity.

What will Students Learn From NCERT Exemplar Chapter 5?

The chapter will help enable the students to understand why states of matter is one of the important chapters in chemistry. The topics included in the NCERT Exemplar solutions chapter 5 will help the students in understanding the different concepts related to the state of matter.

Solving the questions in this exercise will help the students in assessing the different topics and understanding them in a better way. The solutions will provide students the information related to different laws and various kinds of intermolecular forces, thermal energy, and the gaseous state, etc.

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise















Important Topics To Cover For Exams From Chapter 5: States of Matter

Given below are some the important topics that will help the students in studying significantly for the examination:

  • The students will understand why states of matter is one of the important chapters in chemistry. The students will get to learn different concepts related to the state of matter. There are three states: solid, liquid, and gas. So, the topics will help the students in understanding all three forms more accurately.
  • The topics covered in this section are intermolecular forces, thermal energy, gas laws, the ideal gas equation, the kinetic theory of gases, the deviation of gases from ideal behavior, liquefaction of gases, and the liquid state.

Check NCERT Textbook Chapter Solutions

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - Thermodynamics

Chapter-6 - Equilibrium

Chapter-7 - Redox Reaction

Chapter-8 - Organic chemistry- some basic principles and techniques

Chapter-9 - Hydrocarbons

Do follow the links provided in the table to get the idea about the syllabus and the respective books. You can also take a look at subject-wise notes and solutions to make your exam preparations better.


NCERT Exemplar Class 11 Solutions

Follow the links provided in the table below to get hands on exemplar solutions of other subjects as well-


Frequently Asked Questions (FAQs)

1. What determines the state of matter of a substance?

Primarily, the temperature and pressure influence the state of matter. Higher temperatures generally favor states with higher kinetic energy (gases), while higher pressures favor more compact states (solids or liquids). The strength of intermolecular forces also plays a crucial role.


2. What are intermolecular forces (IMFs), and why are they important for understanding states of matter?

Intermolecular forces are the attractive or repulsive forces between molecules. They are weaker than intramolecular forces (e.g., covalent bonds within a molecule). IMFs are critical because they determine how strongly molecules with each other, influencing melting points, boiling points, viscosity, and surface tension. Stronger IMFs generally lead to higher melting and boiling points.


3. How does temperature affect the viscosity of a liquid?

Increasing temperature generally decreases the viscosity of a liquid because the increased kinetic energy allows molecules to move past each other more easily, overcoming the intermolecular forces.


4. What is the difference between boiling and evaporation?

Both involve liquid changing to gas, but,

Boiling: Occurs throughout the liquid when the vapor pressure equals the external pressure. Requires a specific temperature (boiling point). Evaporation: Occurs only at the surface of the liquid at any temperature.


5. What is an ideal gas? Do real gases behave ideally all the time?

An ideal gas is a hypothetical gas that perfectly obeys the assumptions of the KMT. Real gases deviate from ideal behavior, especially at high pressures and low temperatures, because under these conditions: The volume of the gas particles becomes significant compared to the total volume. Intermolecular forces become non-negligible.

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top