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NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

Edited By Vishal kumar | Updated on Jun 24, 2025 10:44 AM IST

Kinetic Theory: Class 11 Physics Chapter 12 is one of the important chapters in Physics of Class 11 and candidates appearing on CBSE Board Examination, JEE, NEET, and other tests are advised to go through this Topics. We have developed NCERT Solutions - Class 11 Physics Chapter 12 Kinetic Theory that will help at the time of exams. The experts of the subject have elaborated the explanation to illustrate the concept to the students which are difficult to understand in Class 11 NCERT Solutions. These NCERT solutions are in accordance with the new CBSE only curriculum and would enable the students to confidently endeavor both the theoretical and numerical questions.

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This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 12: Download PDF
  2. NCERT Solutions for Class 11 Physics Kinetic Theory - Exercise Questions
  3. NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Additional Questions
  4. Class 11 physics NCERT Chapter 12: Higher Order Thinking Skills (HOTS) Questions
  5. NCERT Chapter 12 Kinetic Theory Topics
  6. Kinetic Theory Chapter 12 Physics NCERT: Important Formulae
  7. Approach to Solve Questions of Kinetic Theory of Gases
  8. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  9. NCERT Solutions for Class 11 Physics Chapter-Wise

Did you ever consider why gases seem to expand and how they do apply pressure? The Kinetic Theory of Gases explains the processes by making assumptions that gases consist of small particles that move very rapidly. Boyle, Newton, Maxwell, and Boltzmann were among the prominent scientists who developed the theory, which predicts the actions of gases, how they will respond to pressure and temperature alterations, and their ability to flow and mix.

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NCERT Solutions for Class 11 Physics Chapter 12: Download PDF

The NCERT Solutions to Class 11 Chapter 12 Kinetic Theory includes step-by-step solutions of the textbook questions, which makes the students to have a clear picture of the behavior of gases at molecular scale. One can access these solutions online and download as a PDF solution without any payment making practice and revision convenient.

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NCERT Solutions for Class 11 Physics Kinetic Theory - Exercise Questions

Q12.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V actual is

Vactual=NA43π(d2)3
Vactual=6.023×1023×43π×(3×10102)3
Vactual=8.51×106m3
Vactual=8.51×103litres

The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres

VactualVmolar=8.51×10322.4

VactualVmolar=3.8×104

Q12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 00C ). Show that it is 22.4Litres .

Answer:

As per the ideal gas equation

PV=nRTV=nRTP

For one mole of a gas at STP we have

V=1×8.314×2731.013×105
V=0.0224m3

V=22.4 litres

Q12.3 Figure 13.8 shows plot of PV/T versus P for 1.00×103 kg of oxygen gas at two different temperatures.

PV/T versus P graph

(a) What does the dotted plot signify?
(b) Which is true: T1>T2orT1<T2 ?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00×103 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H2=2.02μ , of O2=32.0μ, R=8.31Jmol1K1.)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T 1 is closer to the horizontal line that the one corresponding to T 2 we conclude T 1 is greater than T 2.

(c) As per the ideal gas equation

PVT=nR

The molar mass of oxygen = 32 g

n=132

R = 8.314

nR=132×8.314
nR=0.256JK1

(d) If we obtained similar plots for 1.00×103 kg of hydrogen we would not get the same value of PV/T at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

m=PVTMR=0.256×28.314=5.48×105Kg

Q12.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 270C . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 170C . Estimate the mass of oxygen taken out of the cylinder ( R=8.31Jmol1K1 , molecular mass of O2=32μ ).

Answer:

Initial volume, V 1 = Volume of Cylinder = 30 l

Initial Pressure P 1 = 15 atm

Initial Temperature T 1 = 27 o C = 300 K

The initial number of moles n 1 inside the cylinder is

n1=P1V1RT1
n1=15×1.013×105×30×1038.314×300
n1=18.28

Final volume, V 2 = Volume of Cylinder = 30 l

Final Pressure P 2 = 11 atm

Final Temperature T 2 = 17 o C = 290 K

The final number of moles n 2 inside the cylinder is

n2=P2V2RT2
n2=11×1.013×105×30×1038.314×290
n2=13.86

Moles of oxygen taken out of the cylinder = n 2 -n 1 = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

m=4.42×32
m=141.44g

Q12.5 An air bubble of volume 1.0cm3 rises from the bottom of a lake 40m deep at a temperature of 120C . To what volume does it grow when it reaches the surface, which is at a temperature of 350C ?

Answer:

Initial Volume of the bubble, V 1 = 1.0 cm 3

Initial temperature, T 1 = 12 o C = 273 + 12 = 285 K

The density of water is ρw=103 kg m3

Initial Pressure is P 1

Depth of the bottom of the lake = 40 m

P1= Atmospheric Pressure + Pressure due to water P1=Patm+ρwghP1=1.013×105+103×9.8×40P1=4.93×105 Pa

Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure =1.013×105Pa

Let the final volume be V 2

As the number of moles inside the bubble remains constant, we have

P1V1T1=P2V2T2

V2=P1T2V1P2T1

V2=4.93×105×308×11.013×105×285

V2=5.26 cm3

Q12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m3 at a temperature of 270C and 1atm pressure.

Answer:

The volume of the room, V = 25.0 m 3

Temperature of the room, T = 27 o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

n=PVRTn=1.013×105×258.314×300n=1015.35

Avogadro's Number, NA=6.022×1023
Number of molecules inside the room is N

N=nNAN=1015.35×6.022×1023N=6.114×1026

Q12.7 Estimate the average thermal energy of a helium atom at (i) room temperature ( 270C ), (ii) the temperature on the surface of the Sun ( 6000K ), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

Answer:

The average energy of a Helium atom is given as 3kT2 since it is monoatomic

(i)

E=3kT2

E=3×1.38×1023×3002

E=6.21×1021 J

(ii)

E=3kT2

E=3×1.38×1023×60002

E=1.242×1019 J

(iii)

E=3kT2

E=3×1.38×1023×1072

E=2.07×1016 J

Q12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v rms the largest?

Answer:

As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.

Root mean square velocity is given as

vrms=3kTm

As we can see, vrms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon as its molar mass is the least.

Q12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at 200C ? (atomic mass of Ar=39.9μ , of He=4.0μ ).

Answer:

As we know root mean square velocity is given as vrms=3RTM

Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at 200C

3R×T39.9=3R×2534

T=2523.7 K

Q12.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0atm and temperature 170C . Take the radius of a nitrogen molecule to be roughly 1.0A . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of N2=28.0μ ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 o C

The radius of the Nitrogen molecule, r=1 Å.

The molecular mass of N 2 = 28 u

The molar mass of N 2 = 28 g

From the ideal gas equation

PV=nRT

nV=PRT

The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as

n=NAnV=6.022×1023×2×1.013×1058.314×(17+273)

n=5.06×1025

The mean free path λ is given as

λ=12πnd2

λ=12×π×5.06×1025×(2×1×1010)2

λ=1.11×107 m

The root mean square velocity v rms is given as

vrms=3RTM

vrms=3×8.314×29028×103

vrms=508.26 m s1

The time between collisions T is given as

T=1Collision Frequency

T=1ν

T=λvrms

T=1.11×107508.26

T=2.18×1010s

Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter

T=dvrms

T=2×1×1010508.26

T=3.935×1013s

The ratio of the average time between collisions to the collision time is

TT=2.18×10103.935×1013TT=554
Thus, we can see that the time between collisions is much larger than the collision time.

NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Additional Questions

Q 1) A metre-long narrow bore held horizontally (and closed at one end) contains a 76cm long mercury thread, which traps a 15cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P 1 = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm 2

The initial volume of the air column, V 1 = 15x cm 3

Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P 2 = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V 2 = (24 + y)x cm 3

Since the temperature of the air column does not change

P1V1=P2V276×15x=y×(24+y)x1140=y2+24yy2+24y1140=0

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm

Q 2) From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7cm3s1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2cm3s1. Identify the gas.

Answer:

As per Graham's Law of diffusion, if two gases of Molar Mass M 1 and M 2 diffuse with rates R 1 and R 2 respectively, their diffusion rates are related by the following equation

R1R2=M2M1

In the given question

R 1 = 28.7 cm 3 s -1

R 2 = 7.2 cm 3 s -1

M 1 = 2 g

M2=M1(R1R2)2

M2=2×(28.77.2)2

M2=31.78g

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n2=n1exp[mg(h2h1)/kbT] where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column: n2=n1exp[mgNA(ρρ)(h2h1)/(ρRT)] where ρ is the density of the suspended particle, and ρ , that of the surrounding medium. [ NA s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes' principle to find the apparent weight of the suspended particle.]

Answer:

n2=n1exp[mg(h2h1)/kbT] (i)

Let the suspended particles be spherical and have a radius r

The gravitational force acting on the suspended particles would be

FG=43πr3ρg

The buoyant force acting on them would be

FB=43πr3ρg

The net force acting on the particles become

Fnet=FGFB

Fnet=43πr3ρg43πr3ρg

Fnet=43πr3g(ρρ)

Replacing mg in equation (i) with the above equation, we get

n2=n1exp[43πr3g(ρρ)(h2h1)/kbT]

n2=n1exp[43πr3g(ρρ)(h2h1)RTNA]

n2=n1exp[mgNA(ρρ)(h2h1)RTρ]

The above is the equation to be derived

Q 4) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance
Atomic Mass (u)
Density (10 3 Kg m-3 )
Carbon (diamond)
12.01
2.22
Gold
197
19.32
Nitrogen (liquid)
14.01
1
Lithium
6.94
0.53
Fluorine
19
1.14

Answer:

Let one mole of a substance of atomic radius r and density ρ have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is NA=6.022×1023

NA43πr3ρ=Mr=(3M4NAπρ)13

For Carbon

NA43πr3ρ=Mr=(3×12.014×6.022×1023×π×2.22×103)13r=1.29 Å.

For gold

NA43πr3ρ=Mr=(3×197.004×6.022×1023×π×19.32×103)13r=1.59 Å.

For Nitrogen

NA43πr3ρ=Mr=(3×14.014×6.022×1023×π×1.00×103)13r=1.77 Å.

For Lithium

NA43πr3ρ=Mr=(3×6.944×6.022×1023×π×0.53×103)13r=1.73 Å.

For Fluorine

NA43πr3ρ=Mr=(3×19.004×6.022×1023×π×1.14×103)13r=1.88 Å.


Class 11 physics NCERT Chapter 12: Higher Order Thinking Skills (HOTS) Questions

Q1:

Two moles of an ideal gas with CpCv=53 are mixed 3 moles of another ideal gas with CpCv=43. The value of CpCv for the mixture is:-

Answer:

For ideal gas:- CpCv=R
For first case:-

Cp1Cv1=53 and Cp1Cv1=R


Cp1=53Cv1 and 53Cv1Cv1=R

23Cv1=R

Cv1=32R

So, Cp1=52R
For second case:-

Cp2Cv2=43 and Cp2Cv2=R

Cp2=43Cv2 and 43Cv2Cv2=R

Cv2=3R and Cp2=4R

Ymix =n1Cp1+n2Cp2n1Cv1+n2Cv2=2×52R+3×4R2×32R+3×3R=1.417=1.42


Q2:

If temperature of the atmosphere varies with height as T=(T0ah), where a and T0 are positive constants, then the pressure as a function of height h is (assume atmospheric pressure at sea level ( h=0 ) is P0 and molecule mass M of the air and acceleration due to gravity g be constant)

Answer:

dPdh=ρg=(PMRT)g=[PMR( T0ah)]g


dPP=(MgR)dh T0ah


P0PdPP=(MgR)0hdh( T0ah)


P=P0( T0ah T0)MsRa


Q3:

In two jars A and B, the pressure, volume and temperature in jar A are respectively P, V, and T and that of B are 2P, V/4 and 2T. Then the ratio of the number of molecules in jar A and B will be,

Answer:

NANB=PAVAPBPB×TBTANANB=P×V×(2T)2P×V4×T=41


Q4:

The temperature of an open room of volume 30 m3 increases from 17oC to 27oC due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nfni will be :

Answer:

PV = nRT

ni=PVRTi,nf=PVRTfnfni=PVR(1Tf1Ti)=105×308.31×(13001290)nfni=105×308.31×10300×290=105290×8.31 Change in the Number of molecules =105×6.023×1023290×8.31=2.5×1025


Q5:

A gas mixture contains 3 moles of oxygen and x mole of monoatomic gas at temperature T. Considering only translational and rotational but not vibrational modes, the total energy of the system is 15 RT then the value of x is.

Answer:

U=nf2RT

According to the question,

15RT=3×52RT+x×32RT15=152+3x2x=5


NCERT Chapter 12 Kinetic Theory Topics

Chapter 12: Kinetic Theory in Class 11 Physics explains the microscopic behavior of gases based on the motion of their molecules. It covers concepts like the kinetic theory of gases, pressure-temperature relationship, and degrees of freedom.

12.1 Introduction
12.2 Molecular Nature Of Matter
12.3 Behaviour Of Gases
12.4 Kinetic Theory Of An Ideal Gas
12.4.1 Pressure Of An Ideal Gas
12.4.2 Kinetic Interpretation Of Temperature
12.5 Law Of Equipartition Of Energy
12.6 Specific Heat Capacity
12.6.1 Monatomic Gases
12.6.2 Diatomic Gases
12.6.3 Polyatomic Gases
12.6.4 Specific Heat Capacity Of Solids
12.7 Mean Free Path

Kinetic Theory Chapter 12 Physics NCERT: Important Formulae

The important formulae from this chapter that help in solving numerical problems related to molecular speeds, kinetic energy, and gas laws, making them essential for board exams and competitive tests like JEE and NEET are listed below:

1. Ideal Gas Equation

PV=nRT

Where:

P= pressure ,V= volume ,n= number of moles ,R= universal gas constant ,T= temperature (in Kelvin) 

2. Average Kinetic Energy per Molecule

KEavg=32kT
Where k= Boltzmann constant, T= temperature in Kelvin

3. Total Kinetic Energy of Gas

KEtotal =32nRT

4. Pressure of Ideal Gas from Kinetic Theory

P=13ρv¯2
Where ρ= mass per unit volume, v¯2= mean square speed

5. Root Mean Square Speed

vrms=3kTm=3RTM
Where:
m= mass of one molecule,
M= molar mass

7. Relation Between Different Speeds

vrms >vavg >vmp 
Where:
- vrms=3RTM
- vavg=8RTπM
- vmp=2RTM

8. Degrees of Freedom and Energy

E=f2kT per molecule 
Where f= degrees of freedom

Etotal =f2nRT for n moles 

Approach to Solve Questions of Kinetic Theory of Gases

  • Take some time reading the question or cross it out, and underline the available data (such as temperature, pressure, volume, the number of moles or mass of gas).

  • See what the question actually asks (e.g. the speed, the kinetic energy, internal energy, pressure).

  • Express everything in terms of SI units:

Temperature in Kelvin

Cube volume in m3

Mass in kg

Molar Mass in kilogram per mol

  • Think of the underlying simple principle (e.g. relation between temperature and kinetic energy or pressure and molecular motion).
  • Choose the proper formula according to what is asked (do not need to memorize too many just identify the formula relating what is offered to what is requested).
  • Put in the known values into the formula cautiously.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT textbooks are almost a must have in studying and preparing competitive exams such as JEE and NEET, but are definitely lacking in preparing the numericals, tricky applications and deeper conceptual advancement. This is a list of what NCERT does not cover and what additional students are expected to learn in the chapter Kinetic Theory of Gases in preparation of JEE/NEET:

Concept NameJEENCERT
States Of Matter
Kinetic Theory Of Gases Assumptions
The Gas Laws
Ideal Gas Equation
Real Gas And Equation
Pressure Of An Ideal Gas
The Maxwell Distribution Laws
Degree Of Freedom
Kinetic Energy Of Ideal Gas
Mean Free Path
Specific Heat Of A Gas
Mayer's Formula

NCERT Solutions for Class 11 Physics Chapter-Wise

NCERT Solutions for Class 11 Subject Wise

Also, check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. What is the molecular nature of matter?

Everything wheather it is solids, liquids, and gases are made of tiny moving particles called atoms and molecules. Their movement decides whether something is solid liquid or gas .

2. How do gases behave in kinetic theory?

Gas molecules move randomly bouncing off each other and their container. This explains why gases expand and fill spaces following rules like Boyle’s and Charles’ laws.

3. What is the law of equipartition of energy?

It’s like sharing energy equally! In a gas, molecules divide energy between movement types—sliding, spinning, and vibrating—all getting a fair share.

4. Why do gases have two specific heat capacities?

Gases absorb heat differently depending on conditions:

Cₚ (at constant pressure) when the gas expands.

Cᵥ (at constant volume) when it stays in the same space.

The extra energy in Cₚ goes into expansion!

What is the mean free path?

It’s how far a gas molecule travels before colliding with another. Like walking in a crowd, some go farther before bumping into someone!

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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

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