NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

 

NCERT solutions for class 11 chemistry chapter 5 States of Matter- The matter is existing in 3 physical states which are solid, liquid and gas. In NCERT solutions for class 11 chemistry chapter 5 States of matter, you will deal with the two states of matter namely liquid state and gaseous state, the laws governing the behaviour of ideal gases and properties associated with liquids. In solutions of NCERT class 11 chemistry chapter 5 States of Matter, there are 23 questions in the exercise. The CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter are prepared and solved by subject experts. These NCERT solutions will help you in preparation of class 11 final examination as well as in the preparation of various competitive exams like NEET, JEE, BITSAT etc.

NCERT solutions for class 11 chemistry chapter 5 States of Matter is an important chapter for class 11 students because the topics of this chapter are the basics to the topics to be studied in class 12. The important topics of this chapter are intermolecular forces, thermal energy, the gaseous state, ideal gas equation, kinetic molecular theory of gases, liquid State and more. After completing the solutions of NCERT  class 11 chemistry chapter 5 states of matter students will be able to explain the existence of different states of matter, explain the laws governing the behaviour of ideal gases; able to apply gas laws in real life situations; describe the conditions required for liquefaction of gases and also able to differentiate between vapours and gaseous state.

 

Some common characteristics of three forms of matter solid liquid and gas are summarised below-

 S.No.

      Gases         Liquids

     Solids

  1

No definite shape No definite shape

Definite shape

  2

Indefinite volume Definite volume

Definite volume

  3

Compressible Slightly compressible

Nearly compressible

  4

Low density Intermediate density

High density

NCERT solutions for class 11 chemistry chapter 5 States of Matter- Exercise Questions

Question 5.1     What will be the minimum pressure required to compress 500\; dm^{3} of air at 1 bar to 200\; dm^{3} at 30^{o}C?

Answer:

Given the volume of air to be compressed is 500dm^3 at 1 bar pressure to 200dm^3 at 30^{\circ}C.

So, as the temperature remains constant at 30^{\circ}C.

We have Boyle's law where, P_{1}V_{1} = P_{2}V_{2}

To calculate the final pressure P_{2}.

We have,

P_{1}V_{1} = P_{2}V_{2}

P_{2} = \frac{P_{1}V_{1}}{V_{2}}

= \frac{1\times 500}{200} bar = 2.5\ bar

Therefore, the minimum pressure required is 2.5\ bar.

Question 5.2     A vessel of 120 \; mL capacity contains a certain amount of gas at 35^{o}C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 \; mL at 35^{o}C. What would be its pressure?

Answer:

Given volume of air to be transferred from the capacity of 120 \; mL vessel at 1.2\ bar pressure to 180 \; mL vessel at 35^{o}C

So, as the temperature remains constant at 35^{o}C.

We have Boyle's law where, P_{1}V_{1} = P_{2}V_{2}

To calculate the final pressure P_{2}.

We have,

P_{1}V_{1} = P_{2}V_{2}

P_{2} = \frac{P_{1}V_{1}}{V_{2}}

= \frac{1.2\times 120}{180} bar = 0.8\ bar

Therefore, the pressure required is 0.8\ bar.

Question 5.3  Using the equation of state pV=nRT; show that at a given temperature density of a gas is proportional to gas pressure p.

Answer:

Given ideal gas equation ;

PV = nRT

So, at given fixed temperature T. we have the

Density\ of\ gas (\rho ) = \frac{mass(m)}{Volume(V)}

And we know n = \frac{Mass\ of\ gas}{Molar\ mass\ of\ gas}

so, we get P = \frac{mRT}{MV}  

P = \frac{\rho RT}{M}

But at constant Temperature, we have P \propto \rho.

Question 5.4   At 0^{o}\; C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Answer:

Given the condition :

Temperature = 0^{\circ}C

Pressure = 2\ bar for oxide gas and  Pressure = 5\ bar for dinitrogen.

As Density of gas is represented by, d = \frac{PM}{RT}

And given that density is the same for both the gases at the same temperature condition.

So, we have M_{1}P_{1} = M_{2}P_{2}  (as R is constant}

or, M_{1}\times 2 =28\times 5                      (\because Molecular\ mass\ of\ N_2 = 28\ u)

or, M_{1} = 70\ u

Question 5.5   Pressure of 1 g of an ideal gas A at 27^{o}C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

Answer:

Given Pressure of 1g of an ideal gas A at 27^{\circ}C is P_{A} = 2\ bar.

When 2g of another ideal gas B is introduced in the same flask at the same temperature,

The pressure becomes P_{A}+P_{B}= 3\ bar.

\implies P_{B }= 1\ bar.

We can assume the molecular masses of A and B be M_{A}\ and\ M_{B} respectively.

 The ideal gas equation, PV=nRT

So, we have P_{A}V=n_{A}RT   and   P_{B}V=n_{B}RT

Therefore, \frac{P_{A}}{P_{B}} = \frac{n_{A}}{n_{B}} = \frac{1M_{B}}{2M_{A}} = \frac{M_{B}}{2M_{A}}

or, \frac{M_{B}}{M_{A}} = 2\times\frac{P_{A}}{P_{B}} = 2\times \frac{2}{1} =4

Hence the relation between the two gases is M_{B} =4M_{A}.

Question 5.6    The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts?

Answer:

We have the chemical reaction:

2 Al +2NaOH +H_2O \rightarrow 2NaAlO_2+3H_{2} 

(Where dihydrogen is being produced.)

So, at 20^{\circ}C and 1\ bar pressure, the volume of gas that will be released when 0.15g of aluminium reacts.

As we can see from the reaction equation that 2 moles of aluminium produce 3 moles of dihydrogen.

i.e., 2\times 27g  reacts to give  3\times 22.4Litres

At STP condition 

273.15K  and  1atm,

54g(2\times 27g) of Al gives 3\times 22400mL of H_{2}.

Therefore 0.15g of Al will give = \frac{3\times22400\times0.15}{54}mL of H_{2}

i.e., 186.67mL of H_{2}.

At STP condition:

273.15K  and  1atm,

Now, P_{1} =1atm , V_{1} = 186.67mLT_{1} = 273.15K

Then we assume the volume of dihydrogen be V_{2} at the pressure P_{2} = 0.987atm

(Since 1 bar = 0.987atm)  and temperature T_{2} = 20^{\circ}C

\implies (273.15+20)K = 293.15K

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

V_{2}= \frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}

= \frac{1\times186.67\times293.15}{0.987\times273.15}

= 202.98mL\approx 203mL

Therefore, 203mL of dihydrogen will be released.

Question 5.7     What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9\; dm^{3} flask at 27 ^{o}C ?

Answer:

Given a mixture of 3.2g of methane and 4.4g of carbon dioxide contained a 9dm^3 flask at 27 ^{o}C.

So, the pressure exerted by the mixture will be

P =\frac{n}{V}RT = \frac{m}{M}\frac{RT}{V}

The pressure exerted by the Methane gas,P_{CH_{4}} =(\frac{3.2mol}{16}) \frac{0.0821 dm^3 atm K^{-1}mol^{-1}\times300K}{9 dm^3} = 0.55\ atm

The pressure exerted by the Carbon dioxide gas,

P_{CO_{2}} =(\frac{4.4mol}{44}) \frac{0.0821 dm^3 atm K^{-1}mol^{-1}\times300K}{9 dm^3} = 0.27\ atm

So, the total pressure exerted = 0.55+0.27 = 0.82\ atm

And in terms of SI units, we get,

R = 8.314p\ m^3\ K^{-1}mol^{-1},\ V= 9\times 10^{-3}\ m^3

P = 5.543\times10^4 Pa +2.771\times 10^4Pa = 8.314\times10^4 Pa.

Question 5.8     What will be the pressure of the gaseous mixture when 0.5\: L of H_{2} at 0.8 bar and 2.0\: L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27^{o}C ?

Answer:

Pressure of the gas mixture will be the sum of partial pressure of H_{2} gas and partial pressure of O_{2} gas.

So, calculating the partial pressures of each gas.

 Partial pressure of H_{2} in a 1L volume vessel.

P_{1} = 0.8\ barP_{2},\ V_{1} = 0.5L,\ V_{2} =1L

As the temperature remains constant, P_{1}V_{1} =P_{2}V_{2}

\implies (0.8\ bar)(0.5L) = P_{2}(1L)

or P_{2} = 0.4\ bar    or   P_{H_{2}} = 0.4\ bar

Now, calculating the partial pressure of O_{2} gas in 1L vessel.

P'_{1}V_{1} = P'_{2} V'_{2}

\implies (0.7\ bar)(2L) = P_{2}(1L)

\implies P'_{2} = 1.4\ bar  or  P_{O_{2}} = 1.4\ bar

Therefore the total pressure  = P_{H_{2}} +P_{O_{2}} = 0.4\ bar+1.4\ bar = 1.8\ bar

Question 5.9     Density of a gas is found to be 5.46 \: g/dm^{3}at 27^{o}C at 2 bar pressure. What will be its density at STP ?

Answer:

Given the density of a gas is equal to 5.46g/dm^3 at 27^{\circ}C and at 2bar pressure.

Density d = \frac{PM}{RT}, for the same gas at different temperatures and pressures we can apply,

\frac{d_{1}}{d_{2}} = \frac{P_{1}}{T_{1}}\times\frac{T_{2}}{P_{2}}.

Here, d_{1} =5.46g\ dm^_{-3} , T_{1} =27^{\circ}C =300KP_{1} =2bar.

then at STP, we have d_{2},\ T_{2} = 0^{\circ}C =273K,\ P_{2} =1bar.

\therefore \frac{5.46g\ dm^{-3}}{d_{2}} = \frac{2bar}{300K}\times\frac{273K}{1bar}    or    d_{2} = 3g\ dm^{-3}

Question 5.10     34.05 \; mL of phosphorus vapour weighs 0.0625 g at 546 ^{o}C and 0.1 bar pressure. What is the molar mass of phosphorus?

Answer:

Given,

The volume of phosphorus vapour = 34.05mL  which weights about 0.0625g .

Temperature = 546^{\circ}C = 273+546 = 819K

Pressure =0.1 bar

Hence we apply the ideal gas equation,

PV=nRT , where the number of moles will be  n =\frac{PV}{RT}

= \frac{1bar\times (34.05\times 10^{-3} dm^3)}{0.083 bar\ dm^3\ K^{-1}Mol^{-1}\times 819K}

= 5\times 10^{-4} mol

\therefore Mass\ of\ 1\ mole = \frac{0.0625}{5\times 10^{-4}}g = 125g

Therefore the molar mass is 125g\ mol^{-1}

Question 5.11     A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

Answer:

Assume the round-bottomed flask has volume = V\ cm^3

so, the volume of air in the flask at 27^{\circ}C = V\ cm^3.

To find out how much air has been expelled out, we use Charle's Law:

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Therefore Volume\ expelled = 2.5V-V =1.5V

Or, Fraction of air expelled =\frac{1.5V}{2.5V} =\frac{3}{5}
 

Question 5.12     Calculate the temperature of 4.0 \: mol of a gas occupying 5\: \: dm^{3} at 3.32 bar.
                 (R = 0.083 \: bar\: dm^{3} K^{-1} mol^{-1}).

Answer:

We have the ideal gas equation,

PV=nRT 

Pressure, p=3.32bar

Volume, V=5 dm^3

Number of moles, n=4 mol

Gas constant R= 0.083

Temperature,  T=?

T = \frac{PV}{nR} = \frac{3.32bar\times 5\ dm^3}{4\ mol\times 0.083 bar\ dm^3\ K^{-1}mol^{-1}} = 50K

so, temperature, T=50K

Question 5.13     Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Answer:

Dinitrogen N_{2}  has a molar mass = 28g\ mol^{-1}.

Given 1.4g of N_{2} gas.

That means \frac{1.4}{28} = 0.05\ mol

= 0.05\times6.02\times10^{23} = 3.01\times 10^{23} number of molecules.

And as 1 molecule of N_{2} has 14 electrons.

So, 3.01\times 10^{23} molecules of N_{2} has 14\times 3.01\times 1023 = 4.214\times 10^{23} electrons.

Question 5.14     How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second ?

Answer:

Given that 10^{10} grains are distributed each second.

Time taken to distribute 10^{10} grains = 1second.

Time taken to distribute one Avogadro number of wheat grains.

=\frac{1s\times 6.022\times 10^{23}grains}{10^{10}grains}

= \frac{6.022\times10^{23}}{60\times 60\times 24\times 365} = 1.9\times 10^{6}\ years.

Question 5.15     Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of  1\: dm^{3} at 27°C. R = 0.083\: bar \: dm^{3} K^{-1} mol^{-1}.

Answer:

Given the mass of oxygen gas and mass of hydrogen gas.

Molar mass of O_{2} =32g\ mol^{-1}

Therefore 8g\ O_{2} = \frac{8}{32}mol = 0.25\ mol

Molar mass of H_{2} = 2g\ mol^{-1}

Therefore 4g H_{2} =\frac{4}{2} =2\ mol

\Rightarrow Total number of moles of mixture n = 2+0.25 =2.25

and V = 1\ dm^3,\ T = 27^{\circ }C = 300K,\ R=0.083\ bar\ dm^{3}K^{-1}\ mol^{-1}

So, Ideal gas equation; PV = nRT

or P = \frac{nRT}{V} = \frac{(2.25mol\times) (0.083 bar\ dm^3K^{-1}mol^{-1})(300K)}{1 dm^3}

Therefore Total pressure is = 56.025\ bar

Question 5.16     Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C.  

(Density of air = 1.2 \: kg\: m^{-3}and R = 0.083 \: bar \: dm^{3} K^{-1} mol^{-1}).

Answer:

The payload can be defined as:

(Mass of the displaced air - Mass of the balloon)

Given the radius of the balloon, r = 10 m.

Mass of the balloon, m = 100 kg.

Therefore, the volume of the balloon will be:

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\times\frac{22}{7}\times10^3 = 4190.5\ m^3

Now, the volume of the air displaced:

V_{d} = 4190.5\ m^3

The mass of the air displaced :

m_{d} = density \times V_{d} = 1.2kgm^{-3}\times 4190.5\ m^3

        = 5028.6kg

Let W be the mass of helium gas filled into the balloon, then

PV=(\frac{W}{m})rt

Or, W = \frac{PVM}{RT}

= \frac{1.66\times4190.5\times10^3\times4}{0.083\times300} = 1117\ kg approximately.

The balloon is filled with He with total mass of = 1117+100= 1217kg

\therefore The payload of the balloon will be:

= 5028.6 -1217 = 3811.6kg

Question 5.17     Calculate the volume occupied by 8.8 g of CO_{2} at 31.1°C and 1 bar pressure.  R = 0.083 \: bar \: L\: K^{-1} mol^{-1}.

Answer:

Given the mass of carbon dioxide is 8.8grams at 31.1^{\circ}C and at 1 bar pressure.

So, the number of moles of 

CO_{2} (n) = \frac{Mass\ of\ CO_{2}}{Molar\ Mass}

= \frac{8.8g}{44g\ mol^{-1}} = 0.2\ mol

Now, the pressure of CO_{2} (P) = 1\ bar

Given R = 0.083 \: bar \: L\: K^{-1} mol^{-1}

also, the Temperature (T) = 273+31.1 = 304.1K

The Ideal gas equation; PV=nRT

V=\frac{nRT}{P} = \frac{0.2\times 0.083\times304.1}{1\ bar} = 5.048L

So volume occupied by 8.8 g of carbon dioxide is 5.048L

Question 5.18     2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Answer:

Let the molar mass of the gas be M_{gas},

Then, given that 2.9g of a gas at 95^{\circ}C occupied the same volume as 

0.184g of H_{2} at 17^{\circ }C

So, we have the relation, P_{1}=P_{2}   and V_{1} = V_{2}

Therefore P_{1}V_{2} = P_{2} V_{2}

wihch gives, n_{1}RT_{1} =n_{2}RT_{2}

\implies n_{1}T_{1} =n_{2}T_{2}

Or, \frac{2.9}{M_{gas}}\times (95+273) = \frac{0.184}{2}\times(17+273)

Or, M_{gas} = \frac{2.9\times368\times 2 }{0.184\times 290} = 40g\ mol^{-1}

Question 5.19     A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Answer:

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen.

So, let us assume the weight of dihydrogen in a total mixture weight of W be 20Wgrams and dioxygen be 80Wgrams.

Then, the number of moles of H_{2} will be, n_{H_{2}} = \frac{20W}{2} = 10W\ moles and the number of moles of O_{2}n_{O_{2}} = \frac{80W}{32} = 2.5W\ moles.

And given that the total pressure of the mixture is P_{Total} = 1\ bar.

Then we have a partial pressure of H_{2},

P_{H_{2}}=(Mol.\ fraction)_{H_{2}}\times P_{Total}

= \frac{n_{H_{2}}}{n_{H_{2}}+n_{O_{2}}}\times P_{Total}

= \frac{10W}{10W+2.5W}\times 1\ bar

=0.8\ bar

Therefore, the partial pressure of H_{2} is 0.8\ bar.

Question 5.20     What would be the SI unit for the quantity pV^{2}T^{2}/n ?

Answer:

Given quantity, pV^{2}T^{2}/n,

The SI units of each factors are,

For pressure p is Nm^{-2}.

For Volume V is m^3.

For Temperaute T isK.

For number of moles, n is mol.

Therefore we have for the quantity 

The SI unit is =\frac{(Nm^{-2})(m^3)^2)(K)^2}{mol}

= Nm^4K^2 mol^{-1}

Question 5.21     In terms of Charles’ law explain why –273 °C is the lowest possible temperature.

Answer:

Charles' Law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Charles found that for all gases, at any given pressure, the graph of volume vs temperature (in Celsius) is a straight line and on extending to zero volume, each line intercepts the temperature axis at – 273.15 ° C.

We can see that the volume of the gas at – 273.15 ° C will be zero. This means that gas will not exist. In fact, all the gases get liquified before this temperature is reached.

Question 5.22     Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

Answer:

The critical temperature we know is the highest temperature at which liquid exists above it is gas.

Given that the critical temperature of Carbon dioxide and Methane are.31.1 ^{\circ}C\ and\ -81.9^{\circ}C 

Higher is the critical temperature of a gas, easier is its liquefaction.

So, as the critical temperature increases the gas is now easier to liquefaction.

That means the intermolecular forces of attraction between the molecules of a gas are directly proportional to its Critical temperature.

Hence, Carbon dioxide has stronger intermolecular forces than Methane.

Question 5.23     Explain the physical significance of van der Waals parameters.

Answer:

The equation of van der Waals after taking into account the corrections for pressure and volume,

\left ( p+\frac{an^2}{V^2} \right )\left ( V-nb \right ) =nRT

Where a and b are called van der Waals constants or parameters.

Here the significance of a and b is important:

Value of 'a' is a measure of the magnitude of intermolecular attractive forces within the gas and is independent of temperature and pressure.

And the value of 'b' is the volume occupied by the molecule and 'nb' is the total volume occupied by the molecules.

NCERT solutions for class 11 chemistry

Chapter 1

NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry

Chapter-2

CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom

Chapter-3

Solutions of NCERT class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties

Chapter-4

NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure

Chapter-5

CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter

Chapter-6

Solutions of NCERT class 11 chemistry chapter 6 Thermodynamics

Chapter-7

NCERT solutions for class 11 chemistry chapter 7 Equilibrium

Chapter-8

CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reaction

Chapter-9

Solutions of NCERT class 11 chemistry chapter 9 Hydrogen

Chapter-10

NCERT solutions for class 11 chemistry chapter 10 The S-Block Elements

Chapter-11

CBSE NCERT solutions for class 11 chemistry chapter 11 The P-Block Elements

Chapter-12

Solutions of NCERT class 11 chemistry chapter 12 Organic chemistry- some basic principles and techniques

Chapter-13

NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons

Chapter-14

CBSE NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry

NCERT solutions for class 11 subject wise

NCERT solutions for class 11 biology

Solutions of NCERT class 11 maths

CBSE NCERT solutions for class 11 chemistry

NCERT solutions for class 11 physics

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