# 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

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 $2.5\ bar$.

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 $0.8\ bar$.

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 $P \propto \rho$.

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, $M_{1} = 70\ u$

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 $M_{B} =4M_{A}$.

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} = 186.67mL$$T_{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.

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,

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

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

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\ bar$$P_{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  $= P_{H_{2}} +P_{O_{2}} = 0.4\ bar+1.4\ bar = 1.8\ bar$

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 =300K$$P_{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    $d_{2} = 3g\ dm^{-3}$

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 $125g\ mol^{-1}$

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 $=\frac{1.5V}{2.5V} =\frac{3}{5}$

We have the ideal gas equation,

$PV=nRT$

Pressure, p=3.32bar

Volume, V=5 $\dpi{100} dm^3$

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

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.

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.

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

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

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 $= 56.025\ bar$

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

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:

$= 5028.6 -1217 = 3811.6kg$

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

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

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 $0.8\ bar$.

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 is$K$.

For number of moles, n is mol.

Therefore we have for the quantity

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

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

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.

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

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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