NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter: Download Questions & Answers PDF

5.28. (b) directly proportional to its absolute temperature

5.29.  (b) increases by three times

5.32.  (c) F₂

5.10. We know,

                        PV=nRT

                         n = PV/RT

                         n= (0.1 x34.05 x 10^-3)/ (0.083 x 819)

                           = 5.01 x 10^-5 mol

Hence, the atomic mass of phosphorus= 0.0625/ (5.01 x10^-5)

                                                              = 1247.5 g/mol

5.30.  (a) decreases

5.1. Initial Pressure, P₁ = 1 bar                           Final Pressure, P₂ =?      

               Initial Volume, V₁= 500 dm³                       Final Volume, V₂=200 dm³
As the value of temperature constant (=30°C)
                  P₁V₁=P₂V₂
1 bar x 500 dm³ = P₂ x 200 dm³ 

Or                   P₂=500/200 bar

                           =2.5 bar

= mRT/MV

Replacing m/V by d (i.e. density), we get

                            P = dRT/M

    P? d

Hence, at a given temperature, the density of a gas is directly proportional to its pressure.

5.42. Dalton's law of partial pressure: When two or more non-reacting gases are enclosed in a vessel, the total pressure of the gaseous mixture is equal to the sum of the partial pressures that each gas will exert when enclosed separately in the same vessel at constant temperature.

P= P1 + P2 + P3

Where, P is the total pressure of the three gases A, B, and C enclosed in a container. P1, P2 and P3 are the partial pressures of the three gases when enclosed separately in the same vessel at a given temperature one by one.

No, the law cannot be applied. Carbon monoxide and oxygen readily combine to form carbon dioxide. The law can be applied only to the non-reacting gases.

5.17. No. of moles of CO₂ = Given mass of CO₂ / Molar mass

                                           = 8.8g / 44g mol^-1 = 0.2 mol

Pressure of CO₂ = 1 bar

RR = 0.083 bar dm³ K^–1 mol^–1

T = 273 + 31.1 K = 304.1 K

According to ideal gas equation,

PV = nRT

Therefore, V = nRT/P

                      = (0.2 x 0.083 x 304.1) / 1 bar = 5.048 L

5.13. Molecular mass of N₂ = 28g
28 g of N₂ has No. of molecules = 6.022 x 10²³ 

1.4 g of N₂ has No. of molecules = (6.022 x 10²³ x 1.4 g)/28 g= 3.011 x 10²² molecules.
Atomic No. of Nitrogen (N) = 7
1 molecule of N₂ has electrons = 7 x 2 = 14
3.011 x 10²² molecules of N₂ have electrons= 14 x 3.011 x 10²²= 4.215 x10²³ electrons.

5.6. The chemical equation for the reaction is
                                    2 Al + 2 NaOH + H₂O? 2 NaAlO₂ + 3H₂ 
i.e. 2 moles of Al give 3 moles of H₂ gas.

Moles of aluminium = 270.15g = 5.56*10^?3 moles

Moles of H₂ produced= 23*5.56*10^?3

           = 8.33*10^?3 moles

           Volume of H₂ = nRT / P

                                   = 8.33*10^?3 x 0.0831*293

                      = 0.203 litres

Therefore, volume of dihydrogen released is 0.203 litres.

5.37.  (i) Surface tension decreases with increase of temperature.

(ii) Viscosity decreases with increase of temperature.

5.39. In liquids, the molecules are more compact in comparison to gases.

5.40. (a) Surface tension: It is defined as the force acting per unit length perpendicular to the line drawn on the surface.

(b) Surface tension of a liquid depends upon following factors.

(i) Temperature: Surface tension decreases with rise in temperature. As the temperature of the liquid increases the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decreases the surface tension.

(ii) Nature of the liquid: Greater the magnitude of intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

5.18. Given, P₁ = P₂ and V₁ = V₂

We know that P₁V₁ = P₂V₂

Or,                n₁RT₁ = n₂RT₂

i.e.,                  n₁T₁ = n₂T₂

 Substituting n = w/M, we get

(W₁/M₁) x T₁ = (W₂/M₂) x T₂

(2.9/M₁) x (95 + 273) = (0.184/2) x (17 + 273)

M₁ = (2.9 x 368 x 2) / (0.184 x 290) = 40 g mol^-1

5.35. When density of liquid and vapours becomes the same; the clear boundary between liquid and vapours disappears. This temperature is called critical temperature.

5.7. Applying PV = nRT

                      We get P = nRT / V

Given:               n_CH4 = 3.2 / 16 mol = 0.2 mol

              n_CO2 = 4.4 /44 mol = 0.1 mol

So,

P_CH4 = (0.2 mol) (0.0821 dm³atm K^-1 mol^-1) (300 K) / (9 dm³)

        = 0.55 atm

 P_CO2= (0.1 mol) (0.0821 dm³atm K^-1 mol^-1) (300 K) / (9 dm³)

         = 0.27 atm

P_total = P_CH4 + P_CO2 = 0.55 + 0.27

         = 0.82 atm

         = 8.314 x 10⁴ Pa

Hence, the pressure exerted by a mixture is 8.314 x 10⁴ Pa

5.12. Given,

P= 3.32 bar

V= 5 dm³ 

n= 4 mol

R= 0.083 bar dm³ K^-1 mol^-1

PV = nRT

Or T = PV / nR = 3.32 x 5 dm³ / (4.0 mol x 0.083 bar dm³ K^-1 mol^-1)

         = 50 K

5.14. Time taken to distribute 10¹⁰ wheat grains = 1s

Time taken to distribute Avogadro number of wheat grains = (1s x 6.022 x 10²³) / 10¹⁰

= 6.022 x 10¹³ s

= (6.022 x 10¹³ / 60 x 60 x 24 x 365) year

= 1.9 x 10⁶ years

5.23.  (a) The dipole-dipole interaction between two HCl molecules is stronger than the London forces but is weaker than ion-ion interaction because only partial charges are involved.

5.26.  (d) Compression brings the molecules in close vicinity and cooling slows down the movement of molecules therefore, intermolecular interactions may hold the closely and slowly moving molecules together and the gas liquifies.

5.19. As the mixture H₂ and O₂ contains 20% by weight of dihydrogen, therefore,

If H₂ = 20g, then O₂ = 80g

No. of moles of H₂ = 20/2 = 10 moles

No. of moles of O₂ = 80/32 = 2.5 moles

Partial pressure of H₂ = [No. of moles of H₂/ (No. of moles of H₂ + No. of moles of O₂)] x

P_total= [10 (10 + 2.5)] x 1 = 0.8 bar

5.31.  (b) frictional resistance

5.15. Molar mass of O₂ = 32 g/mol

It means, 8 g of O₂ has 8/32 mol = 0.25 mol

Molar mass of H₂ = 2 g/mol

It means, 4 g of H₂ has 4/2 mol = 2 mol

Therefore, total number of moles, n = 2 + 0.25 = 2.25 mol

Given, V = 1dm³, T = 27°C = 300 K, R = 0.083 bar dm³ K^-1 mol^-1

Applying PV = nRT,

P = nRT / V

   = (2.25) (0.083 bar dm³ K^-1 mol^-1) (300 K) / (1dm³)

   = 56.025 bar

5.16. Radius of the balloon = 10 m

Therefore, volume of the balloon = (4/3)? r³ = (4/3) x (22/7) x (10 m)³

= 4190.5 m³

Volume of He filled at 1.66 bar and 27 °C = 4290.5 m³

To calculate the mass of He,

PV = nRT = (w/M) RT, where M is molar mass of He i.e. 4 g per mole or 4 x 10^-3 kg mol^-1

=> w =  [ (4 x 10^-3 kg mol^-1) (1.66 bar) (4190.5 m³)] / [ (0.083 bar dm³ K^-1 mol^-1) (300K)]

         = 1117.5 kg

Total mass of the balloon along with He = 100 + 1117.5 = 1217.5 kg

Maximum mass of the air that can be displaced by balloon to go up = volume x density

= 4190.5 m³ x 1.2 kg m^-3 = 5028.6 kg

Hence, pay load = 5028.6 – 1217.5 kg = 3811.1 kg

Answer: According to ideal gas equation
                        PV = nRT

Or                        P= nRT/V

Replacing n by m/M, we get

                            P M₁ x 2 = 28 x 5 (Molecular mass of N₂ = 28 g/mol)
or                                             M₁ = 70 g/mol

5.8. Calculation of partial pressure of H₂ in 1L vessel:

P₁= 0.8 bar, P₂=? V₁= 0.5 L, V₂ = 1.0 L
As temperature remains constant, P₁V₁ = P₂V₂
=> (0.8 bar) (0.5 L) = P₂  (1.0 L)

=>P₂ = 0.40 bar, i.e., P_H2 = 0.40 bar
Calculation of partial pressure of O₂ in 1 L vessel
P₁V₁ = P₂V₂
(0.7 bar) (2.0 L) = P₂  (1L)

=>P₂ = 1.4 bar

=>Po₂= 1.4 bar
Total pressure =P_H2 + P_Q2 = 0.4 bar + 1.4 bar = 1.8 bar

5.36.  Avogadro law states that equal volumes of all gases under same conditions of temperature and pressure contain equal number of molecules.

Dalton's law of partial pressure states that total pressure exerted by a mixture of non-reacting gases is equal to the sum of partial pressures exerted by them.

Boyle's law states that under isothermal condition, pressure of a fixed amount of a gas is inversely proportional to its volume.

Charles' law is a relationship between volume and absolute temperature under isobaric condition. Itstates that volume of a fixed amount of gas is directly proportional to its absolute temperature (V? T).

5.22. Higher the critical temperature, more easily the gas can be liquefied, i.e., greater are the intermolecular forces of attraction. Hence, CO₂ has stronger intermolecular forces than CH₄.

5.27.  (c) At high altitudes atmospheric pressure is low. Therefore, liquids at high altitudes boil at lower temperatures in comparison to that at sea level. Since water boils at low temperature on hills, the pressure cooker is used for cooking food. In hospitals surgical instruments are sterilized in autoclaves in which boiling point of water is increased by increasing the pressure above the atmospheric pressure by using a weight covering the vent.

5.21. At -273°C, volume of the gas becomes equal to zero, i.e., the gas ceases to exist.

5.24.  (b) Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

5.25.  (a) Other than attractive forces, molecules also exert repulsive forces on one another. When two molecules are brought into close contact with each other, the repulsion between the electron clouds and that between the nuclei of two molecules comes into play. Magnitude of the repulsion rises very rapidly as the distance separating the molecules decreases. This is the reason that liquids and solids are hard to compress. In these states molecules are already in close contact; therefore, they resist further compression; as that would result in the increase of repulsive interactions.

5.34. The temperature at which the vapour pressure of a liquid is equal to external pressure is called boiling point of liquid.

5.33. Molecules of liquids are held together byattractive intermolecular forces. Liquids havedefinite volume because molecules do notseparate from each other. However, moleculesof liquids can move past one another freely, therefore, liquids can flow, can be poured andcan assume the shape of the container in whichthese are stored.

5.38. The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure, is called Boyle temperature or Boyle point.

5.41. Ideal Gas: A gas that follows Boyle's law, Charles' law and Avogadro law strictly, is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.

Real Gases: Gases which deviate from ideal gas behaviour are known as real gases. NH₃ is expected to show more deviation. Since NH₃ is polar in nature and it can be liquified easily.

5.20. SI unit of pressure, P = Nm^-2

SI unit of volume = m³

Si unit of temperature, T = K

SI unit of number of moles, n = mol

Thus, SI unit of pV²T²/n = (Nm^-2) (m³)² (K) ²mol = Nm⁴K²mol^-1

5.5. Suppose molecular masses of A and B are M_A and M_B respectively. Then their number of moles will be

n_A= 1/M_A                                n_B = 2/M_B

Given: P_A = 2 bar        and P_A + P_B = 3 bar

          => P_B = 1bar

Since, PV = nRT

P_A= n_ART and P_BV= n_BRT

Therefore, (P_A / P_B) = (n_A / n_B)= (M_B / 2M_A)

=> M_B / M_A = 2 x P_A / P_B = 2 x 2 /1 = 4

        => M_B = 4 M_A