MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

In this article, we will share MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter Pdf, These solutions are solved by subject experts from the latest edition books.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

MP Board Class 11 Chemistry States of Matter Textbook Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30° C?
Solution.
From the given data :
P1 = 1 bar
P2 = ?
V1 = 500 dm3 V2 = 200 dm3
Temperature remains constant According to Boyle’s law,
P1 V1 = P2V2 or P2= \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~V}_{2}}\)
p2 = \(\frac{1(\text { bar }) \times 500\left(\mathrm{dm}^{3}\right)}{200\left(\mathrm{dm}^{3}\right)}\)
= 2.5 bar.

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas at 35° C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?
Solution.
From the given data:
P1 = 1.2 bar P2 = ?
V1 = 120 mL V2 = 180 mL
Temperature is constant.
According to Boyle’s law
P1 V1 = P2 V2 or P2=\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~V}_{2}} \)
P2 = \(\frac{1.2(\mathrm{bar}) \times 120(\mathrm{~mL})}{180(\mathrm{~mL})}\)
= 0.8 bar.

Question 3.
Using the equation of state PV = nRT show that at a given temperature, the density of the gas is proportional to the gas pressure P.
Solution.
According to ideal gas equation: PV = nRT or P = \(\frac{n R T}{\mathrm{~V}}\)
n = \(\frac{\text { Mass of gas }(m)}{\text { Molar mass of gas }(M)}\)
P = \(\frac{m \mathrm{RT}}{\mathrm{MV}}\)
Now density (d) = \(\frac{m}{\mathrm{~V}}\)
∴ P = \(\frac{d \mathrm{RT}}{\mathrm{M}}\)
or d ∝ P (at constant temperature)

Question 4.
At 0°C, the density of a gaseous oxide at 2 bar is the same as that of nitrogen at 5 bar. What is the molecular mass of the oxide?
Solution.
Density (d1) of N2 at 5 bar and 0°C
d1 = \(\frac{\mathrm{PM}}{\mathrm{RT}}\) = \(\frac{5(\text { bar }) \times 28\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}{\mathrm{RT}}\)

Density (d2) of gaseous oxide at 2 bar and 0°C
d2 = \(\frac{\mathrm{PM}}{\mathrm{RT}}\) = \(\frac{2(\mathrm{bar}) \times \mathrm{M}\left(\mathrm{g} \mathrm{mol}^{-1}\right)}{\mathrm{RT}}\)

Now, d1=d2
or \(\frac{5(\text { bar }) \times 28\left(\mathrm{~g} \mathrm{~mol}^{-1}\right.}{\mathrm{RT}}\) = \(\frac{2(\mathrm{bar}) \times \mathrm{M}\left(\mathrm{g} \mathrm{mol}^{-1}\right)}{\mathrm{RT}}\)
or M = \(\frac{5 \times 28}{2}\)
= 70 g mol-1.

Question 5.
Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.
Solution.
Let MA and MB be the molar masses of the two gases A and B. According to available data:
Moles of gas A, (nA) = \(\frac{\text { Mass of gas A }}{\text { Molar mass }}\) = \(\frac{1(\mathrm{~g})}{\mathrm{M}_{\mathrm{A}}\left(\mathrm{g} \mathrm{mol}^{-1}\right)}\)
Moles of gas B, (nB) = \(\frac{\text { Mass of gas B }}{\text { Molar mass }}\) = \(\frac{2(g)}{M_{B}\left(g m o l^{-1}\right)}\)
∴ \(\frac{n_{\mathrm{A}}}{n_{\mathrm{B}}}\) = \(\frac{1(g)}{M_{A}\left(g m o l^{-1}\right)} \times \frac{M_{B}\left(g m o l^{-1}\right)}{2(g)}\) = \(\frac{\mathrm{M}_{\mathrm{B}}}{2 \mathrm{M}_{\mathrm{A}}} \) ……………………………………… (1)

Now, pressure of gas A, (PA) = 2 bar
Total pressure of A and B, (PA + PB) = 3 bar
PB = (3 – 2) = 1 bar
According to ideal gas equation,
PAV = nART and PB V = nB RT
∴ \(\frac{P_{A}}{P_{B}}\) = \(\frac{n_{\mathrm{A}}}{n_{\mathrm{B}}}\) or \(\frac{n_{\mathrm{A}}}{n_{\mathrm{B}}}\) = \(\frac{(2 \text { bar })}{(1 \text { bar })}\) = \(\frac{2}{1} \) …………………….. (ii)
Equating (i) and (ii),
\(\frac{\mathbf{M}_{\mathrm{B}}}{2 \mathrm{M}_{\mathrm{A}}}\) = \(\frac{2}{1}\) or
MB = 4MA.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 6.
The drain cleaner, ‘Drainex’ contains small bits of aluminium which reacts with caustic soda to produce hydrogen. What volume of hydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts.
Solution.
The chemical reaction taking place is
MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter 1
Now, Hydrogen Produced by 54 g of Al = 3 mol
Hydrogen produced by 0.15 of Al = \(\frac{3 \times 0.15}{54}\)
= 8.33 × 10-3 mol

Volume of 8.33 × 10-3mol of hydrogen at 20° C and 1 bar
V = \(\frac{n \mathrm{RT}}{\mathrm{P}}\)
= \(\begin{gathered}
8.33 \times 10^{-3}(\mathrm{~mol}) \times 0.083\left(\operatorname{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) \\
\frac{\times 293.15(\mathrm{~K})}{1 \mathrm{bar}}
\end{gathered}\)
= 0.20268
= 202.68 mL.

Question 7.
What will be the pressure exerted (in pascal) by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27°C?
Solution.
Moles of methane (n1) = \(\frac{\text { Mass of methane }}{\text { Molar mass }}\) = \(\frac{3.2(\mathrm{~g})}{16.0\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 0.2 mol
Moles of carbon dioxide (n2) = \(\frac{\text { Mass of } \mathrm{CO}_{2}}{\text { Molar mass }}\) = \(\frac{4.4(\mathrm{~g})}{44.0\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 0.1 mol
Total number of gas moles = 0.2 + 0.1 = 0.3 mol

Total gas pressure (P) = \(\frac{n \mathrm{RT}}{\mathrm{V}}\)
= \(\frac{0.3(\mathrm{~mol}) \times 8.314\left(\mathrm{~Pa} \mathrm{~m}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times 300.15(\mathrm{~K})}{9 \times 10^{-3}\left(\mathrm{~m}^{3}\right)}\)
= 8.318 × 104 Pa.

Question 8.
What will be the pressure of a gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of 02 at 0.7 bar are introduced in a 1 L empty vessel at 27°C?
Solution.
Moles of H2, (nH2) = \(\frac{\mathrm{PV}}{\mathrm{RT}}\) = \(\frac{0.8(\text { bar }) \times 0.5(\mathrm{~L})}{\mathrm{RT}}\) = \(\frac{0.4}{\mathrm{RT}}\)
Moles of O2 (no2) = \(\frac{\mathrm{PV}}{\mathrm{RT}}\) = \(\frac{0.7(\mathrm{bar}) \times 2.0(\mathrm{~L})}{\mathrm{RT}}\) = \(\frac{1.4}{\mathrm{RT}}\)

Total number of gaseous moles (n)
= \(\frac{0.4}{\mathrm{RT}}+\frac{1.4}{\mathrm{RT}}\) = \(\frac{1.8(\text { bar } \mathrm{L})}{\mathrm{RT}} \)
According to gas equation,
P = \(\frac{n \mathrm{RT}}{\mathrm{V}}\) = \(\frac{1.8 \text { (bar L) RT }}{\mathrm{RT} \times \mathrm{V}} \) = \(\frac{1.8(\text { bar } \mathrm{L})}{1(\mathrm{~L})}\)
= 1.8 bar.

Question 9.
The density of a gas is found to be 5.46 g/dm3 at 27°C and under 2 bar pressure. What will be its density at STP?
Solution:
d1 = \(\frac{\mathrm{P}_{1} \mathrm{M}}{\mathrm{RT}_{1}}\) ; d2 = \(\frac{\mathrm{P}_{2} \mathrm{M}}{\mathrm{R} \mathrm{T}_{2}}\)
or \(\frac{d_{1}}{d_{2}}\) = \(\frac{\mathrm{P}_{1} \mathrm{~T}_{2}}{\mathrm{P}_{2} \mathrm{~T}_{1}} \) thus, d2 = \(\frac{d_{1} \mathrm{~T}_{1} \mathrm{P}_{2}}{\mathrm{~T}_{2} \mathrm{P}_{1}}\)

Substituting the values we have
d2 = \(\frac{5.46\left(\mathrm{~g} \mathrm{dm}^{-3}\right) \times 300.15(\mathrm{~K}) \times 1(\mathrm{bar})}{2(\mathrm{bar}) \times 273.15(\mathrm{~K})}\)
= 2.99 g dm-3
or = 3.0 g dm-3.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 10.
34.05 mL of phosphorus vapours weigh 0.0625 g at 546°C and 1.0 bar pressure. What is the molar mass of phosphorus?
Solution.
According to ideal gas equation.
PV = nRT
Now, n = \(\frac{\mathrm{W}}{\mathrm{M}}\)
∴ PV = \(\frac{\text { WRT }}{\text { M }}\) or M = \(\frac{\text { WRT }}{\text { PV }}\)

Substituting the values from the data
M = \(\frac{0.0625(\mathrm{~g}) \times 0.083\left(\mathrm{bar} \mathrm{L} \cdot \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) \times 819.15(\mathrm{~K})}{1(\mathrm{bar}) \times 34.04 \times 10^{-3}(\mathrm{~L})}\)
= 124.8 g mol-1.

Question 11.
A student forgot to add the reaction mixture to a round-bottomed flask at 27°C but put it on the flame. After a lapse of time, he realised his mistake. By using a pyrometer, he found that the temperature of the flask was 477°C. What frac¬tion of air would have been expelled out?
Solution.
Let the volume of the flask be V mL
T1 = 27 + 273.15 = 300.15 K;
T2 = 477 + 273.15 = 750. 15 K

As P remains constant, thus according to Charles’ law
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\)
or V2 = \(\frac{\mathrm{V}_{1} \mathrm{~T}_{2}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{V}(\mathrm{mL}) \times 750.15(\mathrm{~K})}{300.15(\mathrm{~K})}\)
= 2.499 V or 2.5 V
Volume of air expelled out at 750.15 K = 2.5 V-V= 1.5 V
Fraction of air expelled = \(\frac{1.5 \mathrm{~V}}{2.5 \mathrm{~V}}\) = \(\frac{3}{5}\) = 0.6.

Question 12.
Calculate the temperature of 4.0 moles of a gas occupying 5 dm3 at 3.32 bar (R = 0.083 bar dm3 K-1 mol-1).
Solution.
According to ideal gas equation,
PV = nRT or T = \(\frac{\mathrm{PV}}{n \mathrm{R}}\)
Substituting the values from the given data
T = \(\frac{3.32(\text { bar }) \times 5\left(\mathrm{dm}^{3}\right)}{4.0(\mathrm{~mole}) \times 0.083\left(\mathrm{bar} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)}\)
= 50 K.

Question 13.
Calculate the total number of electrons present in 1.4 g of nitrogen gas.
Solution.
Molar mass of N2 = 28 g mol-1
Number of N2 molecules in 28 g = 6.02 × 1023
Number of N2 molecules in 1.4 g
= 6.02 × 10,sup>23 × \(\frac{1.4(\mathrm{~g})}{28(\mathrm{~g})}\) = 3.01 × 1022
Atomic number of nitrogen (N) = 7
Number of electrons in 1 molecule of N2 = 7 x 2 = 14
Number of electrons in 3.01 × 1022 molecules of N2
= (14 × 3.01 × 1022)
= 4.214 × 1023.

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains if 1016 grains are distributed each second?
Solution.
Time taken to distribute 1010 grains = 1 s
Time taken to distribute 6.22 ×1023 grains
= 1(s) \(\frac{6.02 \times 10^{23} \text { (grains) }}{10^{10} \text { (grains) }}\)
= \(\frac{6.02 \times 10^{13}}{60 \times 60 \times 24 \times 365}\) Years
= 1.9 × 106 Years .

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 15.
Calculate the total pressure of a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 dm3 at 27°C (R = 0.083 bar dm3 K-1 mol-1).
Solution.
Moles of oxygen (n1) = \(\frac{\text { Mass of oxygen }}{\text { Molar mass }}\) = \(\frac{8.0(\mathrm{~g})}{32.0\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 0.25 mol
Moles of hydrogen (n2) = \(\frac{\text { Mass of hydrogen }}{\text { Molar mass }}\) = \(\frac{4.0(\mathrm{~g})}{2.0\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 2 mol
Total number of gas moles = 0.25 + 2 2.25 mol
Total gaseous prestire (P) = \(\frac{n_{\text {total }} \mathrm{RT}}{\mathrm{V}}\)
= \(\frac{2.25(\mathrm{~mol}) \times 0.083\left(\mathrm{bar} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}\right) \times 300.15(\mathrm{~K})}{1\left(\mathrm{dm}^{3}\right)}\)
= 56.05 bar.

Question 16.
Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m and 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 dm3 K-1 mol-1).
Solution.
Calculation of the mass of displaced air Radius of balloon (r) = 10 m
Volume of balloon (V)
= 4/3 πr3
= 4/3 × 22/7 × 103 (m3)
= 4190.48 m3 Mass of the displaced air
= Volume of air (balloon) × Density of air
= 4190.5 (m3) × 1.2 (kg m-3) = 5028.6 kg.

Calculation of the mass of the filled balloon.
Moles of He present (n) = \(\frac{\mathrm{PV}}{\mathrm{RT}}\)
= \(\frac{1.66(\text { bar }) \times 4190.48 \times 10^{3}\left(\mathrm{dm}^{3}\right)}{0.083\left(\text { bar } \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times 300.15(\mathrm{~K})}\)
= 279.22 × 103
Mass of He present
= Moles of He × Molar mass of He
= 279.22 × 103 (mol) × 4 (g mol-1)
= 1116.88 × 103 g = 1116.88 kg
Mass of filled balloon
= 100 + 1116.88 = 1216.88 kg
Mass of air displaced by balloon to go up
= V×d
= 4190.48 (m3) × 1.2 (kg m-3) = 5028.58 kg
Payload, i.e., extra load that can be attached to balloon
= 5028.58- 1216.88
= 3811.7 kg.

Question 17.
Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure (R = 0.083 bar LK-1 mol-1).
Solution.
From the given data :
Mass of CO2 (n) = \(\frac{\text { Mass of } \mathrm{CO}_{2}}{\text { Molar mass }}\) = \(\frac{8.8(\mathrm{~g})}{44\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 0.2 mol
According to ideal gas equation, PV = nRT
or V = \(\frac{n \mathrm{RT}}{\mathrm{P}}\)
= \(\frac{0.2(\mathrm{~mol}) \times 0.083\left(\text { bar } \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1} 1 \times 304.25(\mathrm{~K})\right.}{1(\mathrm{bar})}\)
= 5.05 L.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C and at the same pressure. What is the molar mass of the gas?
Solution.
Let the molar volume of gas be M
∴ n(g) = \(\frac{2.9(\mathrm{~g})}{\mathrm{M}}\) ; n(H2) = \(\frac{0.184(\mathrm{~g})}{2\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 0.092
Volume of H2,
V(H2) = \(\frac{n R \mathrm{~T}}{\mathrm{P}}\) = \(\frac{0.092 \times \mathrm{R} \times 290.15}{\mathrm{P}}\) = \(\frac{26.70 \mathrm{R}}{\mathrm{P}}\)
Volume of gas V(g) = \(\frac{n \mathrm{RT}}{\mathrm{P}}\) = \(\frac{2.9 \times \mathrm{R} \times 368.15}{\mathrm{M} \times \mathrm{P}}\) = \(\frac{1067.63 \mathrm{R}}{\mathrm{MP}}\)
Now , V(H2) = V(g)
or \(\frac{26.70 \mathrm{R}}{\mathrm{P}}\) = \(\frac{1067.63 \mathrm{R}}{\mathrm{MP}}\)
M = \(\frac{1067.63}{26.70}\) = 39.98 = 40 g mol -1.

Question 19.
A mixture of hydrogen and oxygen at one bar pressure contains 20% by mass of hydrogen. Calculate the partial pressure of hydrogen.
Solution.
Let the mass of H2 in the mixture = 20 g
∴ The mass of O2 = 80 g
Moles of hydrogen (nH2 ) = \(\frac{20(\mathrm{~g})}{2\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 10 mol;
Moles of O2 (no2 ) = = 2.5 mol
Mole fraction of H2 (xH2) = \(\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{H}_{2}}+n_{\mathrm{O}_{2}}}\) = \(\frac{10}{10+2.5}\) = \(\frac{10}{12.5}\)
Total pressure = 1 bar
Partial pressure of H2 = Total pressure × xH2
\(\frac{1(\text { bar }) \times 10}{12.5}\)
= 0.8 bar.

Question 20.
What would be SI units of a quantity pV2 T2/n?
Solution.
Putting SI units for different parameters we get
pV2 T2/n = \(\frac{\left(\mathrm{Nm}^{-2}\right)\left(\mathrm{m}^{3}\right)^{2}(\mathrm{~K})^{2}}{\mathrm{~mol}}\) = Nm4K2 mol-1.

Question 21.
In terms of Charles’ law explain why – 273°C is the lowest possible temperature?
Solution.
Theoretical considerations based on Charles’ law suggest that at – 273°C the gases should occupy zero volume. In other words, the gases should cease to exist. If the temperature falls below – 273°C, the volume of the gas according to the law should become -ve which is not possible. This indirectly means that – 273°C is the lowest possible temperature that can be achieved theoretically.

Question 22.
Critical temperatures for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has stronger intermolecular forces and why?
Solution.
The higher the value of critical temperature of the gas is, the more easy will be its liquefaction which in turn reflects the higher magnitude or stronger intermolecular forces. Since critical temperature of CO2 (31.1°C) is higher than that of CH4 (- 81.9°C). It means that interparticle forces in CO2 are stronger than those in methane.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 23.
Explain the physical significance of van der Waal parameters.
Solution.
van der Waal parameter ‘a’ is a measure of the magnitude of intermolecular forces. The van der Waal parameter ‘b’ which is also called co-volume is a measure of effective size of the gas molecules.

MP Board Class 11 Chemistry States of Matter Important Questions and Answers

Question 1.
Why do we add 273 to the temperature while dealing with the problems on gas equation? ,
Answer:
According to Charles’ law, V ∝ (t + 273) at constant P and n.

Question 2.
Is it possible to cool the gas to 0 Kelvin?
Answer:
No because gases condense to liquids and solids before this temperature is reached.

Question 3.
What is the ratio of average molecular K.E. of CO2 to that of S02 at 27°C?
Answer:
1:1.

Question 4.
Critical temperature of .two gases A and B are 30°C and – 50°C respectively, which of them has stronger intermolecular forces and why?
Answer:
Interparticle forces are stronger in A.

Question 5.
What do you mean by vapour?
Answer:
Gases below their critical temperatures are called vapours.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 6.
What is the meaning of pressure of a gas?
Answer:
Gas pressure is due to bombardment of gas molecules against the walls of the container or force with which molecule stroke against wall of container per unit area.

Question 7.
What is the difference between barometer and manometer?
Answer:
Barometer is instrument used to measure atmospheric pressure whereas manometer measures gas pressure.

Question 8.
If a plot of V vs °C at constant pressure is drawn, at what temperatures will it cut the volume and temperature axes?
Answer:
It cuts the volume axis at 0°C but it cuts the temperature axis at – 273.15°C i.e., absolute zero.

Question 9.
Point out the difference between London dispersion forces and dipole-dipole forces.
Answer:
Dispersion forces are the attractive forces among non-polar molecules such as H2, He, etc., while dipole-dipole forces are attractive forces among the polar molecules.

Question 10.
Falling liquid drops are spherical. Why?
Answer:
Because of surface tension, the molecules tend to minimise the surface area and the sphere has minimum surface area.

Question 11.
Can a gas with a = 0 be liquefied?
Answer:
A gas with a = 0 means absence of intermolecular force. Hence such a gas cannot be liquefied.

Question 12.
A bottle of liquid ammonia is cooled before opening the seal, why?
Answer:
If the seal is opened without cooling, the liquid ammonia suddenly vaporises and gushes out of the bottle with force. This may lead to serious accidents.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 13.
Why does ethyl alcohol have lower boiling point than water?
Answer:
Due to hydrogen bonding. Hydrogen bonding in water is more extensive than that in ethanol.

Question 14.
Suppose water has risen up to a height h in a glass capillary. The tube is then broken from a point much below concave meniscus. Predict whether the water will continuously flow or stop.
Answer:
The liquid will rise to the top but will not flow out of the capillary. The decrease in height is compensated by increase in radius of curvature.

Question 15.
What is evaporation?
Answer:
The process of escape of liquid molecules from its surface into the gas phase is called evaporation.

Question 16.
List various factors on which rate of evaporation of a liquid depends.
Answer:
The rate of evaporation of a liquid depends upon the following factors:

  1. Nature of liquid
  2. Temperature
  3. Surface area of liquid
  4. Blowing of current of air across the surface of liquid.

Question 17.
Define vapour pressure.
Answer:
It is the pressure exerted by vapours of the liquid in equilibrium with the liquid at particular temperature. Vapour pressure of a liquid depends upon temperature.

Question 18.
Define surface tension.
Answer:
Surface tension is the energy required to increase the surface area of a liquid by a unit constant. The S.I. unit of surface tension is J m2.

Question 19.
Which property of the liquid accounts for the following:
(i) Spherical shape of liquid drops. ‘
(ii) Slow flow of castor oil?
Answer:
(i) surface tension
(ii) viscosity.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 20.
Explain, why water would completely fill a fine capillary tube which is open at both ends, when one end is immersed in water.
Answer:
The surface tension pulls water into capillary. In a fine capillary, the surface tension is great enough to overcome the attraction of gravity on water.

Question 21.
Why are the falling liquid drops spherical?
Answer:
Due to surface tension the molecules tend to minimise the surface area and surface area is minimum in case of a sphere.

Question 22.
Define viscosity.
Answer:
The force of friction that one part of a liquid offers to another part of the liquid is called viscosity.

Question 23.
What are the units of viscosity?
Answer:
The units of viscosity are dynes cm-2s also called Poise. The SI unit of viscosity is Nm-2 s.

Question 24.
How do the following properties of a liquid change with change in its tem¬perature?
(i) Surface tension
(ii) Viscosity.
Answer:
(i) Surface tension decreases with increase in temperature.
(ii) Viscosity decreases with increase in temperature.

Question 25.
What is meant by excluded volume or co-volume?
Answer:
The van der Waal’s constant V is known as co-volume or excluded volume. It is a measure of the size of gas molecules. Its value is four times the actual volume of the gas molecules (i.e., b = 4 × actual volume).

Question 26.
What would happen if the molecular collisions are non-elastic?
Answer:
If the collisions are non-elastic, there would be constant loss of energy. This would cause molecular motion to slow down and ultimately the mass will settle down at the bottom of the container like a precipitate.

Question 27.
In the plot of Z (compressibility factor) vs. P, Z attains a value of unity at particular pressure. What does it signify?
Answer:
This implies that at this value of pressure attractive and repulsive forces balance each other. Below this pressure attraction dominates and Z < 1. Above this pressure repulsion dominate and Z > 1.

Question 28.
Why aerated water bottles are kept underwater during summer?
Answer:
Aerated water contains CO2 dissolved in water. The solubility of gas decreases with rise in temperature. Thus, amount of free gas in bottle can increase in summer, which may result into increase of pressure and ultimately an explosion. To avoid this they are kept under cold water.

Question 29.
Clearly make a distinction between boiling point and normal boiling point of a liquid.
Answer:
The temperature at which the vapour pressure of a liquid equals the atmospheric pressure is called the boiling point of the liquid. The normal boiling point of a liquid is the boiling point at 1 atmospheric pressure.

Question 30.
What are mesogenic substances?
Answer:
Some substances on melting change into a transition phase in which part of the molecular order is still maintained and a total conversion of the solid state into liquid state does not occur readily. Such substances are known as mesogenic substances (i.e., substances which can produce liquid crystals).

Question 31.
What are liquid crystals?
Answer:
In liquid crystals, the internal structure has an order intermediate between those of solids and liquids. They show the characteristic property of anisotropy and have the mechanical properties of liquids. They show magnetic and optical properties of crystalline solids.

MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter

Question 32.
How can you describe the liquid state in kinetic energy model?
Answer:
In kinetic energy model, the liquid state can be described as follows:

  • The attractive forces between molecules are appreciable.
  • The molecules are relatively close together.
  • The molecules are in constant motion.
  • The average K.E. of molecules is proportional to absolute temperature (i.e., K.E. T).

Question 33.
Differentiate between evaporation and boiling.
Answer:
Difference between Evaporation and Boiling

Evaporation Boiling
1. It occurs at the surface of the liquid. 1. It involves the formation of bubbles even below the surface within the bulk of the liquid.
2. It occurs spontaneously at all temperatures. 2. It occurs only at a specific temperature at which vapour pressure equals the imposed pressure on the liquid surface.
3. It is a slow phenomenon. 3. It is a rapid phenomenon.

Question 34.
Why excluded volume (b’ is four times the actual volume of molecules?
Answer:
Assume molecules to be spheres with radius ‘r’.
∴ The volume of molecule = \(\frac{4 \pi r^{3}}{3}\)
The closest approach between two molecules = 2r (as shown in figure)
Since two molecules cannot come closer than distance 2r, thus, excluded volume per pair of molecules in binary collision
= Volume of sphere of radius 2r
MP Board Class 11th Chemistry Solutions Chapter 5 States of Matter 2
Excluded volume per molecule
= \(\frac{1}{2}\left[8 \times \frac{4 \pi r^{3}}{3}\right]\) = \(\frac{4 \pi r^{3}}{3}\)
= 4 × volume of one molecule.

Question 35.
Discuss the deviation of real gases from ideal gas behaviour with respect to pressure and temperature.
Answer:
Effect of Pressure. The extent of deviation at different pressures can be under¬stood in terms of compressibility factor (Z).
We know, Z = \(\frac{\mathrm{PV}}{n \mathrm{RT}}\)
For ideal gas Z = 1 (PV = nRT)
For real gas Z≠1 (PV ≠nRT)
In case z < 1, then the gas is more compressible than expected from ideal gas behaviour. Such a deviation is considered as negative deviation. In case z > 1, it indicates that the gas is less compressible than expected from ideal gas behaviour. Such a deviation is considered as positive deviation.
Effect of Temperature. It has been observed that the deviation is quite large at low temperatures. As the temperature increases the deviation becomes lesser and lesser. At certain temperature, the deviation almost disappears and the gas becomes behaving like an ideal gas. This temperature is called Boyle’s Temperature.