Course Content
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
Class 11th Chemistry Online Class For 100% Result
About Lesson

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter


Question 1. What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
Answer: P1 = 1 bar,P2 = ?       V1= 500 dm3 ,V2=200 dm3
As temperature remains constant at 30°C,
1 bar x 500 dm3 = P2 x 200 dm3 or P2=500/200 bar=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?
Answer:  V1= 120 mL, P1=1.2 bar,
V2 = 180 mL, P2 = ?
As temperature remains constant, P1V1 = P2V2
(1.2 bar) (120 mL) = P2 (180mL)

States of Matter

Question 4.  At 0°C, the density of a gaseous oxide at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer: Using the expression, d =MP/RT , at the same temperature and for same density,
M1P1 = M2P2 (as R is constant)
(Gaseous oxide) (N2)
M1 x 2 = 28 x 5(Molecular mass of N2 = 28 u)
or M1 = 70u

States of Matter

Question 6. The drain cleaner, Drainex contains small bits of aluminium 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 aluminium reacts?
Answer: The chemical equation for the reaction is
2 Al + 2 NaOH + H20 -> 2 NaAl02 + 3H2 (3 x 22400 mL At N.T.P)
2 x 27 = 54 g.
54 g of Al at N.T.P release
H2 gas = 3 x 22400 0.15 g of Al at N.T.P release

States of Matter Notes &  Solution Class 11 Chemistry

Question 8. What will be the pressure of the gas mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in all vessel at 27 °C?
Answer: 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) or P2 = 0.40 bar, i.e., PH2 = 0.40 bar
Calculation of partial pressure of 02 in 1 L vessel
P1‘ V1 = P2‘V2
(0.7 bar) (2.0 L) = P2 (1L) or P2‘ = 1.4 bar, i.e.,Po2= 1.4 bar
Total pressure =PHz + PQ2 = 0.4 bar + 1.4 bar = 1.8 bar

States of Matter Notes &  Solution Class 11 Chemistry
States of Matter
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter Notes &  Solution Class 11 Chemistry
States of Matter

Question 21. In terms of Charles’ law explain why -273°C is the lowest possible temperature.
Answer: At -273°C, volume of the gas becomes equal to zero, i.e., the gas ceases to exist.

Question 22. Critical temperature for CO2 and CH4 are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?
Answer:  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.


Question 23. Explain the physical significance of vander Waals parameters.
Answer: ‘a’ is a pleasure of the magnitude of the intermolecular forces of attraction, while b is a measure of the effective size of the gas molecules.


Exercise Files
No Attachment Found
No Attachment Found
Wisdom Academy