NIOS Class 12 Chemistry Chapter 7 Solutions

NIOS Class 12 Chemistry Chapter 7 Solutions Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Chemistry Chapter 7 Solutions Notes in English and select need one. NIOS Class 12 Chemistry Solutions English Medium Download PDF. NIOS Study Material of Class 12 Chemistry Notes Paper Code: 313.

NIOS Class 12 Chemistry Chapter 7 Solutions

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Chemistry Notes, NIOS Senior Secondary Course Chemistry Solutions in English for All Chapter, You can practice these here.

Solutions

Chapter: 7

Module – III: States of Matter

INTEXT QUESTIONS 7.1

1. List the various methods of expressing the concentration of a solution?

Ans: Molarity, Molality, Normality, Mole fraction, Mass percentage. Molarity is the number of moles of solute dissolved per litre of the solution.

2. Define the following:

(i) Molarity.

(ii) Molality.

(iii) Normality.

Ans: (i) Molarity is the number of moles of solute dissolved per litre of the solution.

(ii) Molality is the number of moles of solute dissolved per kg of solvent.

(iii) Normality is the number of gram equivalents of solute dissolved per litre of solution.

INTEXT QUESTIONS 7.2

1. State Raoult’s law.

Ans: For a solution of volatile liquids the partial vapour pressure of each liquid is proportional to its mole fraction.

2. State Henry’s law and list the conditions necessary for the validity of Henry’s law.

Ans: The mass of a gas dissolved in a solvent is directly proportional to its partial pressure. Pressure should not be too high, temperature should not be too low. The gas should not associate or dissociate.

INTEXT QUESTIONS 7.3

1. Define colligative property. List two colligative properties.

Ans: Properties that depend upon the number of particles of solute and not on the nature of solute. e.g. Elevation of boiling point, depression of freezing point.

2. What type of liquid pairs show (i) positive deviations (ii) negative deviations.

Ans: For which A–B molecular interactions are:

(i) weaker than A–A and B–B interactions.

(ii) stronger than A–A and B–B interactions.

3. Why is the determination of osmotic pressure a better method as compared to other colligative properties for determining the molar masses of biomolecules.

Ans: At low concentration the magnitude of osmotic pressure is large enough for measurement.

TERMINAL EXERCISE

1. What do you understand by ideal and non-ideal solutions?

Ans: An ideal solution is a theoretical concept where solute and solvent molecules have the same intermolecular interactions, and the solution obeys Raoult’s law at all concentrations and temperatures. A non-ideal solution, on the other hand, deviates from Raoult’s law due to differences in intermolecular forces between the solute and solvent. 

2. Define freezing point and boiling point.

Ans: Freezing Point: The freezing point is the temperature at which a liquid changes state to become a solid at standard atmospheric pressure. It is the temperature at which the solid and liquid phases of a substance are in equilibrium.

Boiling Point: The boiling point is the temperature at which a liquid changes state to become a gas at standard atmospheric pressure. It is the temperature at which the vapor pressure of the liquid equals the surrounding atmospheric pressure.

3. Derive the relationship ΔT_b = K_b m?

Ans: Derivation of ΔTb = Kb m:

The elevation of boiling point (ΔTb) is related to the molality (m) of a solution by the following equation:

ΔTb = Kb m

Where Kb is the boiling-point elevation constant.

Derivation:

(i) Boiling Point Elevation: The boiling point elevation is defined as ΔTb = Tb – T⁰b, where Tb is the boiling point of the solution and T⁰b is the boiling point of the pure solvent.

(ii) Raoult’s Law: For a dilute solution, the vapor pressure of the solvent is given by P = P⁰ Xsolvent, where P⁰ is the vapor pressure of the pure solvent and Xsolvent is the mole fraction of the solvent.

(iii) Boiling Point Elevation and Molality: Using thermodynamic principles, it can be shown that ΔTb ∝ m, where m is the molality of the solution.

(iv) Boiling-Point Elevation Constant (Kb): The constant of proportionality is Kb, which depends on the properties of the solvent. Therefore, ΔTb = Kb m.

This equation shows that the elevation of boiling point is directly proportional to the molality of the solution.

4. A solution containing 7 g of a non-volatile solute in 250 g of water boils at 373.26 K. Find the molecular mass of the solute.

Ans: To find the molecular mass of the solute, we can use the boiling point elevation formula:

Given,

Mass of solute = 7 g

Mass of solvent (water) = 250 g = 0.250 kg

Boiling point of pure water = 373.15 K

Boiling point of solution = 373.26 K

Boiling point elevation,

∆T(b) = 373.26 – 373.15 = 0.11 K

 Use formula:

∆T(b) = K(b) . m

m = {∆T(b)}/{K(b)} = (0.11)/(0.512)

= 0.2148 mol/kg

Molality is also:

m = (moles of solute)/(mass of solvent in kg) = (n)/(0.250)

So,

n = 0.2148 × 0.250 = 0.0537 mol

 Now, find molar mass (M):

M = (mass of solute)/(moles of solute) = (7)/(0.0537)  ≈ 130.3 g/mol

The molecular mass of the solute is approximately 130.3 g/mol.