Chemistry NCERT Exemplar Solutions Class 12th Chapter Two: Overview, Questions, Preparation

1. Define the following modes of expressing the concentration of a solution. Which of these modes are independent of temperature and why?

(i) $\frac{\mathbf{w}}{\mathbf{w}}$  (percentage mass by mass)

Ans: Mass percentage $\frac{\mathbf{w}}{\mathbf{w}}$ : It is a weight by weight relationship. Thus mathematically ,

Mass % of a component = $\mathrm{}\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$

where , W represents weight of the component or solution

(ii) $\frac{\mathbf{V}}{\mathbf{V}}$ (volume by volume percentage)

Ans: Volume percentage $\frac{\mathbf{V}}{\mathbf{V}}$ :

Volume % of a component= $\mathrm{}\frac{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$

where , V represents volume

(iii) $\frac{\mathbf{w}}{\mathbf{V}}$  (mass by volume percentage)

Ans: It is mass by volume percentage $\frac{\mathbf{w}}{\mathbf{V}}$ : Thus mathematically, where W is weight of solute in grams , but measures volume of solution in milliliters (ml).

(iv) ppm. (parts per million)

Ans: Parts per million or ppm is a unit of measure of dissolved solids in solution, in terms of a ratio between the number of parts of solids/solute to a million parts  of total volume. Thus, 1 Parts per million (ppm):- is a concentration of solution that contains 1 gram solute and 1000000 ml solution (same as 1 milligram solute per litre of solution) 

(v) x, (mole fraction)

Ans: Mole fraction of a component (x)= $\frac{\mathrm{N}\mathrm{o}.\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}$

(vi) M (Molarity)

Ans: The no. of moles of solute per litre of solution.

Molarity (M )= $\frac{\left(\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\right)}{\left(\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\right)}$

(vii) m (Molality)

Ans: The no. of moles of solute per kg of solvent.

Molality (m) = $\frac{\left(\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\right)}{\left(\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{k}\mathrm{g}\right)}$

Effect of temperature: mass %, ppm, mole fraction  and molality do not change with temperature whereas molarity, volume percentage and mass by volume percentage changes with temperature because volume of solution (liquid) varies with changes in temperature.

2. Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.

(i) CHCl₃ (l) and CH₂Cl₂ (l)

Ans: For a binary solution having both components as volatile liquids (viz. CHCl₃ and CH₂Cl₂), the total pressure will be

p= $\frac{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{a}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{i}\mathrm{x}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}}{\mathrm{b}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}\mathrm{s}\mathrm{ }}$

p₁= partial vapour pressure of component 1 (ie. CHCl₃)

p₂= partial vapour pressure of component 2 (ie. CH₂Cl₂)

(ii) NaCl(s) and H₂O (l)

Ans: For a solution containing non-volatile solute ie. NaCl (s) and H₂O (l), the Raoult’s law is applicable only to vaporisable component (1) ie. H₂O (l) and total vapour pressure is written as 

P= P₁= x₁P₁⁰+ x₂P₂⁰

= x₁P₁⁰+ (1-x₁) P₂⁰

(P₁⁰-P₂⁰)X¹ + P₂⁰

p= $\frac{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{a}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{i}\mathrm{x}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}}{\mathrm{b}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}\mathrm{s}\mathrm{ }}$

p₁= partial vapour pressure of component 1 (CHCl₃ ) 

p₂= partial vapour pressure of component 2 (CH₂Cl₂)

3. Explain the terms ideal and non-ideal solutions in the light of forces of interactions operating between molecules in liquid solutions.

Ans: Ideal solution: The 'ideal solution' is a binary solution of two volatile liquids that follows Raoult's rule at any concentration and temperature.

Ideal solutions are formed when the intermolecular attractive forces between the solute(A) and the solvent(B) (ie.A-B interaction) are approximately equal to those between the solvent-solvent (A-A) and the solute-solute (BB). Enthalpy of mixing, mixing H=0, in such a perfect solution.

Volume change on mixing, Δ mixing V=0.

Examples: n- hexane and n-heptane.

Non ideal solution:  At any concentration and temperature, these binary solutions of two volatile liquids do not obey Raoult's law.

P_A≠P_oAx_A

P_B≠P_oBx_B

Furthermore, non-ideal solutions are formed when the intermolecular attractive forces between the solute and the solvent (A-B interaction) are not equal (either stronger or weaker) to those between the solvent and the solvent (A-A) and the solute and the solute (B-B). 

Enthalpy of mixing, ΔH is not equal to 0. & Volume change on mixing,   V is not equal to 0.

Example: CS₂ and acetone.

where,P_oA, P_oB denote the vapour pressures of pure solvent and PA, PB denote the partial vapour pressures of components A and B in solution, and X denotes the mole fractions of the two components denoted by the subscripts A and B.

4. Why is it not possible to obtain pure ethanol by fractional distillation? What general name is given to binary mixtures which show deviation from Raoult’s law and whose components cannot be separated by fractional distillation. How many types of such mixtures are there?

Ans: "Azeotropes” is the general term for binary mixes that deviate from Rault's law and whose components cannot be separated by fractional distillation. Because of the following reasons, fractional distillation cannot produce pure ethanol: Azeotropes are binary solutions (liquid mixes) with the same composition in the liquid and vapour phases, therefore fractional distillation cannot separate the components of an azeotrope. On fractional distillation, an ethanol-water mixture (obtained via sugar fermentation) yields a solution containing approximately 95 percent ethanol by volume of ethanol. Because the liquid and vapour phases have the identical composition, they cannot be separated. There are two sorts of binary mixtures that are referred to as-

(1) Minimum boiling azeotrope: At a given composition, non-ideal solutions show a considerable positive divergence from the minimal boiling azeotrope. For instance, 95 percent ethanol and 5% water (by volume): Below are the boiling points of pure ethanol, water, and its azeotrope. Ethanol has a melting point of 351.3 degrees Celsius, while water has a melting point of 373 degrees Celsius and azeotrope has a melting point of 351.1 degrees Celsius.

(2) Maximum boiling azeotrope: Non-ideal solutions with a considerable negative departure from Rault's law from the maximum boiling azeotrope at a given composition. By mass, nitric acid and water have a composition of 68 percent nitric acid and 32 percent water. The boiling point of such an azeotropic HNO₃-H₂O mixture is 393.5 K.