Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

This is a long answer type question as classified in NCERT Exemplar

 (i) When egg is placed in a dilute mineral acid solution (preferably dilute HCl solution), the hard external CaCO₃ layer of the egg dissolves out /removed without damaging its semipermeable membrane.

(ii) Yes, this egg can be inserted into a bottle with a narrow neck without distorting in shape. The process involved utilising phenomenon of osmosis is explained as below -

Egg is placed in a mineral acid solution – after some time the egg is removed and placed in a hypertonic solution- size of the egg gradually decreases after some time and it shrivels due to osmosis. Since the egg has shrivelled it can now be inserted easily into a bottle with a narrow mouth. The egg is, therefore, placed in a bottle with a narrow neck & then a hypotonic solution is filled into this bottle. On adding hypotonic solution, the egg regains shape due to osmosis.

Hypertonic solution- is a solution with  higher salt concentration  than that of  the normal body cells so that the solvent/water is drawn out of the cell by osmosis or any solution with higher osmotic pressure  than another solution is called "Hypertonic solution".    

Hypotonic solution is a solution  with lower salt concentration than that of the normal body  cells so that water/solvent flows into the cell by osmosis or hypotonic solution is a solution which has lower osmotic pressure than the other solution. 

This is a long answer type question as classified in NCERT Exemplar

The process of osmosis is of immense biological and industrial importance as is evident from the following examples:

A. Biological Importance :

(i) Osmosis is responsible for some of the water movement from the soil into the plant roots and then into the top parts of the plant.

(ii) Adding salt to meat to protect it against bacterial action (ie. salting).

(iii) Sugar is used to protect fruits from bacterial attack. Bacterium in canned fruit shrivels and dies as a result of the osmosis process.

(iv) Salt blood cells break due to osmosis when placed in water containing less than 0.9 percent (mass by volume). Osmosis causes water retention in tissue cells and intercellular gaps in those who consume a lot of salt or eat salty foods. Edema is the name given to the puffiness or swelling that results as a result of this.

(v) When withered flowers are immersed in freshwater, they revive.

B. Industrial importance

(i) For desalination of seawater, reverse osmosis is used: when a pressure greater than osmotic pressure is applied to saltwater, clean water is squeezed out through a semipermeable membrane. For this, a number of semipermeable membranes are available.

This is a long answer type question as classified in NCERT Exemplar

"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.

This is a long answer type question as classified in NCERT Exemplar

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.

This is a long answer type question as classified in NCERT Exemplar

(i)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) 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) 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).

$\frac{\mathbf{w}}{\mathbf{V}}$ = $\mathrm{}\frac{\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$

(iv) 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) 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) 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) 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.