CBSE Class 12 Chemistry Solutions Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. What are isotonic solutions ? Give one example. (CBSE 2014)
Answer:
Solutions having equal osmotic pressure are called isotonic solutions, e.g., 0.1 M glucose and 0.1 M sucrose solution.

Teacher's Note:
1. Isotonic solutions at the same temperature have equal molar concentration of solute particles, so no osmosis occurs between them.

Question 2. What is van’t Hoff equation for dilute solution ?
Answer:
πV = nRT where n is the no. of moles of solute present in V litres of solution, π is osmotic pressure, T is temperature and R is gas constant or solution constant.

Teacher's Note:
1. Dividing by V gives π = CRT, the form most often used in numericals.

Question 3. Why does water from the soil rise to the top of a tall tree ?
Answer:
Due to osmosis through the roots because root cell walls are made up of semipermeable membrane.

Teacher's Note:
1. The concentration of sap inside root cells is higher than that of soil water, so water flows in by osmosis.

Question 4. What happens when blood cells are placed in pure water ?
Answer:
Due to osmosis, water molecules move into blood cells through the cell walls. As a result, blood cells swell and may even burst.

Teacher's Note:
1. Pure water is hypotonic to the cell fluid; this inflow is called endosmosis.

Question 5. Define reverse osmosis. Give one use of it. (CBSE 2011, Raj. Board 2012)
Answer:
In osmosis, net flow of solvent across the semipermeable membrane is from solvent to solution. The flow can be reversed by applying pressure greater than osmotic pressure on the solution. This is called reverse osmosis. It is used for desalination of sea water.

Teacher's Note:
1. The applied pressure must exceed the osmotic pressure; equal pressure only stops osmosis.

Question 6. Define osmotic pressure. (CBSE 2007)
Answer:
Osmotic pressure is the excess pressure that must be applied on the solution side to just prevent the passage of solvent into it through a semipermeable membrane.

Teacher's Note:
1. Osmotic pressure is a colligative property: π = CRT.

Question 7. What happens when a pressure greater than osmotic pressure is applied on the solution side separated from the solvent by a semipermeable membrane ? (CBSE 2016, 2020)
Answer:
When the pressure applied on the solution becomes greater than the osmotic pressure of the solution, then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side. The process is called reverse osmosis.

Teacher's Note:
1. Only solvent molecules move in reverse osmosis; the solute stays behind, which is why it purifies water.

Question 8. Give one most important application of the phenomenon of depression in freezing point in everyday life.
Answer:
In lowering the freezing point of water in the car radiator by adding suitable solutes (ethylene glycol) to run it in sub-zero weather.

Teacher's Note:
1. Spreading salt on icy roads is another everyday use of the same effect.

Question 9. What would be the value of van’t Hoff factor for dilute aqueous solution of K₂SO₄? (Uttarakhand 2012)
Answer:
K₂SO₄ → 2 K⁺ + SO₄²⁻   ∴ i = 3.

Teacher's Note:
1. For complete dissociation, i equals the number of ions given by one formula unit.

Question 10. Arrange the following solutions in increasing order of their van’t Hoff factor :
0.1 M CaCl₂, 0.1 M KCl, 0.1 M Al₂(SO₄)₃, 0.1 M C₁₂H₂₂O₁₁ (Raj. Board 2012)
Answer:
Greater the number of ions produced on dissociation, greater is the van’t Hoff factor. Hence, the order is
0.1 M C₁₂H₂₂O₁₁ < 0.1 M KCl < 0.1 M CaCl₂ < 0.1 M Al₂(SO₄)₃

Teacher's Note:
1. Ideal values: sucrose 1, KCl 2, CaCl₂ 3, Al₂(SO₄)₃ 5.

Question 11. When is the value of van’t Hoff factor more than one ?
Answer:
i > 1 when the solute undergoes dissociation in the solution.

Teacher's Note:
1. i < 1 indicates association; i = 1 means neither.

Question 12. Define molal elevation constant or ebullioscopic constant.
Answer:
It is equal to elevation in boiling point that takes place when molality of the solution is unity.

Teacher's Note:
1. For water K_b = 0.52 K kg mol⁻¹; it depends only on the solvent.

Question 13. Between 2 M glucose solution and 1 M glucose solution, which one has a lower freezing point? (Karnataka Board 2012)
Answer:
Higher the concentration, greater is the depression and hence lower is the freezing point. Thus, 2 M glucose solution will have lower freezing point.

Teacher's Note:
1. Greater depression means a lower freezing point; do not confuse the two words.

Question 14. Between 1 M glucose solution and 1 M NaCl solution which one will have higher boiling point and why ?
Answer:
1 M NaCl will have higher boiling point because NaCl dissociates in the solution and hence the number of particles is more and therefore, the elevation in boiling point is more. Glucose is a non-electrolyte and does not dissociate or associate in the solution.

Teacher's Note:
1. Compare i × concentration: NaCl gives about 2 M particles, glucose only 1 M.

Question 15. How much molar mass do you expect for NaCl using colligative properties ? (Assume complete dissociation)
Answer:
As NaCl dissociates to give 2 ions in the solution, observed colligative property will be double than normal value and molar mass will be halved viz. 58.5/2 = 29.25.

Teacher's Note:
1. Observed molar mass = normal molar mass / i.

Short Answer Questions (2 Marks)

Question 16. What are the characteristics of an ideal solution? Why do solutions behave ideally only at low concentration ?
Answer:
Characteristics : (i) It obeys Raoult’s law over the entire range of concentration, i.e., p_A = p_A°x_A and p_B = p_B°x_B. (ii) Δ_mixH = 0. (iii) Δ_mixV = 0. (iv) A–B interactions are nearly the same as A–A and B–B interactions.
At low concentration, solute particles are far apart and the solution is mostly solvent. The solute-solute interactions are negligible and the solute-solvent interactions hardly alter the environment of solvent molecules. Hence, the solution behaves nearly ideally; with increase in concentration these interactions become significant and deviations appear.

Teacher's Note:
1. A perfectly ideal solution is rare; very dilute solutions approach ideal behaviour.

Question 17. State Raoult’s law. Using the law, how would you distinguish between ideal and non-ideal solutions ?
Answer:
Raoult’s law : For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction: p_A = p_A°x_A, p_B = p_B°x_B.
A solution which obeys Raoult’s law over the entire range of concentration is an ideal solution; its plots of p_A, p_B and p_total against mole fraction are straight lines.
A solution which does not obey Raoult’s law is a non-ideal solution; its vapour pressure is either higher (positive deviation) or lower (negative deviation) than predicted, and the plots are curves.

Teacher's Note:
1. On a vapour pressure versus composition graph, ideal means straight lines; any bulge up or down means non-ideal.

Question 18. What type of deviation (positive or negative) from ideal behaviour will be shown by the solution of cyclohexane and ethanol ? Give suitable reason.
Answer:
The solution shows positive deviation.
In pure ethanol, molecules are held together by hydrogen bonds. On adding cyclohexane, its molecules get in between the ethanol molecules and break some hydrogen bonds. The cyclohexane–ethanol interactions are weaker than ethanol–ethanol interactions, so molecules escape more easily and the vapour pressure is higher than predicted by Raoult’s law.

Teacher's Note:
1. A non-polar liquid mixed with a hydrogen-bonded liquid nearly always gives positive deviation.

Question 19. What type of azeotropic mixture will be formed by a solution of acetone and chloroform ? Justify on the basis of strength of intermolecular interactions that develop in the solution. (CBSE 2019)
Answer:
A maximum boiling azeotrope is formed.
Chloroform molecule forms a hydrogen bond with the oxygen of acetone: Cl₃C–H···O=C(CH₃)₂. These chloroform–acetone interactions are stronger than chloroform–chloroform and acetone–acetone interactions. Hence, the escaping tendency of molecules decreases, vapour pressure is lowered (negative deviation) and the boiling point is raised, giving a maximum boiling azeotrope.

Teacher's Note:
1. Negative deviation goes with maximum boiling azeotrope; positive deviation with minimum boiling azeotrope.

Question 20. State Raoult’s law. Write the conditions necessary for a solution to show ideal behaviour.
Answer:
Raoult’s law : For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction: p_A = p_A°x_A.
Conditions for ideal behaviour : (i) It should obey Raoult’s law over the entire range of concentration. (ii) Δ_mixH = 0. (iii) Δ_mixV = 0. (iv) A–B interactions should be nearly the same as A–A and B–B interactions (components of similar structure and polarity).

Teacher's Note:
1. Ideal behaviour also needs the solution to be fairly dilute for real liquids.

Question 21. Why constant boiling mixtures behave like a single component when subjected to distillation ?
Answer:
A constant boiling mixture (azeotrope) has the same composition in the liquid phase and in the vapour phase. Hence, on boiling, the vapour formed has the same composition as the liquid, the composition of the liquid does not change and it boils at a constant temperature. Therefore, it distils over unchanged like a pure single liquid and cannot be separated by fractional distillation.

Teacher's Note:
1. Fractional distillation works only when the vapour is richer in one component than the liquid; for an azeotrope it is not.

Question 22. What is the importance of semipermeable membrane in osmosis? Explain. (UP Board 2012)
Answer:
A semipermeable membrane allows only solvent molecules to pass through it but not the solute particles. Osmosis can take place only through such a membrane. If the membrane allowed solute particles also to pass, diffusion would take place and concentrations on the two sides would simply become equal; no osmotic flow and no osmotic pressure would be observed.
Examples: animal bladder, cell membranes, parchment paper, and copper ferrocyanide, Cu₂[Fe(CN)₆], deposited in the pores of a porous pot (artificial membrane).

Teacher's Note:
1. Without the semipermeable membrane there is no osmotic pressure; this is what separates osmosis from diffusion.

Question 23. (i) What is osmotic pressure ?
(ii) State van’t Hoff-Boyle’s law
(iii) What is an ideal solution ? (Karnataka Board 2012)
Answer:
(i) Osmotic pressure is the excess pressure that must be applied on the solution side to just prevent the passage of solvent into it through a semipermeable membrane.
(ii) At constant temperature, the osmotic pressure of a dilute solution is directly proportional to its molar concentration (or inversely proportional to the volume of solution containing 1 mole of solute): π ∝ C or πV = constant.
(iii) An ideal solution is one which obeys Raoult’s law over the entire range of concentration, with Δ_mixH = 0 and Δ_mixV = 0.

Teacher's Note:
1. van’t Hoff-Boyle’s law is the solution analogue of Boyle’s law for gases (PV = constant).

Question 24. Why the boiling point of a liquid gets raised on dissolution of non-volatile solute into it ?
Answer:
A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. On dissolving a non-volatile solute, the vapour pressure of the solution becomes lower than that of the pure solvent. Hence, the solution has to be heated to a higher temperature to make its vapour pressure equal to the atmospheric pressure. Thus, the boiling point of the solution is higher than that of the pure solvent.

Teacher's Note:
1. Elevation in boiling point is a direct consequence of lowering of vapour pressure.

Question 25. Explain qualitatively the elevation of boiling point of the solution using Raoult’s law. (West Bengal Board 2012)
Answer:
According to Raoult’s law, the vapour pressure of a solution of a non-volatile solute is p_s = p°x₁, which is less than p° because x₁ < 1. At the normal boiling point of the solvent (T_b°), the vapour pressure of the solution is therefore less than 1 atm. To make it equal to 1 atm, the solution must be heated to a higher temperature T_b. The difference T_b − T_b° = ΔT_b is the elevation in boiling point.
As the lowering of vapour pressure is proportional to the mole fraction of solute, ΔT_b is also proportional to it; for dilute solutions ΔT_b = K_bm.

Teacher's Note:
1. More solute particles, lower vapour pressure, higher boiling point.

Question 26. Explain why the freezing point of a solvent is lowered on dissolving a non-volatile solute into it.
Answer:
Freezing point is the temperature at which the vapour pressure of the liquid becomes equal to that of the solid (both phases coexist). On dissolving a non-volatile solute, the vapour pressure of the solution is lowered. As the vapour pressure of the solid solvent falls steeply with decrease in temperature, the vapour pressure of the solution becomes equal to that of the solid solvent only at a lower temperature. Hence, the freezing point of the solution is lower than that of the pure solvent.

Teacher's Note:
1. Only the solvent freezes out; the solid formed is pure solvent, while the solute stays in the liquid.

Question 27. An aqueous solution containing urea was found to have boiling point more than the normal boiling point of water (373.13 K). When the same solution was cooled, it was found that its freezing point is less than the normal freezing point of water (273.13 K). Explain these observations.
Answer:
Urea is a non-volatile solute. On dissolving it in water, the vapour pressure of water is lowered.
(i) As the vapour pressure of the solution is lower, it has to be heated to a higher temperature to make its vapour pressure equal to the atmospheric pressure. Hence, its boiling point is higher than 373.13 K (elevation in boiling point).
(ii) The vapour pressure of the solution becomes equal to that of ice at a temperature lower than that for pure water. Hence, its freezing point is lower than 273.13 K (depression in freezing point).

Teacher's Note:
1. Both effects come from the same cause, lowering of vapour pressure, so they always occur together.

Question 28. What is van’t Hoff factor? What possible values can it have if the solute molecules undergo
(i) Association and
(ii) Dissociation, in solution ? (HP Board 2011)
Answer:
van’t Hoff factor is the ratio of the normal molar mass to the observed molar mass of the solute, or of the observed to the calculated colligative property: \( i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} \)
(i) For association, i < 1 (e.g. about 0.5 for complete dimerisation).
(ii) For dissociation, i > 1 (e.g. about 2 for KCl, 3 for K₂SO₄).

Teacher's Note:
1. If the solute neither associates nor dissociates, i = 1.

Question 29. Define
(i) Abnormal molar mass
(ii) van’t Hoff factor. (CBSE 2017)
Answer:
(i) When the molar mass of a substance determined by colligative property measurement is lower or higher than the expected (true) value, due to dissociation or association of the solute in solution, it is called abnormal molar mass.
(ii) van’t Hoff factor is the ratio of the normal (calculated) molar mass to the observed (abnormal) molar mass of the solute, or the ratio of the observed colligative property to the calculated colligative property: \( i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} \)

Teacher's Note:
1. Note the order of the ratio: normal over observed for molar mass, but observed over calculated for colligative property.

Short Answer Questions (3 Marks)

Question 30. Define the terms ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions ? (CBSE 2010)
Answer:
Osmosis : The spontaneous flow of solvent molecules from pure solvent (or a less concentrated solution) into the solution (or a more concentrated solution) through a semipermeable membrane is called osmosis.
Osmotic pressure : The excess pressure that must be applied on the solution side to just prevent osmosis is called osmotic pressure.
Advantages : (i) Osmotic pressure is measured at room temperature, so it is useful for biomolecules (proteins, polymers) which are unstable at higher temperatures. (ii) Molarity of the solution is used instead of molality. (iii) Even for very dilute solutions the osmotic pressure is large and can be measured accurately, so it is well suited for macromolecules (polymers) whose solutions have very small ΔT_b or ΔT_f.

Teacher's Note:
1. For a polymer of molar mass 10⁵ g/mol, ΔT_f is too tiny to measure, but π is a few hundred pascals, easily measured.

Question 31. Illustrate elevation in boiling point with the help of vapour pressure-temperature curve of a solution. Show that elevation in boiling point is a colligative property.

Answer:
The vapour pressure curve of the solution lies below that of the pure solvent at all temperatures (figure). The solvent boils at T_b° where its vapour pressure becomes 1 atm, but the solution reaches 1 atm only at a higher temperature T_b. Hence, elevation in boiling point, ΔT_b = T_b − T_b°.
Colligative nature : Elevation in boiling point is proportional to the lowering of vapour pressure, which is proportional to the mole fraction of the solute: ΔT_b ∝ Δp ∝ x₂.
For a dilute solution, \( x_2 \approx \frac{n_2}{n_1} = \frac{n_2 M_1}{W_1} \), which is proportional to molality m. Hence ΔT_b ∝ m or ΔT_b = K_b m.
Since ΔT_b depends only upon the number of moles of solute in a given mass of solvent and not upon its nature, it is a colligative property.

Teacher's Note:
1. In the graph, both curves must be drawn with the solution curve below the solvent curve throughout.

Question 32. Show graphically how the vapour pressure of a solvent and a solution of non-volatile solute change with temperature ? Show on this graph the boiling points of the solvent and the solution. Which is higher and why ?

Answer:
The graph shows vapour pressure against temperature for the pure solvent and for the solution. Both increase with temperature, but the solution curve lies below the solvent curve. T_b° (solvent) and T_b (solution) are the temperatures at which the vapour pressures become equal to 1 atm.
The boiling point of the solution (T_b) is higher. This is because the vapour pressure of the solution is lower than that of the pure solvent at every temperature (some surface is occupied by non-volatile solute particles), so it must be heated to a higher temperature to make its vapour pressure equal to the atmospheric pressure.

Teacher's Note:
1. Label the 1 atm line, T_b° and T_b clearly; unlabelled graphs lose marks.

Question 33. Explain how the measurement of depression in freezing point can be used for the determination of molecular masses of non-volatile solutes.
Answer:
For a dilute solution, depression in freezing point is proportional to molality: ΔT_f = K_f m
If w₂ g of solute of molar mass M₂ is dissolved in w₁ g of solvent, \( m = \frac{w_2/M_2}{w_1} \times 1000 \)
∴ \( \Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1} \) or \( M_2 = \frac{1000\, K_f\, w_2}{\Delta T_f\, w_1} \)
Knowing K_f of the solvent and measuring ΔT_f experimentally (e.g. by Beckmann’s method or Rast’s camphor method), M₂ can be calculated.

Teacher's Note:
1. Use w₁ in grams with the factor 1000; if w₁ is in kg, drop the 1000.

Question 34. With the help of a suitable diagram, show that the lowering of vapour pressure of a solution than the pure solvent causes a lowering of freezing point for the solution compared to that of the pure solvent.

Answer:
At the freezing point, the liquid and solid phases are in equilibrium, so their vapour pressures are equal. In the diagram, the vapour pressure curve of the solid solvent meets the liquid solvent curve at T_f°, the freezing point of the pure solvent.
Because a non-volatile solute lowers the vapour pressure, the curve of the solution lies below that of the pure liquid solvent. It meets the solid solvent curve at a lower temperature T_f. Hence, the freezing point of the solution is lower than that of the pure solvent, and ΔT_f = T_f° − T_f.

Teacher's Note:
1. The solid solvent curve is much steeper than the liquid curves; that is why the crossing point shifts to lower temperature.

Question 35. Why do electrolytes show abnormal molecular masses ? Name the factors responsible for the abnormality.
Answer:
Colligative properties depend upon the number of solute particles. Electrolytes dissociate into ions in solution, so the number of particles increases and the observed colligative property is higher than the calculated value. As molar mass is inversely proportional to the colligative property, the observed molar mass is lower than the true value; such molar masses are called abnormal molar masses.
Factors responsible : (i) Dissociation of the solute (number of particles increases, molar mass found is lower), e.g. NaCl, KCl in water. (ii) Association of the solute (number of particles decreases, molar mass found is higher), e.g. acetic acid or benzoic acid dimerising in benzene.

Teacher's Note:
1. Remember the inverse relation: more particles, larger colligative effect, smaller observed molar mass.

Question 36. What is Abnormal Molecular Mass ? Discuss its being in Molecular Association/Dissociation.
Answer:
When the molar mass of a substance determined from a colligative property is different from the theoretical (true) value, it is called abnormal molecular mass.
Association : If n molecules of a solute associate, nA → (A)_n, the number of particles decreases. The colligative property becomes less and the observed molar mass is more than the true value. E.g. ethanoic acid dimerises in benzene through hydrogen bonding, so its observed molar mass is nearly 120 g mol⁻¹ (true value 60). For degree of association α, i = 1 − α + α/n.
Dissociation : If a solute dissociates into n ions, the number of particles increases. The colligative property becomes more and the observed molar mass is less than the true value. E.g. KCl in water gives K⁺ and Cl⁻; its observed molar mass is nearly half of 74.5. For degree of dissociation α, i = 1 + (n − 1)α.

Teacher's Note:
1. Association gives i < 1 and higher observed molar mass; dissociation gives i > 1 and lower observed molar mass.

Question 37. What is van’t Hoff factor ? How does it help in the determination of degree of association or dissociation of a solute in solution ?
Answer:
van’t Hoff factor (i) is the ratio of the observed colligative property to the calculated (normal) colligative property, or the ratio of normal molar mass to observed molar mass:
\( i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} = \frac{\text{Normal molar mass}}{\text{Observed molar mass}} \)
Degree of dissociation : If one molecule gives n ions and α is the degree of dissociation, particles after dissociation = 1 − α + nα. Hence i = 1 + (n − 1)α or \( \alpha = \frac{i - 1}{n - 1} \)
Degree of association : If n molecules associate to form one and α is the degree of association, particles = 1 − α + α/n. Hence i = 1 − α + α/n or \( \alpha = \frac{1 - i}{1 - 1/n} \)
Thus, knowing i from colligative property measurement, α can be calculated.

Teacher's Note:
1. Find i first from observed and calculated values, then substitute in the formula for association or dissociation.

Question 38. Define osmotic pressure. Arrange the following solutions in the increasing order of their osmotic pressure :
(a) 34.2 g/lit sucrose
(b) 60 g/lit urea
(c) 90 g/lit glucose
(d) 58.5 g/lit sodium chloride.
Give reason in support of your answer.
Answer:
Osmotic pressure is the excess pressure that must be applied on the solution side to just prevent the passage of solvent into it through a semipermeable membrane.
Molar concentrations of particles: (a) sucrose = 34.2/342 = 0.1 M (b) urea = 60/60 = 1 M (c) glucose = 90/180 = 0.5 M (d) NaCl = 58.5/58.5 = 1 M, which gives 2 M particles (Na⁺ and Cl⁻, i = 2).
As π = iCRT, the osmotic pressure is proportional to the concentration of particles. Hence the increasing order is :
sucrose < glucose < urea < sodium chloride

Teacher's Note:
1. Convert g/L to mol/L first, then multiply by i for electrolytes before comparing.

Long Answer Questions (5 Marks)

Question 39. (a) Define Azeotropes and explain briefly minimum boiling azeotrope by taking suitable example.
(b) The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition of the mixture in the vapour phase.
Answer:
(a) Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boiling at a constant temperature. Solutions showing large positive deviation from Raoult’s law have a maximum in the vapour pressure curve at a particular composition; at this composition the boiling point is minimum and the mixture boils without change in composition. It is called a minimum boiling azeotrope. Example : ethanol-water mixture containing about 95% ethanol by volume boils at 351.1 K.
(b) If x_A is mol fraction of A in liquid phase, then
x_A (450) + (1 − x_A) (700) = 600
or x_A = 0.4, x_B = 0.6.
Hence, p_A = 0.4 × 450 = 180 mm, p_B = 0.6 × 700 = 420 mm,
\( y_A = \frac{180}{600} = 0.3 \), \( y_B = \frac{420}{600} = 0.7 \) i.e. y_A = 0.3, y_B = 0.7

Teacher's Note:
1. The vapour is always richer in the more volatile component (B here: 0.7 in vapour versus 0.6 in liquid).

Question 40. Define the solubility of a solid in liquid. Briefly describe the various factors on which the solubility of a solid in a liquid depends.
Answer:
Solubility of a solid in a liquid is the maximum amount of the solid that can be dissolved in 100 g (or 1 kg) of the solvent at a given temperature to form a saturated solution.
Factors :
(i) Nature of solute and solvent : “Like dissolves like”. Polar (ionic) solutes dissolve in polar solvents (NaCl, sugar in water) and non-polar solutes dissolve in non-polar solvents (naphthalene, anthracene in benzene), because the solute-solvent interactions are similar to solute-solute and solvent-solvent interactions.
(ii) Temperature : In a nearly saturated solution, dissolution is a dynamic equilibrium: Solute + Solvent ⇄ Solution. By Le Chatelier’s principle, if dissolution is endothermic (Δ_solH > 0), solubility increases with temperature; if it is exothermic (Δ_solH < 0), solubility decreases with temperature.
(iii) Pressure : Pressure has no significant effect on the solubility of solids in liquids, because solids and liquids are highly incompressible.

Teacher's Note:
1. Give the Le Chatelier reasoning for temperature; simply saying “solubility increases with temperature” is not always true.

Question 41. Define osmotic pressure and describe Berkeley and Hartley’s method for the determination of osmotic pressure. (J & K Board 2012)
Answer:
Osmotic pressure is the excess pressure that must be applied on the solution side to just prevent the passage of solvent into it through a semipermeable membrane.
Berkeley and Hartley’s method : The apparatus consists of a porous porcelain tube in whose walls a semipermeable membrane of copper ferrocyanide is deposited. The tube is fitted with a capillary tube at one end and a water reservoir at the other, and is placed inside a gun-metal jacket fitted with a piston and a pressure gauge.
Pure water is taken in the porcelain tube and the solution in the outer jacket. Water tends to pass into the solution by osmosis, which is shown by a fall in the level of water in the capillary. Pressure is applied on the solution through the piston so that the level in the capillary remains stationary. The pressure recorded on the gauge is the osmotic pressure of the solution.
Advantages : (i) The measurement is quick. (ii) The concentration of the solution does not change during the measurement because osmosis is prevented. (iii) The membrane is not strained because equal pressure acts from both sides.

Teacher's Note:
1. The principle is “just prevent osmosis”: the pressure that keeps the capillary level steady equals the osmotic pressure.

Question 42. State and explain Raoult’s law for (a) volatile solute (b) non-volatile solute.
Answer:
(a) For a solution of volatile liquids : The partial vapour pressure of each component of the solution is directly proportional to its mole fraction. For a solution of components 1 and 2, p₁ = p₁°x₁ and p₂ = p₂°x₂.
By Dalton’s law, total vapour pressure p = p₁ + p₂ = p₁°x₁ + p₂°x₂ = p₁° + (p₂° − p₁°)x₂. Thus the total vapour pressure varies linearly with the mole fraction of either component. Composition of vapour: y₁ = p₁/p, y₂ = p₂/p.
(b) For a solution of a non-volatile solute : Only the solvent contributes to the vapour pressure. The vapour pressure of the solution is directly proportional to the mole fraction of the solvent: p_s = p°x₁.
As x₁ = 1 − x₂, \( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} = x_2 \), i.e., the relative lowering of vapour pressure is equal to the mole fraction of the solute.

Teacher's Note:
1. Both forms are the same law; for a non-volatile solute, the second term p₂°x₂ is simply zero.

Question 43. What do you understand by colligative properties of a solution? Explain briefly osmosis and osmotic pressure. (J & K Board 2012)
Answer:
Colligative properties are those properties of dilute solutions of non-volatile solutes which depend only upon the number of solute particles present in a given amount of solvent or solution and not upon the nature of the solute. They are: (i) relative lowering of vapour pressure, (ii) elevation in boiling point, (iii) depression in freezing point and (iv) osmotic pressure.
Osmosis : When a solution and pure solvent (or two solutions of different concentrations) are separated by a semipermeable membrane, solvent molecules flow from the solvent (or dilute solution) into the solution (or more concentrated solution). This spontaneous flow of solvent is called osmosis, e.g. a raw mango placed in concentrated salt solution shrivels, dried raisins swell in water.
Osmotic pressure : The excess pressure that must be applied on the solution side to just prevent the flow of solvent into it through the semipermeable membrane is called osmotic pressure (π). For dilute solutions, π = CRT (van’t Hoff equation), so it depends only on the molar concentration of solute particles; hence it is a colligative property.

Teacher's Note:
1. Osmosis is the process; osmotic pressure is the pressure needed to stop it. Keep the two definitions separate.