CBSE Class 12 Chemistry Solutions Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. Calculate the volume of 75% of H₂SO₄ by weight (d = 1.8 g/mL) required to prepare 1 L of 0.2 M solution.
Answer:
M₁ = \( \frac{\text{Mass \%} \times d \times 10}{98} \) = \( \frac{75 \times 1.8 \times 10}{98} \) = 13.78 M
M₁V₁ = M₂V₂
13.78 × V₁ = 0.2 × 1000 mL
V₁ = 14.5 mL

Teacher's Note:
1. Find the molarity of the stock acid first; only then can the dilution formula be used.

Question 2. Why water cannot be completely separated from aqueous solution of ethyl alcohol?
Answer:
Due to formation of azeotrope at (95.4%).

Teacher's Note:
1. Ethanol–water is a minimum boiling azeotrope; distillation stops at this composition.

Question 3. Why anhydrous salts like NaCl or CaCl₂ are used to clear snow from roads on hills?
Answer:
They depress freezing point of water.

Teacher's Note:
1. CaCl₂ gives 3 ions per formula unit, so it lowers the freezing point even more than NaCl.

Question 4. What is the effect on boiling and freezing point of a solution on addition of NaCl?
Answer:
Boiling point increases and freezing point decreases.

Teacher's Note:
1. Both changes follow from the lowering of vapour pressure; NaCl gives i ≈ 2, so the effect is double that of a non-electrolyte.

Question 5. Why osmotic pressure is considered as colligative property?
Answer:
It depends upon number of moles of solute present in solution.

Teacher's Note:
1. π = iCRT contains only concentration and temperature, nothing about the nature of the solute.

Question 6. Liquid A and B on mixing produce a warm solution. Which type of deviation does this solution show?
Answer:
Negative deviation.

Teacher's Note:
1. Heat is released (ΔH_mix < 0) because A–B attractions are stronger than A–A and B–B.

Question 7. Give an example of a compound in which hydrogen bonding results in the formation of a dimer.
Answer:
Carboxylic acids, e.g. ethanoic acid (acetic acid) or benzoic acid in benzene.

Teacher's Note:
1. Dimerisation halves the number of particles, so i ≈ 0.5 and the observed molar mass is nearly doubled.

Question 8. What role does the molecular interaction play in solution containing chloroform and acetone?
Answer:
H-bonding formed, results in negative deviation from Raoult’s law.

Teacher's Note:
1. The H of CHCl₃ bonds with the O of acetone; new attractions mean lower vapour pressure than ideal.

Question 9. How is the molality of a solution different from its molarity?
Answer:
Molality of a solution is the number of moles of the solute present in 1 kg of the solvent and does not change with temperature. Molarity of a solution is the number of moles of the solute present in 1 litre of the solution and changes with temperature.

Teacher's Note:
1. Note both differences: kg of solvent versus litre of solution, and temperature-independent versus temperature-dependent.

Question 10. How are substitutional solids formed?
Answer:
Substitutional solids are formed when some atoms, ions or molecules of one solid occupying the lattice site are replaced by atoms, ions or molecules of another solid having similar nature and nearly same size.

Teacher's Note:
1. Brass (Zn atoms replacing some Cu atoms) is a good example to quote.

Question 11. How is the solubility of gases in water related with their Henry's constants at the same pressure and temperature?
Answer:
According to Henry's law p_A = K_H x_A. At constant partial pressure, greater the K_H of the gas, smaller is x_A, i.e., less is the solubility.

Teacher's Note:
1. Higher K_H, lower solubility – this inverse relation is often tested.

Question 12. How does Henry's constant (K_H) of a gas in a particular solvent vary with temperature?
Answer:
According to Henry's law p_A = K_H x_A. With increase of temperature, solubility of a gas decreases at the same partial pressure, i.e., x_A decreases. Hence, at constant p_A, K_H increases. Thus, K_H increases with increase of temperature.

Teacher's Note:
1. Reason through the formula: falling solubility at fixed pressure forces K_H up.

Question 13. Why oxygen mixed with helium is used by deep sea divers?
Answer:
At great depths the pressure is high, so more nitrogen of air dissolves in blood. When the diver comes up, the pressure decreases, the dissolved nitrogen bubbles out and blocks capillaries, causing a painful condition called bends.
To avoid this, the tanks used by divers are filled with air diluted with helium (11.7% He, 56.2% N₂ and 32.1% O₂), because helium is much less soluble in blood.

Teacher's Note:
1. Link the answer to Henry's law: solubility of a gas rises with pressure, so less soluble helium is safer.

Question 14. Why does vapour pressure of a liquid decrease when a non-volatile solute is added into it? [UP Board 2012]
Answer:
Some liquid molecules at the surface are replaced by the molecules of the solute which are non-volatile.

Teacher's Note:
1. Fewer solvent molecules at the surface means fewer can escape into the vapour per unit time.

Short Answer Questions (2 Marks)

Question 15. Calculate the freezing point of a solution containing 8.1 g of HBr in 100 g of water, assuming the acid to be 90% ionized.
[Given: Molar mass Br = 80 g/mol, K_f water = 1.86 K kg/mol] [CBSE SQP 2018-2019]
Answer:
HBr → H⁺ + Br⁻
i = 1 − α + nα,   n = 2
i = 1 + α
ΔT_f = iK_fm
ΔT_f = (1 + α) 1.86 K kg mol⁻¹ × \( \frac{8.1 \text{ g} \times 1000 \text{ g kg}^{-1}}{81 \text{ g mol}^{-1} \times 100 \text{ g}} \)
ΔT_f = 3.53 °C
T_f° = 0 °C
ΔT_f = T_f° − T_f′
T_f′ = −3.534 °C

Teacher's Note:
1. With α = 0.9, i = 1.9; students often use i = 2 or forget i altogether. Molar mass of HBr = 1 + 80 = 81 g mol⁻¹.

Question 16. 18 g of glucose, C₆H₁₂O₆ (Molar Mass = 180 g mol⁻¹) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil?
(K_b for water = 0.52 K kg mol⁻¹, boiling point of pure water = 373.15 K) [CBSE Delhi 2013; NCERT]
Answer:
ΔT_b = K_b × m
T_b − T_b° = 0.52 K kg mol⁻¹ × \( \frac{18 \text{ g}}{180 \text{ g mol}^{-1}} \) × \( \frac{1}{1 \text{ kg}} \)
⇒ T_b − 373.15 = \( \frac{0.52}{10} \)
⇒ T_b − 373.15 = 0.052 K
⇒ T_b = 0.052 + 373.15
∴ T_b = 373.202 K

Teacher's Note:
1. The question asks for the boiling point, so add ΔT_b to 373.15 K; stopping at 0.052 K loses the final mark.

Question 17. Out of the following three solutions, which has the highest freezing point and why?
(a) 0.1 M urea   (b) 0.1 M BaCl₂   (c) 0.1 M Na₂SO₄
Answer:
Depression in freezing point is a colligative property and depends on the number of solute particles (ΔT_f = iK_fm).
Urea does not dissociate (i = 1), while BaCl₂ and Na₂SO₄ each give 3 ions (i = 3).
Hence 0.1 M urea has the least depression and therefore the highest freezing point.

Teacher's Note:
1. Highest freezing point means the smallest ΔT_f, i.e. the fewest particles.

Question 18. Which of the following solutions have highest boiling point and why?
(a) 1M glucose   (b) 1M KCl   (c) 1M aluminium nitrate
Answer:
Elevation in boiling point depends on the number of solute particles (ΔT_b = iK_bm).
Glucose: i = 1; KCl → K⁺ + Cl⁻: i = 2; Al(NO₃)₃ → Al³⁺ + 3NO₃⁻: i = 4.
Hence 1 M aluminium nitrate gives the most particles and has the highest boiling point.

Teacher's Note:
1. At equal concentration, rank solutions by their van’t Hoff factor.

Question 19. Equal moles of liquid P and Q are mixed. What is the ratio of their moles in the vapour phase? Given that p_P⁰ = 2 × p_Q⁰.
Answer:
Since equal moles are mixed, x_P = x_Q = 0.5
By Raoult’s law, p_P = p_P⁰ x_P and p_Q = p_Q⁰ x_Q
Mole ratio in vapour = \( \frac{y_P}{y_Q} \) = \( \frac{p_P}{p_Q} \) = \( \frac{2p_Q^0 \times 0.5}{p_Q^0 \times 0.5} \) = 2
Ratio of moles of P and Q in the vapour phase = 2 : 1

Teacher's Note:
1. By Dalton’s law, the mole fraction of a component in the vapour is proportional to its partial pressure.

Question 20. On mixing liquid X and Y, volume of the resulting solution decreases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
Answer:
Decrease in volume (ΔV_mix < 0) shows that X–Y attractions are stronger than X–X and Y–Y attractions, so the solution shows negative deviation from Raoult’s law.
Heat is released on mixing (ΔH_mix < 0), so the temperature rises.

Teacher's Note:
1. For negative deviation, ΔV_mix and ΔH_mix are both negative; for positive deviation both are positive.

Question 21. Explain the significance of Henry’s constant (K_H). At the same temperature, hydrogen is more soluble in water than helium. Which of them will have higher value of K_H and why?
Answer:
According to Henry’s law, p = K_Hx. At a given pressure, the higher the value of K_H, the lower is the solubility of the gas in the liquid. K_H depends on the nature of the gas and on temperature.
Since hydrogen is more soluble than helium, helium will have the higher value of K_H.

Teacher's Note:
1. K_H and solubility are inversely related – a common slip is to say the more soluble gas has the higher K_H.

Question 22. How many grams of KCl should be added to 1 kg of water to lower its freezing point to −8.0°C? (K_f = 1.86 K kg/mol)
Answer:
Since KCl dissociate in water completely, i = 2.
ΔT_f = i K_f × m
m = \( \frac{\Delta T_f}{iK_f} \)
m = \( \frac{8}{2 \times 1.86} \) = 2.15 mol/kg
Grams of KCl = 2.15 × 74.5 = 160.2 g/kg

Teacher's Note:
1. Remember to include i = 2 for KCl; without it the answer doubles.

Question 23. With the help of diagram, show the elevation in boiling point colligative properties?

Answer:
A liquid boils when its vapour pressure becomes equal to the atmospheric pressure (1 atm).
On adding a non-volatile solute the vapour pressure of the solution is lowered, so the vapour pressure curve of the solution lies below that of the pure solvent.
Hence the solution has to be heated to a higher temperature T_b to make its vapour pressure equal to 1 atm.
Elevation in boiling point, ΔT_b = T_b − T_b⁰ = K_bm

Teacher's Note:
1. In the diagram, read both boiling points where the curves cut the 1 atm line; the gap between them is ΔT_b.

Question 24. What do you mean by colligative properties? Which colligative property is used to determine molar mass of polymer and why?
Answer:
Colligative properties are those properties of dilute solutions which depend only on the number of solute particles and not on their nature, e.g. relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure.
Osmotic pressure is used to determine the molar mass of polymers because it can be measured at room temperature and its value is appreciable even for very dilute solutions, while polymers are often unstable at higher temperatures.

Teacher's Note:
1. For a polymer solution ΔT_b and ΔT_f are too small to measure accurately, but π is still measurable.

Question 25. Define reverse osmosis. Write its one use.
Answer:
When a pressure larger than the osmotic pressure is applied on the solution side, the solvent flows from the solution into the pure solvent through the semipermeable membrane. This is called reverse osmosis.
Use: Desalination of water.

Teacher's Note:
1. In ordinary osmosis solvent moves into the solution; reverse osmosis pushes it the other way using external pressure.

Question 26. Why does an azeotropic mixture distills without any change in composition?
Answer:
It has same composition of components in liquid and vapour phase.

Teacher's Note:
1. Since the vapour has the same composition as the liquid, condensing it gives back the same mixture.

Question 27. Under what condition Van’t Hoff factor is:
(a) equal to 1?   (b) less than 1?   (c) more than 1?
Answer:
(a) i = 1 when the solute neither associates nor dissociates in the solution.
(b) i < 1 when the solute undergoes association in the solution.
(c) i > 1 when the solute undergoes dissociation in the solution.

Teacher's Note:
1. Association lowers the number of particles and dissociation raises it – that sets the direction of i.

Question 28. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer:
\( \frac{p_A^0 - p_A}{p_A^0} \) = \( \frac{w_B \times M_A}{M_B \times w_A} \)
\( \frac{1.013 - 1.004}{1.013} \) = \( \frac{2 \times 18}{M_B \times 98} \)
M_B = 41.35 g/mol

Teacher's Note:
1. At the normal boiling point the vapour pressure of pure water is 1.013 bar, and 2% means 2 g solute in 98 g water.

Short Answer Questions (3 Marks)

Question 29. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C assuming that it is completely dissociated.
Answer:
If K₂SO₄ is completely dissociated,
K₂SO₄ → 2K⁺ + SO₄²⁻; i = 3
Mol. mass of K₂SO₄ = 2 × 39 + 32 + 4 × 16 = 174 g mol⁻¹
π = iCRT = i \( \frac{W_B \times RT}{M_B \times V} \)
= \( \frac{3 \times 25 \times 10^{-3} \times 0.082 \times 298}{174 \times 2.0} \)
= 5.27 × 10⁻³ atm

Teacher's Note:
1. Convert mg to g (25 mg = 25 × 10⁻³ g) before substituting.

Question 30. If the solubility product of CuS is 6 × 10⁻¹⁶, calculate the maximum molarity of CuS in aqueous solution.
Answer:
K_sp of CuS = 6 × 10⁻¹⁶
If S is the solubility, then CuS → Cu²⁺ + S²⁻
[Cu²⁺] = S, [S²⁻] = S
K_sp = [Cu²⁺][S²⁻] = S × S = S²
Solubility S = \( \sqrt{K_{sp}} \) = \( \sqrt{6 \times 10^{-16}} \) = 2.45 × 10⁻⁸ M
Highest molarity = 2.45 × 10⁻⁸ M

Teacher's Note:
1. For a 1 : 1 salt, solubility is simply the square root of K_sp; the maximum molarity is the saturated concentration.

Question 31. Suggest the most important type of intermolecular attractive interaction in the following pairs:
(a) n-hexane and n-octane   (b) I₂ and CCl₄   (c) NaClO₄ and water
Answer:
(a) Van der Waals interaction (London dispersion forces)
(b) Van der Waals interaction (London dispersion forces)
(c) Ion-dipole interaction

Teacher's Note:
1. Two non-polar species attract only by dispersion forces; an ionic solid in a polar solvent gives ion-dipole attraction.

Question 32. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
Mole fraction of solute = \( \frac{1}{1 + \frac{1000}{18}} \) = 0.0177
\( \frac{p^0 - p_A}{p^0} \) = 0.0177
\( \frac{12.3 - p_A}{12.3} \) = 0.0177
p_A = 12.08 kPa

Teacher's Note:
1. 1 molal means 1 mol solute in 1000 g water, i.e. 1000/18 = 55.5 mol of water.

Question 33. 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. (Molar mass of KOH = 56 g mol⁻¹)
Answer:
Mass of KOH = 30 g (in 100 g of solution)
M = \( \frac{n_B}{V\,(mL)} \) × 1000 = \( \frac{W_B}{M_B \times V\,(mL)} \) × 1000 = \( \frac{30}{56 \times V} \) × 1000
6.90 = \( \frac{30 \times 1000}{56 \times V} \)
V = \( \frac{30 \times 1000}{56 \times 6.90} \) = 77.64 mL
D = \( \frac{\text{Mass of solution}}{V} \) = \( \frac{100}{77.64} \) = 1.29 g mL⁻¹

Teacher's Note:
1. Take 100 g of solution as the basis: it contains 30 g KOH, and its volume follows from the molarity.

Question 34. An anti-freeze solution is prepared from 222.6 g of ethylene glycol C₂H₄(OH)₂ and 200 g of water. Calculate the molality of the solution. If the density of this solution be 1.072 g mL⁻¹, what will be the molarity of the solution?
Answer:
M_B of C₂H₄(OH)₂ = 62 g mol⁻¹
Molality = \( \frac{n_B}{W_A} \) × 1000 = \( \frac{W_B}{M_B \times W_A} \) × 1000 = \( \frac{222.6 \times 1000}{62 \times 200} \) = 17.95 m
Density = \( \frac{\text{Mass}}{\text{Volume}} \)
So, Volume = \( \frac{\text{Mass}}{\text{Density}} \) = \( \frac{422.6}{1.072} \) = 394.22 mL
M = \( \frac{n_B}{V} \) × 1000 = \( \frac{222.6}{394.22 \times 62} \) × 1000 = 9.11 M

Teacher's Note:
1. For molarity, the volume is of the whole solution (solute + solvent mass = 422.6 g divided by density), not of the water alone.

Question 35. What would be the molar mass of compound if 6.21 g of it is dissolved in 24.0 g of CHCl₃ to form a solution that has a boiling point of 68.04°C. The boiling point of pure chloroform is 61.7°C and the boiling point elevation constant K_b for chloroform is 3.63°C/m.
Answer:
Elevation in boiling point ΔT_b = 68.04 − 61.7 = 6.34°C
Mass of substance W_B = 6.21 g
Mass of CHCl₃ W_A = 24.0 g
K_b = 3.63°C/m
M_B = \( \frac{K_b \times W_B \times 1000}{\Delta T_b \times W_A} \) = \( \frac{3.63 \times 6.21 \times 1000}{6.34 \times 24} \)
= 148.15 g mol⁻¹

Teacher's Note:
1. The factor 1000 is there because W_A is in grams while molality is per kilogram of solvent.

Question 36. A solution of glycerol (C₃H₈O₃) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C while pure water boils at 100°C. What mass of glycerol was dissolved to make the solution? (K_b = 0.512 K kg mol⁻¹)
Answer:
ΔT_b = 100.42 − 100 = 0.42 K; M_B of glycerol = 92 g mol⁻¹
W_B = \( \frac{\Delta T_b \times M_B \times W_A}{K_b \times 1000} \) = \( \frac{0.42 \times 92 \times 500}{0.512 \times 1000} \)
= 37.73 g

Teacher's Note:
1. Glycerol is a non-electrolyte, so i = 1; a temperature difference has the same value in °C and K.

Question 37. Differentiate between molality and molarity of a solution. What is the effect of change in temperature of a solution on its molality and molarity ? (CBSE 2009, Jharkhand Board 2011)
Answer:
Molarity (M) is the number of moles of the solute dissolved per litre of the solution: M = \( \frac{n_2}{V\ (\text{in L})} \). Unit: mol L⁻¹.
Molality (m) is the number of moles of the solute dissolved per kilogram of the solvent: m = \( \frac{n_2}{W_1\ (\text{in kg})} \). Unit: mol kg⁻¹.
Molarity is based on volume of the solution while molality is based on mass of the solvent.
Effect of temperature : Molarity decreases with increase in temperature because volume of the solution increases with temperature. Molality does not change with temperature because masses do not change with temperature.

Teacher's Note:
1. Write the units with the definitions; examiners often give a mark for mol L⁻¹ versus mol kg⁻¹.

Long Answer Questions (5 Marks)

Question 38. (i) State Raoult’s law for a solution containing volatile components. How does Raoult’s law becomes a special case of Henry’s law?
(ii) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (K_f for benzene = 5.12 K kg mol⁻¹). [CBSE OD 2013]
Answer:
(i) Partial vapour pressure of a liquid component is directly proportional to its mole fraction in its solution.
The partial pressure of the volatile component or gas is directly proportional to its mole fraction in its solution. Only the proportionality constant K_H differs from p_A°. Thus, Raoult’s law becomes a special case of Henry’s law in which K_H becomes equal to p_A°.
(ii) Given, W_B = 1.00 g, W_A = 50 g, K_f = 5.12 K kg mol⁻¹; ΔT_f = 0.40 K
ΔT_f = K_f \( \frac{W_B \times 1000}{M_B \times W_A (\text{in grams})} \)
M_B = K_f \( \frac{W_B \times 1000}{\Delta T_f \times W_A} \)
= \( \frac{5.12 \times 1 \times 1000}{0.40 \times 50} \) g mol⁻¹
= 256 g mol⁻¹

Teacher's Note:
1. Compare the two laws side by side: p_A = p_A°x_A (Raoult) and p = K_Hx (Henry); only the constant is different.

Question 39. (i) Define the following terms:
(a) Ideal solution
(b) Osmotic pressure.
(ii) Calculate the boiling point elevation for a solution prepared by adding 10 g CaCl₂ to 200 g of water, assuming that CaCl₂ is completely dissociated. (K_b for water = 0.512 K kg mol⁻¹; Molar mass of CaCl₂ = 111 g mol⁻¹) [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
(i) (a) The solution which obey Raoult’s law over the entire range of concentration.
(b) It is the excess pressure that must be applied to a solution to prevent osmosis.
(ii) ΔT_b = iK_bm
Here, m = w_B × 1000/M_B × w_A
ΔT_b = [3 × 0.512 K kg mol⁻¹ × 1000 × 10 g/[111 g mol⁻¹ × 200 g]
= 0.69 K

Teacher's Note:
1. CaCl₂ → Ca²⁺ + 2Cl⁻, so i = 3. The question asks only for the elevation, so 0.69 K is the final answer (the boiling point would be 100.69 °C).

Question 40. (i) Calculate the freezing point of solution when 1.9 g of MgCl₂ (M = 95 g mol⁻¹) was dissolved in 50 g of water, assuming MgCl₂ undergoes complete ionization. (K_f for water = 1.86 K kg mol⁻¹)
(ii) (a) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(b) What happens when the external pressure applied becomes more than the osmotic pressure of solution?
OR
(i) When 2.56 g of sulphur was dissolved in 100 g of CS₂, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_x). (K_f for CS₂ = 3.83 K kg mol⁻¹, Atomic mass of Sulphur = 32 g mol⁻¹).
(ii) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing;
(a) 1.2% sodium chloride solution?
(b) 0.4% sodium chloride solution? [CBSE Delhi 2016]
Answer:
(i) ΔT_f = i \( \frac{K_f W_b \times 1000}{M_b \times W_a} \)
ΔT_f = 3 × \( \left( \frac{1.86 \times 1.9}{95 \times 50} \right) \) × 1000
= 2.23 K
T_f = T_f° − ΔT_f = 273.15 − 2.23
T_f = 270.92 K
(ii) (a) 2 M glucose has a higher boiling point because higher the number of particles lesser is the vapour pressure.
(b) Reverse osmosis.
OR
(i) ΔT_f = \( \frac{K_f W_b \times 1000}{M_b \times W_a} \)
0.383 = \( \left( \frac{3.83 \times 2.56}{M \times 100} \right) \) × 1000
M = 256
S × x = 256
32 × x = 256
x = 8, i.e., the formula of sulphur is S₈.
(ii) (a) If we place blood cells in solution containing 1.2% sodium chloride solution, they shrink.
(b) It we place blood cells in solution containing 0.4% sodium chloride solution, they swell.

Teacher's Note:
1. MgCl₂ → Mg²⁺ + 2Cl⁻ gives i = 3. For blood cells: more concentrated than 0.9% (hypertonic) ⇒ water leaves and cells shrink; less concentrated (hypotonic) ⇒ cells swell.

Question 41. How can you remove the hard calcium carbonate layer of the egg without damaging its semipermeable membrane? Can this egg be inserted into a bottle with a narrow neck without distorting its shape? Explain the process involved. [NCERT Exemplar]
Answer:
This can be achieved as under :
(i) Place the egg in a mineral acid solution for about 2 hours. The outershell of the egg dissolves. Remove any portion left with your fingers.
(ii) Place the egg in a saturated solution (hypertonic) of sodium chloride for about 3 hours. Size of the egg is reduced as the egg shrivels due to osmosis.
(iii) Insert the egg in a bottle with a narrow neck. Add water to the bottle. Water will act as hypotonic solution. Egg regains shape due to osmosis.

Teacher's Note:
1. The acid reacts with CaCO₃ (CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂) but leaves the membrane intact; the rest is osmosis out of and then into the egg.

Question 42. Why is the mass determined by measuring a colligative property in case of some solutes abnormal? Discuss it with the help of Van’t Hoff factor. [NCERT Exemplar]
Answer:
Certain compounds when dissolved in suitable solvents either dissociate or associate.
For example, ethanoic acid dimerizes in benzene due to hydrogen bonding, while in water, it dissociates and forms ions. As a result, the number of chemical species in solution increases or decreases as compared to the number of chemical species of solute added to form the solution. Since, the magnitude of colligative property depends on the number of solute particles, it is expected that the molar mass determined on the basis of colligative properties will be either higher or lower than the expected value or the normal value and is called abnormal molar mass.
In order to account for the extent of dissociation or association of molecules in solution, Van’t Hoff introduced a factor, i, known as the Van’t Hoff factor.
i = \( \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} \) = \( \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} \)

Teacher's Note:
1. Remember the direction: dissociation ⇒ i > 1 and observed molar mass lower; association ⇒ i < 1 and observed molar mass higher.