CBSE Class 12 Chemistry Solutions Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Why is glycol and water mixture used in car radiators in cold countries? [HOTS]
Answer:
Ethylene glycol lowers the freezing point of water. Due to this, coolant in radiators will not freeze. Otherwise, radiator will burst due to freezing of coolant (water).

Teacher's Note:
1. Water expands on freezing, which is what would crack the radiator – mention it for full marks.

Question 2. How is it that measurement of osmotic pressures is more widely used for determining molar masses of macromolecules than the elevation in boiling point or depression in freezing point of their solutions? [CBSE (AI) 2009]
Answer:
The osmotic pressure method has the advantage over elevation in boiling point or depression in freezing point for determining molar masses of macromolecules because
(i) Osmotic pressure is measured at the room temperature and the molarity of solution is used instead of molality.
(ii) Compared to other colligative properties, its magnitude is large even for very dilute solutions.

Teacher's Note:
1. Room-temperature measurement also matters because many macromolecules (proteins) are unstable at higher temperatures.

Question 3. What is the van’t Hoff factor for a compound which undergo dimerisation in an organic solvent?
Answer:
2X → X₂
i = \( \frac{\text{Number of moles of particles after association}}{\text{Number of moles of particles before association}} \) = \( \frac{1}{2} \)

Teacher's Note:
1. For any association i < 1; for complete dimerisation the number of particles is exactly halved.

Question 4. What would be the value of van’t Hoff factor for a dilute solution of K₂SO₄ in water?
Answer:
In dilute solution, K₂SO₄ → 2K⁺ + SO₄²⁻
van’t Hoff factor, i = \( \frac{\text{Number of moles of particles after dissociation}}{\text{Number of moles of particles before dissociation}} \) = \( \frac{3}{1} \) = 3.

Teacher's Note:
1. “Dilute” is the hint that dissociation is complete, so i equals the number of ions.

Question 5. Out of two 0.1 molal aqueous solutions of glucose and of potassium chloride, which one will have a higher boiling point and why? [CBSE (F) 2013, 2019 (56/3/2)]
Answer:
0.1 m KCl solution will have higher boiling point as KCl dissociates in the solution.

Teacher's Note:
1. KCl gives K⁺ and Cl⁻, so i ≈ 2 and ΔT_b is nearly double that of glucose.

Question 6. What will happen to freezing point of a potassium iodide aqueous solution when mercuric iodide is added to solution? [HOTS]
Answer:
HgI₂ + 2KI → K₂HgI₄
Mercuric iodide forms a complex with potassium iodide, therefore, the number of solute particles (KI) in the solution decreases resulting in the decrease in the value of ΔT_f, i.e., depression in the freezing point. As a result, the freezing point (T_f = T_f° − ΔT_f) of the solution will increase.

Teacher's Note:
1. Count ions: 2KI give 4 ions, but K₂[HgI₄] gives only 3 (2K⁺ + [HgI₄]²⁻), so particles decrease.

Question 7. Why is osmotic pressure of 1 M KCl higher than 1 M urea solution?
Answer:
This is because KCl dissociates to give K⁺ and Cl⁻ ions while urea being a molecular solid does not dissociate into ions in the solution.

Teacher's Note:
1. With π = iCRT, i ≈ 2 for KCl and 1 for urea, so π(KCl) is nearly double.

Teacher's Note:
1. Remember “hypo” = lower and “hyper” = higher osmotic pressure, always compared with another solution.

Question 9. State Henry’s law about partial pressure of a gas in a mixture. [CBSE OD 2012]
Answer:
Henry’s law states that the mass of a gas dissolved in given volume of the liquid at constant temperature depends upon the pressure applied.
In terms of partial pressure, the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution: p = K_Hx.

Teacher's Note:
1. Write the mathematical form p = K_Hx along with the statement – it makes the answer complete.

Teacher's Note:
1. Note that p = K_Hx is the form of Henry’s law; Raoult’s law is the special case where K_H equals the vapour pressure of the pure component.

Question 11. Define an ideal solution. [CBSE OD 2013, 2012]
Answer:
Solutions which obey Raoult’s law over the entire range of concentration at specific temperature.

Teacher's Note:
1. For an ideal solution Δ_mixH = 0 and Δ_mixV = 0; these are useful to quote when a second mark is asked.

Question 12. Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’? [CBSE Delhi 2014]
Answer:
Azeotropes: Binary mixtures having same composition in liquid and vapour phase and boil at a constant temperature.

Teacher's Note:
1. Because liquid and vapour have the same composition, an azeotrope cannot be separated into its components by fractional distillation.

Question 13. What type of intermolecular attractive interaction exists in the pair of methanol and acetone? [CBSE Delhi 2014]
Answer:
Hydrogen bonding.

Teacher's Note:
1. The O–H of methanol can hydrogen bond with the C=O oxygen of acetone, but these bonds are weaker than the methanol–methanol H-bonds broken, so the mixture shows positive deviation.

Teacher's Note:
1. Note that ΔT_b and ΔT_f use molality (m), while osmotic pressure uses molarity (C).

Short Answer Questions (2 Marks)

Question 15. Derive the relationship between relative lowering of vapour pressure and molar mass of the solute. [CBSE Chennai 2015]
Answer:
From Raoult’s law we have, p = p_A + p_B
If solute B is non-volatile, then
p = p_A or p = p_A°x_A
p = p_A°(1 − x_B) = p_A° − p_A°x_B
p_A°x_B = p_A° − p
\( \frac{p_A^\circ - p}{p_A^\circ} = x_B \)   or   \( \frac{\Delta p}{p_A^\circ} = \frac{n_B}{n_A + n_B} \)
For a dilute solution n_B << n_A, so n_B can be neglected in denominator in comparison to n_A.
\( \frac{\Delta p}{p_A^\circ} = \frac{n_B}{n_A} = \frac{W_B/M_B}{W_A/M_A} = \frac{W_B\times M_A}{M_B\times W_A} \)
\( M_B = \frac{p_A^\circ\times W_B\times M_A}{\Delta p\times W_A} \)

Teacher's Note:
1. The step n_A + n_B ≈ n_A is valid only for dilute solutions – state this assumption in your derivation.

Question 16. When 1.5 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point of benzene raised from 353.23 K to 353.93 K. Calculate the molar mass of the solute. (K_b for benzene = 2.52 K kg mol⁻¹) [CBSE Chennai 2015]
Answer:
ΔT_b = 353.93 K − 353.23 K = 0.7 K
Substituting W_B = 1.5 g, W_A = 90 g, ΔT_b = 0.7 K, K_b = 2.52 K kg mol⁻¹ in the expression
\( M_B = \frac{K_b\times W_B\times 1000}{\Delta T_b\times W_A} \), we get
\( M_B = \frac{2.52\times1.5\times1000}{0.7\times90} \)
M_B = 60 g mol⁻¹

Teacher's Note:
1. The factor 1000 converts the mass of solvent from grams to kilograms; leave it out only if W_A is already in kg.

Question 17. Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol⁻¹) in 250 g of water. (K_f of water = 1.86 K kg mol⁻¹) [CBSE Delhi 2018]
Answer:
Here, W_B = 60 g, M_B = 180 g mol⁻¹, W_A = 250 g, K_f = 1.86 K kg mol⁻¹
Substituting these values in the expression
\( \Delta T_f = \frac{K_f\times W_B\times 1000}{M_B\times W_A} \), we get
\( \Delta T_f = \frac{1.86\times60\times1000}{180\times250} = 2.48\text{ K} \)
Freezing point of solution, T_f = T_f° − ΔT_f = 273.15 K − 2.48 K = 270.67 K

Teacher's Note:
1. The question asks for the freezing point, not the depression – do not stop at 2.48 K.

Question 18. Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure. [CBSE Delhi 2012]
Answer:
Osmotic pressure (π) is defined as the extra pressure that must be applied to the solution side in order to prevent the flow of solvent molecules into it through a semipermeable membrane.
According to van’t Hoff equation
\( \pi = \frac{n_B}{V}RT \)
where π is the osmotic pressure, R is the gas constant and V is the volume of solution in litres containing n_B moles of the solute.
If W_B grams of the solute of molar mass, M_B is present in the solution, then
\( n_B = \frac{W_B}{M_B} \) and we can write
\( \pi = \frac{W_B\times R\times T}{M_B\times V} \)   or   \( M_B = \frac{W_B\times R\times T}{\pi\times V} \)
Thus, knowing W_B, T, π and V, the molecular mass of the solute, M_B can be calculated.

Teacher's Note:
1. Use R = 0.0821 L atm K⁻¹ mol⁻¹ with π in atm and V in litres (or 0.083 L bar K⁻¹ mol⁻¹ with π in bar).

Question 19. A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass. [CBSE (AI) 2011]
Answer:
\( M_B = \frac{W_B\times R\times T}{\pi\times V} \)   ...(i)
Here, W_B = 8.95 mg = 8.95 × 10⁻³ g, R = 0.0821 L atm mol⁻¹ K⁻¹
T = 25°C = (25 + 273) K = 298 K, π = 0.335 torr = \( \frac{0.335}{760} \) atm
V = 35 mL = 35 × 10⁻³ L
Substituting these values in the equation (i), we get
\( M_B = \frac{8.95\times10^{-3}\text{ g}\times0.0821\text{ L atm mol}^{-1}\text{K}^{-1}\times298\text{ K}\times760}{0.335\text{ atm}\times35\times10^{-3}\text{ L}} \) = 14193.3 g mol⁻¹

Teacher's Note:
1. Convert all three units first: mg to g, torr to atm (÷ 760) and mL to L – most mistakes here are unit slips.

Question 20. Give reasons:
(i) Cooking is faster in pressure cooker than in cooking pan.
(ii) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water. [CBSE 2019 (56/2/1)]
Answer:
(i) The use of pressure cooker reduces cooking time because the weight over the lid does not allow the steam to go out. As a result, pressure inside the cooker becomes high. Higher the pressure, higher is the boiling point and faster is the cooking.
(ii) As the concentration of saline solution is higher than the concentration inside the cell. Thus water will move outside the cytoplasm and the cell will shrink while, distilled water is hypotonic, when RBCs are placed in distilled water, water will enter the cell through simple diffusion and lead to cell swelling.

Teacher's Note:
1. Use the words hypertonic (saline water) and hypotonic (distilled water) and name the process as osmosis.

Question 21. (i) On mixing liquid X and liquid 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?
(ii) What happens when we place the blood cell in water (hypotonic solution)? Give reason. [CBSE Allahabad 2015]
Answer:
(i) The solution will show negative deviation from Raoult’s law. Temperature will rise.
(ii) Due to osmosis water enters into the cell and blood cell will swell.

Teacher's Note:
1. Volume decrease and heat release both come from stronger X–Y attractions, which is why they go together with negative deviation.

Question 22. Define the following terms:
(i) Abnormal molar mass
(ii) van’t Hoff factor (i) [CBSE Delhi 2017]
Answer:
(i) When the molar mass of a substance determined by using any of the colligative properties comes out to be different than the theoretically expected molar mass, the substance said to show abnormal molar mass.
(ii) van’t Hoff factor (i) gives the extent of association or dissociation of the solute particles in the solution. It may be defined as the ratio of observed colligative property to calculated colligative property.
i = \( \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} \)

Teacher's Note:
1. Equivalent form: i = normal molar mass ÷ abnormal (observed) molar mass – note that the ratio is inverted for molar masses.

Question 23. Will the elevation in boiling point be same if 0.1 mol of sodium chloride or 0.1 mol of sugar is dissolved in 1 L of water? Explain. [CBSE Sample Paper 2016] [HOTS]
Answer:
No, the elevation in boiling point is not the same. NaCl, being an electrolyte, dissociates almost completely to give Na⁺ and Cl⁻ ions whereas sugar, being non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double than 0.1 M sugar solution. Elevation in boiling point being a colligative property, is therefore, nearly twice for 0.1 M NaCl solution than for 0.1 M sugar solution.

Teacher's Note:
1. With 1 L of water (about 1 kg), 0.1 mol of solute gives about 0.1 m; then ΔT_b(NaCl) ≈ 2 × ΔT_b(sugar).

Question 24. Define the following terms :
(i) Mole fraction (x)
(ii) Molality of a solution (m) [CBSE Comptt. OD 2015]
Answer:
(i) Mole fraction of a component = \( \frac{\text{Number of moles of the component}}{\text{Total number of moles of all the components}} \)
(ii) Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent.
OR Molality (m) = \( \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \)

Teacher's Note:
1. Molality uses mass of solvent (not solution), which is why it does not change with temperature, unlike molarity.

Question 25. Calculate the molality of ethanol solution in which the mole fraction of water is 0.88. [CBSE SQP 2018-19]
Answer:
Mole fraction of water, X_H2O = 0.88
Mole fraction of ethanol, X_C2H5OH = 1 − 0.88 = 0.12
X_C2H5OH = \( \frac{n_2}{n_1 + n_2} \)   ...(1)
n₂ = number of moles of ethanol, n₁ = number of moles of water.
Molality of ethanol means the number of moles of ethanol present in 1000 g of water.
n₁ = \( \frac{1000}{18} \) = 55.5 moles
Substituting the value of n₁ in equation (1): \( \frac{n_2}{55.5 + n_2} \) = 0.12
n₂ = 7.57 moles
Molality of ethanol (C₂H₅OH) = 7.57 m
Alternatively,
Mole fraction of water = 0.88; Mole fraction of ethanol = 1 − 0.88 = 0.12
Therefore 0.12 moles of ethanol are present in 0.88 moles of water.
Mass of water = 0.88 × 18 = 15.84 g of water.
Molality = number of moles of solute (ethanol) present in 1000 g of solvent (water) = 0.12 × 1000 / 15.84 = 7.57 m

Teacher's Note:
1. Students often confuse molarity and molality: molarity is volume based (per litre of solution), molality is mass based (per kg of solvent).

Question 26. State Henry’s law. What is the effect of temperature on the solubility of a gas in a liquid? [CBSE Delhi 2014]
Answer:
The relationship between pressure and solubility is guided by Henry’s law. According to this law, “The mass of a gas dissolved in given volume of the liquid at a constant temperature depends upon the pressure which is applied.”
It can also be stated as the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution.
p = K_Hχ, where K_H = Henry’s constant.

Effect of Temperature on the Solubility: The solubility of a gas decreases with increase in temperature. This is because dissolution of gases in the liquids is an exothermic process. Therefore, according to Le-Chatelier’s principle, the increase in temperature results in decrease in the solubility of the gas.

Teacher's Note:
1. A higher K_H means lower solubility at the same pressure, and K_H increases with temperature – that is the same fact as gases being less soluble when warm.

Question 27. State Henry’s law and mention two of its important applications. [CBSE Comptt. OD 2012]
Answer:
Henry’s law: The mass of a gas dissolved in given volume of the liquid at a constant temperature depends upon the pressure which is applied. It can also be stated as the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution, p = K_Hx.
Applications of Henry’s Law:
(i) To increase the solubility of CO₂ in soda water and soft drinks, the bottle is sealed under high pressure.
(ii) At high altitude, low blood oxygen causes climber to become weak and make them unable to think clearly, which are symptoms of a condition known as anoxia.

Teacher's Note:
1. Carefully note the units of Henry’s constant in numericals – K_H has pressure units (bar, kbar or torr) when written as p = K_Hx.

Question 28. State Raoult’s law. How is it formulated for solutions of non-volatile solutes? [CBSE Comptt. Delhi 2012]
OR
Derive expression for Raoult’s law when the solute is non-volatile. [CBSE Comptt. Delhi 2013]
Answer:
Raoult’s law for solution of non-volatile solute: The relative lowering of vapour pressure for a solution is equal to the mole fraction of solute when solvent alone is volatile.
\( \frac{p_A^\circ - p_A}{p_A^\circ} \) = χ_B
where, p_A° → Vapour pressure of pure component ‘A’
p_A → Partial vapour pressure of component ‘A’
χ_B → Mole fraction of solute
p_A° − p_A → Lowering of vapour pressure
\( \frac{p_A^\circ - p_A}{p_A^\circ} \) → Relative lowering of vapour pressure.

Teacher's Note:
1. The derivation starts from p_A = χ_Ap_A° (only the solvent is volatile) and uses χ_A = 1 − χ_B to reach the relative lowering form.

Short Answer Questions (3 Marks)

Question 29. Given alongside is the sketch of a plant for carrying out a process.
(i) Name the process occurring in the given plant.
(ii) To which container does the net flow of solvent take place?
(iii) Name one SPM which can be used in this plant.
(iv) Give one practical use of the plant.

Answer:
(i) Reverse osmosis.
(ii) Fresh water container.
(iii) Cellulose acetate placed on a suitable support.
(iv) Desalination of sea water.

Teacher's Note:
1. The clue is the applied pressure being greater than π on the sea-water side – that reverses the normal direction of osmosis.

Question 30. The partial pressure of ethane over a saturated solution containing 6.56 × 10⁻² g of ethane is 1 bar. If the solution were to contain 5.0 × 10⁻² g of ethane, then what will be the partial pressure of the gas? [CBSE Comptt. Delhi 2013]
Answer:
According to Henry’s law, m = K_Hp
⇒ K_H = \( \frac{m}{p} \) = \( \frac{6.56\times10^{-2}\,(\text{g})}{1\,(\text{bar})} \) = 6.56 × 10⁻² g bar⁻¹
In the second case,
p = \( \frac{m}{K_H} \) = \( \frac{5.0\times10^{-2}\,(\text{g})}{6.56\times10^{-2}\,(\text{g bar}^{-1})} \) = 0.762 bar

Teacher's Note:
1. Write each step while solving a numerical and always mention appropriate units. Since mass of gas dissolved is proportional to pressure, a quick ratio check (5.0/6.56 ≈ 0.76) confirms the answer.

Question 31. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. [NCERT]
Answer:
Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
∴ Mass of carbon tetrachloride = (100 − 30) g = 70 g
Molar mass of benzene (C₆H₆) = (6 × 12 + 6 × 1) g mol⁻¹ = 78 g mol⁻¹
∴ Number of moles of C₆H₆ = \( \frac{30}{78} \) mol = 0.3846 mol
Molar mass of carbon tetrachloride (CCl₄) = 1 × 12 + 4 × 35.5 = 154 g mol⁻¹
∴ Number of moles of CCl₄ = \( \frac{70}{154} \) mol = 0.4545 mol
Thus, the mole fraction of C₆H₆ is given as:
\( \frac{\text{Number of moles of C}_6\text{H}_6}{\text{Number of moles of C}_6\text{H}_6 + \text{Number of moles of CCl}_4} \) = \( \frac{0.3846}{0.3846 + 0.4545} \) = 0.458

Teacher's Note:
1. For mass percentage problems, assume 100 g of solution – the percentages then become grams directly.

Question 32. Components of a binary mixture of two liquids A and B were being separated by distillation. After sometimes separation of components stopped and composition of vapour phase became same as that of liquid phase. Both the components started coming in the distillate. Explain why this happened. [NCERT Exemplar]
Answer:
Both components are appearing in the distillate and composition of liquid and vapour is same, this shows that liquids have formed azeotropic mixture. Therefore, it cannot be separated at this stage by distillation.
Azeotropes are binary mixtures having the same composition in fluid, vapour phase and boils at a constant temperature. The components of azeotropic mixtures are separated by fractional distillation.

Teacher's Note:
1. Fractional distillation can only bring the mixture up to the azeotropic composition; to go beyond it other methods (e.g. adding a third substance) are needed.

Question 33. The vapour pressure of pure liquids A and B at 400 K are 450 and 700 mm Hg respectively. Find out the composition of liquid mixture if total pressure at this temperature is 600 mm Hg. [CBSE Comptt. Delhi 2017]
Answer:
p_total = p₁° + (p₂° − p₁°)x₂
600 = 450 + (700 − 450)x₂
x₂ = 0.6
x₁ = 1 − 0.6 = 0.4
Detailed Answer:
p_A° = 450 mm Hg, p_B° = 700 mm Hg, p_Total = 600 mm Hg
According to Raoult’s law,
p_A = χ_A × p_A°,   p_B = χ_B × p_B°
p_Total = p_A + p_B = (1 − χ_B)p_A° + χ_Bp_B° = p_A° + (p_B° − p_A°)χ_B
600 = 450 + (700 − 450)χ_B
600 − 450 = 250χ_B ⇒ 150 = 250χ_B
χ_B = \( \frac{150}{250} \) = 0.6 ⇒ χ_A = 1 − 0.6 = 0.4
Composition: χ_A = 0.4, χ_B = 0.6

Teacher's Note:
1. Check your answer: 0.4 × 450 + 0.6 × 700 = 180 + 420 = 600 mm Hg, which matches the total pressure.

Question 34. Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.
(a) CHCl₃(l) and CH₂Cl₂(l)
(b) NaCl(s) and H₂O(l) [NCERT Exemplar]
Answer:
(a) p = p⁰_Ax_A + p⁰_Bx_B; where p⁰_A, p⁰_B are the vapour pressure of pure components (A) and (B) while x_A and x_B are the mole fractions of the components in the solution.
(b) NaCl(s) is a non-volatile solute. When it is dissolved in water, the vapour pressure is lowered. Vapour pressure of solution can be calculated using the following relations:
p = p⁰x_a   ...(i)
where, x_a = mole fraction of solvent, p⁰ = vapour pressure of pure solvent, p = vapour pressure of solution
Similarly, \( \frac{\Delta p}{p^0} \) = x_b,   Δp = (p⁰ − p),   x_b = mole fraction of solute   ...(ii)

Teacher's Note:
1. In (a) both liquids are volatile so both contribute to the total vapour pressure; in (b) only water is volatile, so only the solvent term remains.

Question 35. 45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate
(i) the freezing point depression and
(ii) the freezing point of the solution
(Given : K_f of water = 1.86 K kg mol⁻¹) [CBSE Comptt. Delhi 2015]
Answer:
(i) ΔT_f = K_fm
⇒ ΔT_f = K_f \( \frac{W_B \times 1000}{M_B \times W_A} \)
⇒ ΔT_f = \( \frac{1.86 \text{ K kg mol}^{-1} \times 45 \text{ g} \times 1000 \text{ g kg}^{-1}}{62 \text{ g mol}^{-1} \times 600 \text{ g}} \)
ΔT_f = 2.25 K or 2.25 °C
(ii) T_f° − T_f = 2.25 °C
⇒ 0 °C − T_f = 2.25 °C
⇒ T_f = −2.25 °C or 270.75 K

Teacher's Note:
1. Ethylene glycol is HOCH₂CH₂OH, C₂H₆O₂ (molar mass 62 g mol⁻¹); it is a non-electrolyte, so i = 1.

Question 36. Calculate the freezing point of an aqueous solution containing 10.5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. (Molar mass of Magnesium bromide = 184 g mol⁻¹, K_f for water = 1.86 K kg mol⁻¹). [CBSE Comptt. Delhi/OD 2018]
Answer:
Moles for MgBr₂ = \( \frac{10.5}{184} \) = 0.0571 mol
Molality = \( \frac{0.0571 \text{ g mol}}{200 \text{ g}} \) × 1000 g kg⁻¹ = 0.2855 mol kg⁻¹
i = 3
ΔT_f = i K_f m
= 3 × 1.86 K kg mol⁻¹ × 0.2855 mol kg⁻¹
= 1.59 K
Freezing point = 273 − 1.59
= 271.41 K or −1.59 °C

Teacher's Note:
1. MgBr₂ → Mg²⁺ + 2Br⁻ gives 3 ions, so i = 3; leaving out i gives only 0.53 K.

Question 37. The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C₆H₅COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van’t Hoff factor and the percentage association of benzoic acid. (K_f for benzene = 5.12 K kg mol⁻¹) [CBSE SQP 2017]
Answer:
ΔT_f = i K_f m
2.12 K = i × \( \frac{5.12 \text{ K kg mol}^{-1} \times 2.5 \text{ g} \times 1000 \text{ g kg}^{-1}}{122 \text{ g mol}^{-1} \times 25 \text{ g}} \)
i = 0.505
for association
i = 1 − \( \frac{\alpha}{2} \)
α = 0.99
Percentage association of benzoic acid is 99.0%

Teacher's Note:
1. For dimerisation, 2A ⇄ A₂, so i = 1 − α + α/2 = 1 − α/2; remember to convert α to a percentage at the end.

Long Answer Questions (5 Marks)

Question 38. (i) Why a person suffering from high blood pressure is advised to take minimum quantity of common salt?
(ii) 2 g of benzoic acid (C₆H₅COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol⁻¹. What is the percentage association of acid if it forms dimer in solution? [CBSE 2019 (56/5/2)]
Answer:
(i) Osmotic pressure is directly proportional to the concentration of the solutes. Our body fluid contains a number of solutes. On taking large amount of common salt, Na⁺ and Cl⁻ ions enter into the body fluid thereby raising the concentration of the solutes. As a result, osmotic pressure increases which may rupture the blood cells.
(ii) The given quantities are
W_B = 2 g, W_A = 25 g, ΔT_f = 1.62 K, K_f = 4.9 K kg mol⁻¹
Substituting these values in equation \( M_B = \frac{K_f\times W_B\times1000}{\Delta T_f\times W_A} \), we get
\( M_B = \frac{4.9\times2\times1000}{1.62\times25} \) = 241.98 g mol⁻¹
Thus, observed molecular mass of benzoic acid in benzene = 241.98 g mol⁻¹
Normal molecular mass of C₆H₅COOH = 122 g mol⁻¹
∴ i = \( \frac{\text{Normal molecular mass}}{\text{Observed molecular mass}} \) = \( \frac{122\text{ g mol}^{-1}}{241.98\text{ g mol}^{-1}} \) = 0.504
i = 0.504   ...(i)
If α represents the degree of association of solute then we would have (1 − α) mole of benzoic acid left in unassociated form and corresponding \( \frac{\alpha}{2} \) as associated moles of benzoic acid at equilibrium.
Now, consider the following equilibrium for the acid:
2(C₆H₅COOH) ⇄ (C₆H₅COOH)₂
Initial: 1 mol, 0; at equilibrium: (1 − α) mol, \( \frac{\alpha}{2} \) mol
Thus, total number of moles of particles at equilibrium
= 1 − α + \( \frac{\alpha}{2} \) = 1 − \( \frac{\alpha}{2} \)
i = \( \frac{\text{Total number of moles of particles after association}}{\text{Number of moles of particles before association}} \)
i = \( \frac{1-\frac{\alpha}{2}}{1} = 1-\frac{\alpha}{2} \)   ...(ii)
From (i) and (ii), we have
0.504 = 1 − \( \frac{\alpha}{2} \)
\( \frac{\alpha}{2} \) = 1 − 0.504 = 0.496
or α = 0.496 × 2 = 0.992
Therefore, degree of association of benzoic acid in benzene is 99.2%.

Teacher's Note:
1. For dimerisation, i = 1 − α/2; for association into n molecules in general, i = 1 − α(1 − 1/n).

Question 39. (i) Give reasons for the following:
(a) At higher altitudes, people suffer from a disease called anoxia. In this disease, they become weak and cannot think clearly. [CBSE 2019 (56/4/1)]
(b) When mercuric iodide is added to an aqueous solution of KI, the freezing point is raised.
(ii) 0.6 mL of acetic acid (CH₃COOH), having density 1.06 g mL⁻¹, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.
Answer:
(i) (a) At higher altitudes, partial pressure of oxygen is less than that at ground level, so that oxygen concentration becomes less in blood or tissues. Hence, people suffer from anoxia.
(b) Due to the formation of complex K₂[HgI₄], number of particles in the solution decreases and hence the freezing point is raised.
(ii) Mass = Density × Volume
∴ Mass of acetic acid = 1.06 g mL⁻¹ × 0.6 mL = 0.636 g
Number of moles of acetic acid = \( \frac{\text{Mass of acetic acid}}{\text{Molar mass}} \) = \( \frac{0.636\text{ g}}{60\text{ g mol}^{-1}} \) = 0.0106 mol
Mass of water = 1 g mL⁻¹ × 1000 mL = 1000 g
Molality = \( \frac{\text{Number of moles of acetic acid}}{\text{Mass of water in grams}} \) × 1000
= \( \frac{0.0106}{1000} \) × 1000 = 0.0106 mol kg⁻¹
ΔT_f = K_f . m
= 1.86 K kg mol⁻¹ × 0.0106 mol kg⁻¹
= 0.0197 K
van’t Hoff Factor (i) = \( \frac{\text{Observed depression in freezing point}}{\text{Calculated depression in freezing point}} \) = \( \frac{0.0205\text{ K}}{0.0197\text{ K}} \)
i = 1.041   ...(i)
If α is the degree of dissociation of acetic acid, then we would have n(1 − α) moles of undissociated acid, nα moles of CH₃COO⁻ and nα moles of H⁺ ions at equilibrium.
CH₃COOH ⇄ H⁺ + CH₃COO⁻
Initial: n mol, 0, 0; at equilibrium: n(1 − α), nα mol, nα mol
Thus, total moles of particles = n − nα + nα + nα = n(1 + α)
i = \( \frac{\text{Total number of moles of particles after dissociation}}{\text{Number of moles of particles before dissociation}} \)
i = \( \frac{n(1+\alpha)}{n} \) = 1 + α   ...(ii)
From (i) and (ii), we have
1.041 = 1 + α   ⇒   α = 1.041 − 1 = 0.041
\( K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \)
[CH₃COOH] = n(1 − α) = 0.0106 (1 − 0.041) = 0.0106 × 0.959
[H⁺] = nα = 0.0106 × 0.041
[CH₃COO⁻] = nα = 0.0106 × 0.041
∴ \( K_a = \frac{0.0106\times0.041\times0.0106\times0.041}{0.0106\times0.959} \)
= 1.86 × 10⁻⁵

Teacher's Note:
1. Since the solution volume is about 1 L, moles can be used directly as concentrations in K_a; this matches the standard value for acetic acid (about 1.8 × 10⁻⁵).

Teacher's Note:
1. For a dilute solution n_B is neglected in the denominator, so χ_B ≈ n_B/n_A; using the exact χ_B = 0.5/47.5 gives practically the same 23.55 mm Hg.

Question 41. (i) Define the following terms :
(a) Ideal solution
(b) Azeotrope
(c) Osmotic pressure
(ii) A solution of glucose (C₆H₁₂O₆) in water is labelled as 10% by weight. What would be the molality of the solution?
(Molar mass of glucose = 180 g mol⁻¹) [CBSE OD 2013]
Answer:
(i) (a) Ideal solution : Those solutions which follow Raoult’s law at specific temperature over the entire range of concentration are called ideal solution.
(b) Azeotrope : A liquid mixture which distills at constant temperature without undergoing any changes in composition is called azeotrope.
(c) Osmotic pressure : The minimum excess pressure that has to be applied on the solution side to prevent the entry of the solvent into the solution through the semipermeable membrane is called osmotic pressure.
(ii) Given : Mass of solute, w = 10 g, Mass of solvent, W = 90 g, Molar mass of solute, M = 180
Molality = \( \frac{w \times 1000}{M \times W} \)
= \( \frac{10 \times 1000}{90 \times 180} \)
= 0.6172 = 0.62 mol kg⁻¹

Teacher's Note:
1. 10% by weight means 10 g glucose in 100 g of solution, so the solvent is 90 g; using 100 g gives the wrong molality (0.56).

Question 42. (i) Define the following terms :
(a) Mole fraction,
(b) Ideal solution.
(ii) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at −0.34 °C. What is the molar mass of the material? [K_f for water = 1.86 K kg mol⁻¹] [CBSE OD 2012]
Answer:
(i) (a) The ratio of number of moles of one component to the total number of moles of solution is known as mole fraction.
(b) The solution which follows Raoult’s law over entire range of concentration at specific temperature is called ideal solution.
(ii) W_B = 15 g, W_A = 450 g, ΔT_f = 0.34 K, M_B = ?
M_B = \( \frac{1000 \times K_f \times W_B}{\Delta T_f \times W_A} \)
= \( \frac{1000 \text{ g kg}^{-1} \times 1.86 \text{ K kg mol}^{-1} \times 15 \text{ g}}{0.34 \text{ K} \times 450 \text{ g}} \)
= 182.35 g/mol

Teacher's Note:
1. A freezing point of −0.34 °C means ΔT_f = 0.34 K (a difference of 1 °C equals 1 K).