CBSE Class 12 Chemistry Solutions Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. What is meant by term reverse osmosis? [CBSE OD 2013]
Answer:
Reverse Osmosis: 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.
(RO) is a water purification technology that uses a semipermeable membrane to remove larger particles from drinking water in reverse osmosis.

Teacher's Note:
1. The key words for the mark are “pressure greater than osmotic pressure” and “solvent flows from solution to pure solvent”.

Question 2. Define the following terms :
(i) Isotonic solutions
(ii) Van’t Hoff factor [CBSE Delhi/OD 2012]
Answer:
(i) The solutions having same osmotic pressure at a given temperature are called isotonic solutions.
(ii) Van’t Hoff factor is expressed as : i = \( \frac{\text{Normal molecular mass}}{\text{Observed molecular mass}} \)
(or any other definition)

Teacher's Note:
1. i can also be written as observed colligative property ÷ calculated colligative property; both forms earn the mark.

Question 3. Explain boiling point elevation constant for a solvent or ebullioscopic constant. [CBSE OD/Foreign 2012]
Answer:
We know that, ΔT_b = K_bm.
where m = 1, ΔT_b = K_b. Thus, boiling point elevation constant is equal to the elevation in boiling point when 1 mole of a solute is dissolved in 1 kg of solvent. It is also called ebullioscopic constant.

Teacher's Note:
1. K_b depends only on the solvent (unit K kg mol⁻¹); for water it is 0.512 K kg mol⁻¹.

Question 4. The property which depends on number of particles of solute is called .............
Answer:
Colligative property

Teacher's Note:
1. Colligative properties depend on how many solute particles are present, not on what they are.

Question 5. Azeotrope mixture cannot be separated by .............
Answer:
Fractional distillation

Teacher's Note:
1. An azeotrope boils with the same composition in liquid and vapour, so distillation cannot change its composition.

Question 6. What is Van’t Hoff factor?
Answer:
It is the ratio of normal molecular mass to observed molecular mass. It is denoted as i.
i = \( \frac{\text{normal molecular mass}}{\text{observed molecular mass}} \)
= \( \frac{\text{no. of particles after association or dissociation}}{\text{no. of particles before association or dissociation}} \)

Teacher's Note:
1. i can also be found as observed colligative property ÷ calculated colligative property.

Question 7. What is the Van’t Hoff factor in K₄[Fe(CN)₆] and BaCl₂?
Answer:
5 and 3

Teacher's Note:
1. K₄[Fe(CN)₆] gives 4K⁺ + [Fe(CN)₆]⁴⁻; the complex ion counts as one particle.

Question 8. Why the molecular mass becomes abnormal?
Answer:
Due to association or dissociation of solute in given solvent.

Teacher's Note:
1. Dissociation gives a lower observed molar mass; association gives a higher one.

Question 9. What role does the molecular interaction play in the solution of alcohol and water?
Answer:
Positive deviation from ideal behaviour.

Teacher's Note:
1. Water molecules break some of the hydrogen bonds between alcohol molecules, and the new attractions are weaker, so vapour pressure is higher than Raoult’s law predicts.

Question 10. What is van’t Hoff factor? How is it related with:
(a) degree of dissociation    (b) degree of association
Answer:
Van’t Hoff factor (i) is the ratio of normal molecular mass to observed molecular mass of the solute.
(a) α = \( \frac{i-1}{n-1} \)
(b) α = \( \frac{i-1}{\frac{1}{n}-1} \)
where n is the number of particles formed by dissociation of one molecule (or the number of molecules that associate).

Teacher's Note:
1. For association i < 1 and 1/n − 1 is negative, so α still comes out positive.

Question 11. Why NaCl is used to clear snow from roads?
Answer:
It lowers freezing point of water.

Teacher's Note:
1. Brine formed with NaCl freezes well below 0°C, so the ice melts at road temperature.

Question 12. Why the boiling point of solution is higher than pure liquid?
Answer:
Due to lowering in vapour pressure.

Teacher's Note:
1. A solution must be heated more to raise its lowered vapour pressure up to the external pressure.

Question 13. Henry law constant for two gases are 21.5 and 49.5 atm, which gas is more soluble?
Answer:
K_H is inversely proportional to solubility.
Therefore, the gas with K_H = 21.5 atm is more soluble.

Teacher's Note:
1. Lower K_H means higher solubility at the same partial pressure.

Question 14. Define azeotrope. Give an example of maximum boiling azeotrope.
Answer:
Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boiling at a constant temperature.
Example of maximum boiling azeotrope: a mixture of about 68% nitric acid and 32% water by mass (boils at 393.5 K).

Teacher's Note:
1. Maximum boiling azeotropes come from solutions showing large negative deviation from Raoult’s law.

Short Answer Questions (2 Marks)

Question 15. What is meant by positive deviations from Raoult’s law? Give an example. What is the sign of Δ_mixH for positive deviation? [CBSE Delhi 2015]
Answer:
In case of positive deviation from Raoult’s law, the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules.
Example: Mixture of ethanol and acetone.
Sign for Δ_mixH is positive.

Teacher's Note:
1. Weaker A–B attractions mean heat is absorbed on mixing (Δ_mixH > 0) and the volume usually increases (Δ_mixV > 0).

Question 16. Why a mixture of Carbon disulphide and acetone shows positive deviation from Raoult’s law? What type of azeotrope is formed by this mixture? [CBSE Comptt. Delhi/OD 2018]
Answer:
Intermolecular forces of attraction between carbon disulphide and acetone are weaker than the pure components.
Minimum boiling azeotrope at a specific composition.
Detailed Answer:
When carbon disulphide is added to the acetone, the dipolar interaction between molecule of solute solvent (A – B) than the respective interaction between solute-solute (A – B) and solvent – solvent (B – B) weakens, which leads to increase in vapour pressure. Hence, shows positive deviation in vapour pressure from Raoult’s Law.
Minimum boiling azeotrope at a specific composition is formed by this mixture.

Teacher's Note:
1. Higher vapour pressure means the mixture boils at a lower temperature, which is why positive deviation gives a minimum boiling azeotrope.

Question 17. Explain why on addition of 1 mol of NaCl to 1 litre of water, the boiling point of water increases, while addition of 1 mol of methyl alcohol to one litre of water decreases its boiling point. [NCERT Exemplar]
Answer:
Sodium Chloride (NaCl) is a non-volatile solute, therefore, addition of NaCl to water lowers the vapour pressure of water. As a result, boiling point of water increases.
Methyl alcohol on the other hand is more volatile than water, hence, its addition increases the total vapour pressure over the solution and a decrease in boiling point of water results.

Teacher's Note:
1. Boiling point is reached when vapour pressure equals atmospheric pressure: lower vapour pressure ⇒ higher boiling point, and vice versa.

Question 18. Define the following terms:
(i) Colligative properties
(ii) Molality (m) [CBSE Delhi Set-1 2017]
Answer:
(i) Properties that are independent of nature of solute and depend on number of moles of solute only.
(ii) Number of moles of solute dissolved per kg of the solvent.

Teacher's Note:
1. Molality is per kg of solvent, not of solution; this is the most common slip in this definition.

Question 19. Define the following terms:
(i) Ideal solution
(ii) Molarity (M) [CBSE Delhi Set-2 2017]
Answer:
(i) The solution that obeys Raoult’s Law over the entire range of concentration.
(ii) Number of moles of solute dissolved per litre of solution or M = \( \frac{W_b \times 1000}{M_b \times V(\text{mL})} \)

Teacher's Note:
1. Molarity is per litre of solution, so it changes with temperature (volume expands), unlike molality.

Question 20. A 1.00 molar aqueous solution of trichloroacetic acid (CCl₃COOH) is heated to its boiling point. The solution has the boiling point of 100.18 °C. Determine the van’t Hoff factor for trichloroacetic acid. (K_b for water = 0.512 K kg mol⁻¹).
OR
Define the following terms :
(i) Mole fraction   (ii) Isotonic solutions   (iii) Van’t Hoff factor   (iv) Ideal solution [CBSE Delhi 2012]
Answer:
ΔT_b = iK_bm
(373.18 − 373) K = i × 0.512 K kg mol⁻¹ × 1 m
0.18 K = i × 0.512 K kg mol⁻¹ × 1 m,
i = 0.35
OR
(i) Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
χ_A = \( \frac{n_A}{n_A + n_B} \) where n_A = number of moles of component A, n_B = number of moles of component B.
(ii) Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
(iii) Van’t Hoff factor is the ratio of normal molecular mass and observed molecular mass.
(iv) Ideal solution: The solution which follows Raoult’s law over entire range of concentrations at specific temperature is called ideal solution.

Teacher's Note:
1. The question gives 1.00 molar; for a dilute aqueous solution molarity is taken as molality. i < 1 here (0.35), which signals association of the acid in this data.

Question 21. What are colligative properties? Write the colligative property which is used to find the molecular mass of macromolecules. [CBSE Foreign Set-1 2017]
Answer:
Properties that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.
Osmotic pressure.

Teacher's Note:
1. Remember the four colligative properties: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

Question 22. What is meant by elevation in boiling point? Why is it a colligative property? [CBSE Foreign Set-3 2017]
Answer:
The increase in boiling point of the solvent in a solution when a non-volatile solute is added.
Because it depends upon molality/the number of solute particles rather than their nature/ ΔT_B ∝ m
Detailed Answer:
Elevation in boiling point can be defined as an increase in boiling point of the solvent in a solution on addition of a non-volatile solute.
It is considered as a colligative property because it depends upon the number of solute particles dissolved in a definite amount of the solvent and not on the nature of the solute.

Teacher's Note:
1. Mention “non-volatile solute”: a volatile solute can actually lower the boiling point.

Question 23. (i) Write the colligative property which is used to find the molecular mass of macromolecules.
(ii) In non-ideal solution, what type of deviation shows the formation of minimum boiling azeotropes? [CBSE OD Set-2 2016]
Answer:
(i) Osmotic pressure is the colligative property which is used to find the molecular mass of macromolecules.
(ii) In non-ideal solution, positive deviation shows the formation of minimum boiling azeotropes.

Teacher's Note:
1. Osmotic pressure is measured at room temperature and gives a sizeable value even for very dilute solutions, which suits heat-sensitive macromolecules.

Question 24. Define osmotic pressure. How is the osmotic pressure related to the concentration of a solute in a solution? [CBSE Comptt. OD 2016; Delhi 2015]
Answer:
The external pressure applied on the solution side to stop the flow of solvent across the semipermeable membrane i.e., osmosis is known as osmotic pressure.
The osmotic pressure is directly proportional to the concentration of the solution i.e., π = CRT.

Teacher's Note:
1. In π = CRT, C is molarity (mol L⁻¹), so use R = 0.0821 L atm K⁻¹ mol⁻¹ to get π in atm.

Question 25. Which of the following solutions has higher freezing point?
0.05 M Al₂(SO₄)₃, 0.1 M K₃[Fe(CN)₆] Justify. [CBSE SQP 2017]
Answer:
0.05 M Al₂(SO₄)₃ has higher freezing point.
0.05 M Al₂(SO₄)₃ : i = 5, ΔT_f ∝ No. of particles; ΔT_f = i × concentration = 5 × 0.05 = 0.25 moles of ions
0.1 M K₃[Fe(CN)₆]: i = 4, = 4 × 0.1 = 0.4 moles of ions

Teacher's Note:
1. Fewer dissolved particles mean a smaller depression, i.e., a higher freezing point; compare i × concentration, not concentration alone.

Question 26. Calculate the mass of compound (molar mass = 256 g mol⁻¹) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (K_f = 5.12 K kg mol⁻¹). [CBSE Delhi 2014]
Answer:
Given : ΔT_f = \( \frac{K_f \times W_2 \times 1000}{W_1 \times M_2} \)
0.48 K = 5.12 K kg mol⁻¹ × \( \frac{W_2 \times 1000 \text{ g kg}^{-1}}{75 \text{ g} \times 256 \text{ g mol}^{-1}} \)
W₂ = \( \frac{0.48 \times 75 \times 256}{5.12 \times 1000} \) g
W₂ = 1.8 g

Teacher's Note:
1. Rearrange the formula for W₂ before putting in numbers; it avoids most arithmetic slips.

Question 27. Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases. [CBSE Delhi Set-1, 2, 3 2017]
Answer:
Vapour pressure of the solvent decreases in the presence of non-volatile solute (glucose) hence boiling point increases.

Teacher's Note:
1. Link the two ideas: lower vapour pressure means a higher temperature is needed for it to reach atmospheric pressure.

Short Answer Questions (3 Marks)

Question 28. 1.00 g of a non electrolyte solute when 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:
Given, W₂ = 1.00 g, W₁ = 50 g, K_f = 5.12 K kg mol⁻¹, ΔT_f = 0.40 K
0.40 K = \( \frac{5.12 \text{ K kg mol}^{-1} \times 1 \text{ g} \times 1000 \text{ g kg}^{-1}}{M_2 \times 50 \text{ g}} \)
⇒ M₂ = \( \frac{K_f \times W_2 \times 1000}{W_1 \times \Delta T_f} \)
= \( \frac{5.12 \times 1 \times 1000}{50 \times 0.40} \) g mol⁻¹
∴ M₂ = 256 g mol⁻¹

Teacher's Note:
1. The factor 1000 converts the solvent mass from grams to kilograms; forgetting it gives 0.256 g mol⁻¹, an obviously impossible molar mass.

Question 29. Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (K_f for water = 1.86 K kg mol⁻¹) [CBSE Delhi 2012]
Answer:
Since one mole of KCl gives 2 mole particles, the value of i = 2
ΔT_f = 2 K
K_f = 1.86 K kg mol⁻¹
Applying equation, ΔT_f = iK_fm
m = \( \frac{\Delta T_f}{iK_f} \) = \( \frac{2}{2 \times 1.86} \)
≈ 0.5376 = 0.54 mol kg⁻¹
Therefore, 0.54 mole of KCl should be added to one kg of water.

Teacher's Note:
1. If the mass is asked, multiply by 74.5 g mol⁻¹: 0.5376 × 74.5 ≈ 40 g of KCl.

Question 30. Calculate the boiling point of a 1M aqueous solution (density 1.04 g mL⁻¹) of Potassium chloride (K_b for water = 0.52 K kg mol⁻¹, Atomic masses : K = 39 u, Cl = 35.5 u). Assume, Potassium chloride is completely dissociated in solution. [CBSE SQP 2016]
Answer:
Molar mass of KCl = 39 + 35.5 = 74.5 g mol⁻¹
As KCl dissociates completely, number of ions produced are 2.
Therefore, van’t Hoff factor, i = 2
Mass of KCl solution = 1000 × 1.04 = 1040 g
Mass of solvent = 1040 − 74.5 = 965.5 g = 0.9655 kg
Molality of the solution :
\( \frac{\text{No. of moles of solute}}{\text{Mass of solvent in kg}} \) = \( \frac{1 \text{ mol}}{0.9655 \text{ kg}} \) = 1.0357 m
ΔT_b = i × K_b × m
= 2 × 0.52 × 1.0357 = 1.077° C
Therefore, boiling point of solution = 100 + 1.077 = 101.077°C

Teacher's Note:
1. Convert molarity to molality using density: subtract the solute mass from the mass of 1 L of solution to get the mass of solvent.

Question 31. A 5 percent solution (by mass) of cane-sugar (M.W. 342) is isotonic with 0.877% solution of substance X. Find the molecular weight of X. [CBSE Comptt. OD 2015]
Answer:
Since, \( \frac{W_{\text{cane sugar}}}{M_{\text{cane sugar}}} \) = \( \frac{W_X}{M_X} \)
\( \frac{5 \text{ g}}{342 \text{ g mol}^{-1}} \) = \( \frac{0.877}{M_X} \)
⇒ M_X = \( \frac{0.877 \times 342}{5} \) g mol⁻¹
⇒ M_X = 59.9 or 60 g mol⁻¹

Teacher's Note:
1. Isotonic solutions have equal molar concentrations; for the same mass of solution, equal moles of solute is the condition used here.

Question 32. Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10⁻² g of K₂SO₄ in 2L of water at 25°C, assuming that it is completely dissociated.
(R = 0.0821 L atm K⁻¹ mol⁻¹, molar mass of K₂SO₄ = 174 g mol⁻¹) [CBSE Delhi 2013]
Answer:
When K₂SO₄ is dissolved in water, ions are produced.
Total number of ions produced = 3
i = 3
π = iCRT = \( \frac{i \times n \times R \times T}{V} \)
⇒ π = 3 × \( \frac{2.5 \times 10^{-2} \text{ g}}{174 \text{ g mol}^{-1}} \) × \( \frac{0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}}{2 \text{ L}} \) × 298 K
= 5.27 × 10⁻³ atm

Teacher's Note:
1. Convert 25°C to 298 K; using 25 in place of T is a frequent mistake in osmotic pressure numericals.

Question 33. A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute.
(Vapour pressure of pure water at 308 K = 32 mm Hg) [CBSE OD 2015]
Answer:
\( \frac{p^0 - p}{p^0} \) = \( \frac{W_s \times M_{\text{solvent}}}{M_s \times W_{\text{solvent}}} \), s = solute
⇒ (32 − 31.84)/32 = 10 × 18/M_s × 200
M_s = 180 g/mol

Teacher's Note:
1. Relative lowering = 0.16/32 = 0.005; then M_s = (10 × 18)/(0.005 × 200) = 180 g mol⁻¹.

Question 34. Give reasons for the following:
(a) When 2g of benzoic acid is dissolved in 25 g of benzene, the experimentally determined molar mass is always greater than the true value.
(b) Mixture of ethanol and acetone shows positive deviation from Raoult’s Law.
(c) The preservation of fruits by adding concentrated sugar solution protects against bacterial action. [CBSE SQP 2018-2019]
Answer:
(a) Molecules of benzoic acid dimerise in benzene, the number of particles are reduced.
(b) The intermolecular interactions between ethanol and acetone are weaker/ the escaping tendency of ethanol and acetone molecules increases on mixing / the vapour pressure increases.
(c) Due to osmosis, a bacterium on fruit loses water, shrivels and dies.
Detailed Answer:
(a) As benzoic acid on dissolving in benzene has molecular mass double of its normal molecular mass. This is caused due to number of particles. Thus, the experimentally determined molar mass is always greater than the true value.
(b) Hydrogen bonding is present in the molecules of ethanol. When acetone is added to ethanol, molecules of acetone enter between the molecules of ethanol breaking the bonds. This weakens the intermolecular attractive forces resulting in increase in vapour pressure of mixture than that in pure state.
(c) Sugar has an osmotic effect i.e., when fruits are placed in a concentrated sugar solution, water is drawn out of the cells of micro-organisms so that they no longer survive.

Teacher's Note:
1. Molar mass from a colligative property is inversely proportional to the number of particles, so association (fewer particles) gives a higher observed molar mass.

Question 35. Give reasons for the following :
(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
(b) Aquatic animals are more comfortable in cold water than in warm water.
(c) Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution. [CBSE Delhi/OD 2018]
Answer:
(a) As compared to other colligative properties, its magnitude is large even for very dilute solutions / macromolecules are generally not stable at higher temperatures and polymers have poor solubility / pressure measurement is around the room temperature and the molarity of the solution is used instead of molality.
(b) Because oxygen is more soluble in cold water or at low temperature.
(c) Due to dissociation of KCl / KCl (aq) → K⁺ + Cl⁻, i is nearly equal to 2.
Detailed Answer:
(a) In osmotic pressure method, pressure is measured at room temperature and instead of molality, molarity of the solution is used. It is preferred for the macromolecules like proteins as they are unstable at high temperature and polymers have poor solubility. Also, due to their high molecular mass, the only colligative property which has a measurable magnitude is osmotic pressure. Therefore, osmotic pressure method is preferred.
(b) In cold water, the solubility of oxygen is more. The amount of dissolved oxygen in water decreases with increase in the temperature of water. Therefore, aquatic animals are more comfortable in cold water than in warm water.
(c) 1M KCl solution dissociates into 1M K⁺ and 1M Cl⁻ whereas 1M sugar solution does not dissociates to give ions. So, the van’t Hoff factor for KCl solution is 2 whereas that for sugar solution is 1. As we know, greater is the van’t Hoff factor, greater would be the colligative property. Hence, elevation of boiling point of 1M KCl solution is nearly double than that of 1M sugar solution.

Teacher's Note:
1. Part (b) is about Henry’s law: K_H of a gas increases with temperature, so gas solubility falls as water warms.

Question 36. Determine the amount of CaCl₂ dissolved in 2.5 L at 27°C such that its osmotic pressure is 0.75 atm at 27°C. (i for CaCl₂ = 2.47)
Answer:
For CaCl₂, i = 2.47
π = iCRT = i \( \frac{n_B}{V} \) × RT
0.75 = \( \frac{2.47 \times n_B \times 0.082 \times 300}{2.5} \)
n_B = \( \frac{0.75 \times 2.5}{2.47 \times 0.082 \times 300} \) = 0.0308 mol
Amount = 0.0308 mol × 111 g mol⁻¹ = 3.42 g

Teacher's Note:
1. Always convert temperature to kelvin (27°C = 300 K) and use R = 0.082 L atm K⁻¹ mol⁻¹ when π is in atm.

Long Answer Questions (5 Marks)

Question 37. (i) Explain the following :
(a) Henry’s law about dissolution of a gas in a liquid,
(b) Boiling point elevation constant for a solvent.
(ii) 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. What mass of glycerol was dissolved to make this solution? [K_b for water = 0.512 K kg mol⁻¹] [CBSE Delhi & OD 2012]
Answer:
(i) (a) The partial pressure of the gas above the liquid (vapour phase) is directly proportional to the mole fraction of the gas dissolved in the liquid.
(b) Boiling Point Elevation Constant : It is equal to elevation in boiling point of 1 molal solution, i.e., mole of solute is dissolved in 1 kg of solvent. (or mathematical expression)
(ii) W_B = ? W_A = 500 g ΔT_b = 100.42°C − 100°C = 0.42°C or 0.42 K
K_b = 0.512 K kg/mol M_B = 92 g/mol
ΔT_b = K_b \( \frac{W_B \times 1000}{M_B \times W_A (\text{in grams})} \)
W_B = \( \frac{\Delta T_b \times M_B \times W_A (\text{in grams})}{1000 \times K_B} \)
= \( \frac{0.42 \text{ K} \times 92 \text{ g mol}^{-1} \times 500}{1000 \text{ g kg}^{-1} \times 0.512 \text{ K kg mol}^{-1}} \)
= 37.73 g

Teacher's Note:
1. Work out the molar mass of glycerol first: C₃H₈O₃ = 3(12) + 8(1) + 3(16) = 92 g mol⁻¹.

Question 38. (i) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K. Given:
(Molar mass of sucrose = 342 g mol⁻¹)
(Molar mass of glucose = 180 g mol⁻¹)
(ii) Define the following terms:
(a) Molality (m)
(b) Abnormal molar mass [CBSE OD Set-1, 2, 3 2017]
Answer:
(i) ΔT_f = K_fm
Here, m = w₂ × 1000/M₂ × w₁
273.15 − 269.15 = K_f × 10 × 1000/342 × 90
K_f = 12.3 K kg/mol
ΔT_f = K_fm
= 12.3 × 10 × 1000/180 × 90
= 7.6 K
T_f = 273.15 − 7.6 = 265.55 K
(or any other correct method)
(ii) (a) Number of moles of solute dissolved in per kilo gram of the solvent.
(b) Abnormal molar mass: If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass.

Teacher's Note:
1. A shortcut check: for the same mass % and solvent, ΔT_f is inversely proportional to molar mass, so ΔT_f(glucose) = 4 × 342/180 ≈ 7.6 K.

Question 39. (i) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(ii) A solution of glucose (molar mass = 180 g mol⁻¹) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL⁻¹) [CBSE OD 2014]
Answer:
(i) It shows positive deviation.
It is due to weaker interaction between acetone and ethanol than ethanol-ethanol interactions.
(ii) Given : W_B = 10 g, W_S = 100 g, W_A = 90 g, M_B = 180 g/mol and d = 1.2 g/mL
M = \( \frac{\text{Wt\%} \times \text{density} \times 10}{\text{Mol.wt.}} \)
M = \( \frac{10 \times 1.2 \times 10}{180} \)
= 0.67 M or 0.67 mol/L
m = \( \frac{W_B \times 1000}{M_B \times W_A (\text{in g})} \)
m = \( \frac{10 \times 1000}{180 \times 90} \)
= 0.62 m or 0.62 mol/kg
(or any other suitable method)

Teacher's Note:
1. For molarity you need the volume of solution: 100 g ÷ 1.2 g mL⁻¹ = 83.3 mL, which is what the factor “density × 10” takes care of.

Question 40. (i) The vapour pressure of benzene and toluene at 293 K are 75 mm Hg and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapour pressures are in equilibrium with the liquid mixture at this temperature.
(ii) What is meant by +ve and −ve deviations from Raoult’s law and how is the sign of ΔH solution related to +ve and −ve deviations from Raoult’s law? [CBSE Comptt. OD 2013]
Answer:
(i) Calculation of the total vapour pressure
Moles of benzene (n_A) = \( \frac{\text{Mass of benzene}}{\text{Molar mass of benzene (C}_6\text{H}_6)} \) = \( \frac{(23.4 \text{ g})}{(78 \text{ g mol}^{-1})} \) = 0.3 mol
Moles of toluene (n_B) = \( \frac{\text{Mass of toluene}}{\text{Molar mass of toluene (C}_7\text{H}_8)} \) = \( \frac{(64.4 \text{ g})}{(92 \text{ g mol}^{-1})} \) = 0.7 mole
Mole fraction of benzene (χ_A) = \( \frac{n_A}{n_A + n_B} \) = \( \frac{(0.3 \text{ mole})}{(0.3 \text{ mole} + 0.7 \text{ mole})} \) = 0.3
Mole fraction of toluene (χ_B) = \( \frac{n_B}{n_A + n_B} \) = \( \frac{(0.7 \text{ mole})}{(0.3 \text{ mole} + 0.7 \text{ mole})} \) = 0.7
Vapour pressure of benzene in solution (p_A) = p°_A × χ_A = (75 mm) × 0.3 = 22.5 mm
Vapour pressure of toluene in solution (p_B) = p°_B × χ_B = (22 mm) × 0.7 = 15.4 mm
Total vapour pressure of solution = p_A + p_B = 22.5 + 15.4 = 37.9 mm
Calculation of mole fraction of benzene in the vapour phase
Mole fraction of benzene in vapour phase = \( \frac{\text{Partial vapour pressure of benzene}}{\text{Total vapour pressure}} \) = \( \frac{(22.5 \text{ mm})}{(37.9 \text{ mm})} \) = 0.59
(ii) Solutions showing positive deviations, the partial vapour pressure of each component of solution is greater than the vapour pressure as expected according to Raoult’s law.
Solutions showing negative deviations, the partial vapour pressure of each component of solution is less than vapour pressure as expected according to Raoult’s law.
For positive deviation Δ_mixH = +ve
For negative deviation Δ_mixH = −ve

Teacher's Note:
1. The vapour is richer in the more volatile component: benzene is 0.3 in the liquid but 0.59 in the vapour. Use Dalton’s law (y = p_A/p_total) for the vapour phase.

Question 41. (i) A 5% solution (by mass) of cane – sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K. [Molecular masses : Glucose C₆H₁₂O₆ : 180 amu; Cane – sugar C₁₂H₂₂O₁₁ : 342 amu]
(ii) State Henry’s law and mention two of its important applications. [CBSE Comptt. OD 2013]
Answer:
(i) 5% by mass cane sugar = 5 g cane sugar (Mol. wt = 342 amu in 100 g solution)
n_B = \( \frac{5}{342} \) and w_A = 95 g
ΔT_f = 273.15 − 271 = 2.15 K
K_f = \( \frac{\Delta T_f \times w_A}{n_B \times 1000} \) = \( \frac{2.15 \times 95}{\frac{5}{342} \times 1000} \)
= 13.97 K kg mol⁻¹
5% by mass glucose = 5 g glucose (Mol. wt = 180 amu) in 100 g solution
ΔT_f = K_f × \( \frac{n_B 1000}{w_A} \) = 13.97 × \( \frac{5}{180} \) × \( \frac{1000}{95} \)
= 4.085 K
T_f = 273.15 − 4.085
Freezing point of glucose solution, T_f = 269.065 K
(ii) According to Henry’s law, the partial pressure of a gas in vapour phase (p) is directly proportional to the mole fraction (x) of the gas in the solution.
Henry’s law is applied in the production of carbonated beverages and for the deep-sea-divers.

Teacher's Note:
1. The K_f value found (13.97) is not the real K_f of water (1.86); it is only a working constant from the given data, so carry it through without questioning it.