CBSE Class 12 Chemistry Solutions Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. What is the sum of mole fractions of all the components in a three component system ? (CBSE 2010)
Answer:
x₁ + x₂ + x₃ = 1.

Teacher's Note:
1. The sum of mole fractions of all components is always 1, whatever the number of components.

Question 2. Why does the molality of a solution remain unchanged with temperature ? (Assam Board 2013)
Answer:
Molality of a solution involves masses of the solute and solvent which do not change with temperature.

Teacher's Note:
1. Molarity changes with temperature because it involves volume, which expands or contracts on heating or cooling.

Question 3. Calculate the molality of H₂SO₄ if the density of 10% (w/w) aqueous solution of H₂SO₄ is 1.84 g cm⁻³ (Molar mass of H₂SO₄ = 98 g mol⁻¹). (Raj. Board 2012)
Answer:
10% (w/w) H₂SO₄ solution means 10 g H₂SO₄ are present in 100 g of solution, i.e., water present = 90 g
Molality = \( \frac{10/98}{90} \times 1000 \) = 1.13 m

Teacher's Note:
1. Density is not needed for molality from a w/w percentage; it is needed only if molarity is asked.

Question 4. What is the effect of temperature on the solubility of sodium sulphate decahydrate ?
Answer:
The solubility first increases upto 32.4°C (called transition temperature) and then decreases.

Teacher's Note:
1. Above 32.4°C the decahydrate changes to anhydrous Na₂SO₄, whose dissolution is exothermic, so solubility falls.

Question 5. Define transition temperature in solubility of a solid in a liquid.
Answer:
The temperature at which the trend of solubility changes (e.g., first increases and then decreases) is called the transition temperature.

Teacher's Note:
1. On a solubility curve the transition temperature shows up as a sharp break in the curve.

Question 6. State the formula relating pressure of a gas with its mole fraction in a liquid solution in contact with it.
Answer:
According to Henry’s law,
Partial pressure of a gas above the solution = K_H × mole fraction of the gas in the solution
or p_A = K_H × x_A, where K_H = Henry’s constant.

Teacher's Note:
1. Higher K_H means lower solubility of the gas at the same partial pressure.

Question 7. Define azeotropic mixture. (HP Board 2013)
Answer:
Azeotropic mixtures are binary mixtures (liquid mixtures) having the same composition in liquid and vapour phase and boiling at a constant temperature. They cannot be separated into their components by fractional distillation.

Teacher's Note:
1. Azeotropes are also called constant boiling mixtures; for example 95% ethanol + 5% water by volume.

Question 8. What is the boiling point of an azeotrope of non-ideal solution showing positive deviations as compared to the boiling points of its components ?
Answer:
The boiling point of such an azeotrope is lower than that of its components.

Teacher's Note:
1. Positive deviation means higher vapour pressure, hence a minimum boiling azeotrope.

Question 9. What type of non-ideal solution is formed when ethyl alcohol is mixed with water ?
Answer:
It forms a non-ideal solution showing positive deviations.

Teacher's Note:
1. Ethanol-water forms a minimum boiling azeotrope (about 95% ethanol by volume).

Question 10. When 50 mL of liquid A is mixed with 50 mL of liquid B total volume of the solution after mixing is found to be less than 100 mL. What type of non-ideal solution is formed ?
Answer:
As ΔV_mixing is negative, it forms a non-ideal solution with negative deviations.

Teacher's Note:
1. Volume contraction on mixing means stronger A–B attractions pull the molecules closer, which is the sign of negative deviation.

Question 11. What happen when acetone is added to pure ethanol ? (CBSE 2020)
Answer:
Acetone molecules get inbetween the ethanol molecules breaking hydrogen bonds thereby weakening ethanol-ethanol attractions. Consequently, vapour pressure of ethanol/solution will be greater than that expected from Raoult’s law, i.e., for ideal solution. In other words, the solution will show positive deviation from Raoult’s law.

Teacher's Note:
1. Breaking of hydrogen bonds on mixing is the usual cause of positive deviation.

Question 12. Define colligative properties. (CBSE 2017)
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 the solution and not upon the nature of the solute.
They are: relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure.

Teacher's Note:
1. Remember the key phrase “number of particles, not nature”; that is why electrolytes give larger values.

Question 13. What is the difference between lowering of vapour pressure and relative lowering of vapour pressure ?
Answer:
Lowering of V.P. = p° − p_s.
Relative lowering of V.P. = (p° − p_s)/p°.

Teacher's Note:
1. Only the relative lowering is a colligative property; it equals the mole fraction of the solute.

Question 14. Write the mathematical form of Raoult’s law of relative lowering of vapour pressure. (Karnataka Board 2012)
Answer:
\( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} \)

Teacher's Note:
1. For very dilute solutions n₂ is small and the right side can be taken as n₂/n₁.

Short Answer Questions (2 Marks)

Question 15. 2 g each of the solutes A and B (Mol mass of A > B) are dissolved separately in 20 g each of the same solvent C. Which will show greater lowering of vapour pressure and why?
Answer:
For dilute solutions, \( \frac{p^\circ - p_s}{p^\circ} \) = \( \frac{w_2 M_1}{w_1 M_2} \) or \( \frac{\Delta p}{p^\circ} \) = \( \frac{w_2 M_1}{w_1 M_2} \)
For same solvent, p° = constant, M₁ = constant. Also, given that w₂ = w_A = w_B = 2 g, w₁ = w_C = 20 g.
Hence, \( \frac{\Delta p_A}{\Delta p_B} \) = \( \frac{M_B}{M_A} \). As M_A > M_B, therefore, Δp_B > Δp_A, i.e., B will show greater lowering of vapour pressure.

Teacher's Note:
1. Equal masses of a lighter solute contain more moles, so they produce a bigger colligative effect.

Question 16. Why is freezing point depression of 0.1 M sodium chloride solution nearly twice that of 0.1 M glucose solution? [Manipur Board 2011]
Answer:
NaCl, being an electrolytes, dissociates almost completely to give Na⁺ and Cl⁻ ions whereas glucose, being non-electrolyte, does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double than in 0.1 M glucose solution. Freezing point depression, being a colligative property, is therefore, nearly twice for NaCl solution than for glucose solution of same molarity.

Teacher's Note:
1. 'Nearly' twice because interionic attractions keep i slightly below 2 in real solutions.

Question 17. Explain why equimolar aqueous solutions of sodium chloride and sodium sulphate are not isotonic? [Manipur Board 2011]
Answer:
NaCl dissociates to give 2 ions (Na⁺ and Cl⁻). Na₂SO₄ dissociates to give 3 ions (2 Na⁺ and SO₄²⁻). Thus, equimolar solutions of NaCl and Na₂SO₄ have different concentrations of ions in the solution. As osmotic pressure depends upon concentration of particles in the solution, they have different osmotic pressures.

Teacher's Note:
1. Isotonic means equal osmotic pressure, which needs equal concentration of particles, i.e., equal iC.

Question 18. The experimentally determined molar mass for what type of substances is always lower than the true value when water is used as solvent. Explain. Give one example of a such a substance and one example of a substance which does not show a large variation from the true value. [CBSE Sample Paper 2019]
Answer:
In dilute solutions when a substance undergoes dissociation, number of particles of the solute increases. As colligative property depends upon the number of particles, the value of colligative property will be higher and as molar mass is inversely proportional to value of colligative property, molar mass will be lower, e.g., KCl.
For substances like glucose which do not dissociate or associate in the solution, does not show a large variation from true value.

Teacher's Note:
1. Dissociation lowers the observed molar mass (i > 1); association raises it (i < 1).

Question 19. To what type of solution an alloy belongs ? Give one example of a solution of liquid in solid.
Answer:
An alloy is a solid solution (solid in solid), e.g., brass (copper and zinc) or bronze (copper and tin).
Solution of liquid in solid: amalgam of mercury with sodium; hydrated salts like CuSO₄.5H₂O.

Teacher's Note:
1. In a solid solution the solvent is a solid; the physical state of the solvent decides the type of solution.

Question 20. Explain the terms ‘Mass fraction’ and ‘Mole fraction’ ?
Answer:
Mass fraction : Mass fraction of a component in a solution is the mass of that component divided by the total mass of the solution. For a solution of A and B,
Mass fraction of A, \( w_A = \frac{W_A}{W_A + W_B} \). Mass fraction × 100 = mass percentage.
Mole fraction : Mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all the components in the solution.
\( x_A = \frac{n_A}{n_A + n_B} \), \( x_B = \frac{n_B}{n_A + n_B} \) and x_A + x_B = 1.

Teacher's Note:
1. Both are ratios and have no units; both are independent of temperature.

Question 21. Which out of molality, molarity and mole fraction of a solution will remain unchanged on raising the temperature and why ? (Bihar Board 2011)
Answer:
Molality and mole fraction remain unchanged on raising the temperature.
This is because molality and mole fraction involve only masses (or moles) of solute and solvent, which do not change with temperature. Molarity involves volume of the solution, which increases with rise in temperature; hence molarity decreases.

Teacher's Note:
1. Any concentration term with volume in it (molarity, normality, % w/V) depends on temperature.

Question 22. State Henry’s law correlating the pressure of a gas and its solubility in a solvent and mention two applications for the law. What helps in existence of aquatic life ? (CBSE 2008)
Answer:
Henry’s law : The solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid. Alternatively, the partial pressure of the gas in vapour phase is proportional to its mole fraction in the solution: p = K_H x.
Applications : (i) To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure. (ii) Tanks used by scuba divers are filled with air diluted with helium to avoid bends (the painful effect of N₂ bubbles in blood).
Aquatic life exists because O₂ from air dissolves in water to a small extent according to Henry’s law; solubility of O₂ is higher in cold water, so aquatic species are more comfortable in cold water.

Teacher's Note:
1. Gas solubility falls as temperature rises (K_H increases), which is why fish find less oxygen in warm water.

Question 23. State Henry’s law and mention its two important applications. (Assam Board 2012)
Answer:
Henry’s law : At a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. In terms of mole fraction, 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_H x.
Applications : (i) Soft drink and soda water bottles are sealed under high pressure to increase the solubility of CO₂. (ii) Scuba divers use air diluted with helium (11.7% He, 56.2% N₂, 32.1% O₂) to avoid bends. (iii) At high altitudes, low partial pressure of O₂ leads to low O₂ in blood (anoxia).

Teacher's Note:
1. K_H has pressure units (bar or kbar) and depends on the nature of gas, solvent and temperature.

Question 24. State Henry’s law and explain why are the tanks used by scuba divers filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen) ? (CBSE Sample Paper 2022-23)
Answer:
Henry’s law : The partial pressure of a gas in vapour phase is proportional to the mole fraction of the gas in the solution: p = K_H x.
Scuba divers breathe air at high pressure under water. At high pressure, the solubility of atmospheric gases (especially N₂) in blood increases. When the divers come towards the surface, pressure decreases and the dissolved N₂ is released, forming bubbles in blood. These block capillaries and cause a medical condition called bends, which is painful and dangerous. To avoid bends and the toxic effects of high concentration of N₂, the tanks are filled with air diluted with helium, which is much less soluble in blood.

Teacher's Note:
1. The key point is that helium has very low solubility in blood, so little gas comes out of solution as the diver rises.

Question 25. How can you justify the observation that the vapour pressure of solution of a non-volatile solute in a given solvent is less than that of the pure solvent ? Also state the law concerning this observation.
Answer:
In a pure solvent, the whole surface is occupied by solvent molecules. When a non-volatile solute is dissolved, some of the surface sites are occupied by solute particles. Hence, the fraction of surface covered by solvent molecules decreases, fewer solvent molecules escape into the vapour phase and the vapour pressure is lowered.
Raoult’s law : For a solution of a non-volatile solute, the vapour pressure of the solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction: p_s = p° x₁, or the relative lowering of vapour pressure is equal to the mole fraction of the solute, \( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} = x_2 \).

Teacher's Note:
1. Since x₁ < 1 in a solution, p_s = p°x₁ is always less than p°.

Question 26. Vapour pressure of a solution is different from that of pure solvent
(i) Name the law which helps us to determine partial vapour pressure of a volatile component in solution.
(ii) State the above law. (Kerala Board 2012)
Answer:
(i) Raoult’s law.
(ii) For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction: p₁ = p₁°x₁ and p₂ = p₂°x₂. Total vapour pressure, p = p₁ + p₂ (Dalton’s law).

Teacher's Note:
1. The proportionality constant in Raoult’s law is the vapour pressure of the pure component; in Henry’s law it is K_H.

Question 27. Define vapour pressure of a liquid. What happens to the vapour pressure when (a) a volatile solute dissolves in the liquid and (b) the dissolved solute is non-volatile.
Answer:
Vapour pressure of a liquid is the pressure exerted by its vapours in equilibrium with the liquid at a given temperature.
(a) When a volatile solute dissolves, both components contribute to the vapour, so the total vapour pressure of the solution is p = p₁°x₁ + p₂°x₂; it may be greater or less than that of the pure liquid depending on whether the solute is more or less volatile than the solvent.
(b) When the solute is non-volatile, the vapour pressure of the solution is lowered, because part of the surface is occupied by non-volatile solute particles: p_s = p°x₁ < p°.

Teacher's Note:
1. Vapour pressure of a liquid depends only on its nature and temperature, not on the amount of liquid taken.

Question 28. Show that the relative lowering of vapour pressure for a solution is equal to the mole fraction of the solute when solvent alone is volatile.
Answer:
According to Raoult’s law, p_s = p°x₁, where x₁ is the mole fraction of the solvent.
As x₁ = 1 − x₂, p_s = p°(1 − x₂) = p° − p°x₂
or p° − p_s = p°x₂ or \( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} = x_2 \)

Teacher's Note:
1. Relative lowering has no units; it is equal to a mole fraction, which is a pure number.

Short Answer Questions (3 Marks)

Question 29. What are azeotropic mixtures? What are their different types? Explain with examples. (Chhatisgarh Board 2012)
Answer:
Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boiling at a constant temperature. They cannot be separated into components by fractional distillation.
(i) Minimum boiling azeotropes : formed by non-ideal solutions showing large positive deviation from Raoult’s law. They boil at a temperature lower than either component. Example : ethanol (about 95% by volume) + water, boiling at 351.1 K.
(ii) Maximum boiling azeotropes : formed by non-ideal solutions showing large negative deviation. They boil at a temperature higher than either component. Example : nitric acid (68% by mass) + water, boiling at 393.5 K; chloroform + acetone.

Teacher's Note:
1. This is why absolute ethanol cannot be obtained from dilute ethanol by fractional distillation alone.

Question 30. State Raoult’s law for a solution containing volatile liquids. Explain with suitable example the concept of maximum boiling azeotropes.
Answer:
Raoult’s law : For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution: p₁ = p₁°x₁, p₂ = p₂°x₂; total p = p₁°x₁ + p₂°x₂.
Maximum boiling azeotrope : Solutions showing large negative deviation from Raoult’s law have a minimum in the vapour pressure–composition curve at a particular composition. At this composition the vapour pressure is lowest, so the boiling point is highest, and the liquid and vapour have the same composition. Such a mixture boils at constant temperature without change in composition and is called a maximum boiling azeotrope.
Example : Nitric acid and water form a maximum boiling azeotrope containing about 68% HNO₃ by mass, boiling at 393.5 K.

Teacher's Note:
1. Minimum in vapour pressure corresponds to maximum in boiling point.

Teacher's Note:
1. Keep the pairing straight: positive deviation, maximum vapour pressure, minimum boiling point.

Question 32. What is a colligative property ? Write down the different types of colligative properties. Show that relative lowering of vapour pressure is a colligative property.
Answer:
Colligative properties are those properties of dilute solutions of non-volatile solutes which depend only upon the number of solute particles and not upon their nature.
Types : (i) Relative lowering of vapour pressure (ii) Elevation in boiling point (iii) Depression in freezing point (iv) Osmotic pressure.
According to Raoult’s law, \( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} = x_2 \)
For a dilute solution, \( \frac{p^\circ - p_s}{p^\circ} \approx \frac{n_2}{n_1} \). Thus, relative lowering of vapour pressure depends only on the number of moles (particles) of the solute present in a given amount of solvent and not on the nature of the solute. Hence, it is a colligative property.

Teacher's Note:
1. Lowering of vapour pressure itself depends on p° (nature of solvent); it is the relative lowering that is colligative.

Question 33. What is Raoult’s law ? How can molar mass of non-volatile solute be determined with its help ? (MP Board 2013)
Answer:
Raoult’s law : For a solution of a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute: \( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1 + n_2} = x_2 \)
Determination of molar mass : For a dilute solution, n₁ + n₂ ≈ n₁. If w₂ g of solute (molar mass M₂) is dissolved in w₁ g of solvent (molar mass M₁),
\( \frac{p^\circ - p_s}{p^\circ} = \frac{w_2/M_2}{w_1/M_1} \) or \( M_2 = \frac{w_2 M_1 p^\circ}{w_1(p^\circ - p_s)} \)
Knowing p° and p_s experimentally, M₂ can be calculated.

Teacher's Note:
1. Check units: the ratio of pressures cancels, so p° and p_s only need to be in the same unit.

Question 34. What is osmotic pressure ? Prove that it is a colligative property. (HP Board 2011)
Answer:
Osmotic pressure is the excess pressure that must be applied on the solution side to just prevent the passage of pure solvent into it when the two are separated by a semipermeable membrane.
According to van’t Hoff equation for dilute solutions, πV = nRT or π = \( \frac{n}{V} \)RT = CRT
where n is the number of moles of solute in V litres of solution and C is the molar concentration. At a given temperature, π ∝ C, i.e., osmotic pressure depends only on the number of moles (particles) of solute per unit volume and not on the nature of the solute. Hence, it is a colligative property.

Teacher's Note:
1. Equimolar solutions of different non-electrolytes (glucose, urea, sucrose) have the same osmotic pressure at the same temperature.

Teacher's Note:
1. The direction trap: in osmosis the solvent moves towards the more concentrated solution.

Question 36. Derive van’t Hoff equation for dilute solutions.
Answer:
Experimentally, osmotic pressure (π) of a dilute solution is found to be :
(i) directly proportional to the molar concentration (C) at constant temperature (van’t Hoff-Boyle’s law): π ∝ C, i.e., π ∝ 1/V (where V is the volume containing 1 mole of solute).
(ii) directly proportional to the absolute temperature at constant concentration (van’t Hoff-Charles’ law): π ∝ T.
Combining, π ∝ CT or π = CRT, where R is the gas constant (solution constant), found to have the same value as the gas constant.
As C = n/V, π = \( \frac{n}{V} \)RT or πV = nRT. This is van’t Hoff equation for dilute solutions; it is similar to the ideal gas equation.

Teacher's Note:
1. This equation also leads to M₂ = w₂RT/(πV) for molar mass determination.

Question 37. Define osmotic pressure. How can molar mass of a substance be determined from the measurement of osmotic pressure of a solution ? (Assam Board 2013)
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.
According to van’t Hoff equation, πV = n₂RT. If w₂ g of solute of molar mass M₂ is dissolved in V litres of solution, n₂ = w₂/M₂
∴ \( \pi V = \frac{w_2}{M_2} RT \) or \( M_2 = \frac{w_2 RT}{\pi V} \)
Knowing w₂, T, V and measuring π, the molar mass M₂ 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 R = 0.083 L bar K⁻¹ mol⁻¹ with π in bar.

Long Answer Questions (5 Marks)

Question 38. (a) Differentiate between molarity and molality of a solution. How does a change in temperature influence their values ?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr₂ in 200 g of water (Molar mass of MgBr₂ = 184 g, K_f for water = 1.86 K kg mol⁻¹) (CBSE 2011)
Answer:
(a) Molarity is the number of moles of solute dissolved per litre of solution (mol L⁻¹), whereas molality is the number of moles of solute dissolved per kg of solvent (mol kg⁻¹). Molarity depends on volume, so it decreases with rise in temperature (volume increases). Molality depends only on masses, so it does not change with temperature.
(b) MgBr₂ → Mg²⁺ + 2Br⁻, ∴ i = 3
Molality, \( m = \frac{10.50/184}{200} \times 1000 \) = 0.2853 mol kg⁻¹
ΔT_f = i K_f m = 3 × 1.86 × 0.2853 = 1.59°
Freezing point of solution = 0 − 1.59 = −1.59°C

Teacher's Note:
1. Do not forget i = 3 for MgBr₂; without it the answer is only −0.53°C.

Question 39. (a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property ? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.0 g of NaCl to 250.0 g of water (K_b for water = 0.512 K kg mol⁻¹, Molar mass of NaCl = 58.44 g) (CBSE 2011)
Answer:
(a) Osmosis is the spontaneous flow of solvent molecules from pure solvent (or a dilute solution) into the solution (or a more concentrated solution) through a semipermeable membrane. Osmotic pressure is the excess pressure that must be applied on the solution side to just prevent osmosis.
Yes, osmotic pressure is a colligative property. By van’t Hoff equation, π = CRT = (n/V)RT, i.e., at a given temperature it depends only on the number of moles of solute per litre of solution and not on the nature of the solute.
(b) NaCl → Na⁺ + Cl⁻, ∴ i = 2
Molality, \( m = \frac{15.0/58.44}{250.0} \times 1000 \) = 1.027 mol kg⁻¹
ΔT_b = i K_b m = 2 × 0.512 × 1.027 = 1.05°
Boiling point of solution = 100 + 1.05 = 101.05°C

Teacher's Note:
1. Assume complete dissociation of NaCl (i = 2) unless the degree of dissociation is given.

Question 40. (a) State the following : (i) Henry’s law about partial pressure of a gas in a mixture,
(ii) Raoult’s law in its general form in reference to solutions.
(b) What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g of chloroform 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. (CBSE 2011)
Answer:
(a) (i) Henry’s law: The partial pressure of a gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution: p = K_H x.
(ii) Raoult’s law (general form): For any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction: p_i = p_i° x_i.
(b) ΔT_b = 68.04 − 61.7 = 6.34°C
\( M_2 = \frac{1000 \times K_b \times w_2}{\Delta T_b \times w_1} = \frac{1000 \times 3.63 \times 6.21}{6.34 \times 24.0} \) = 148.1 g mol⁻¹

Teacher's Note:
1. Here K_b is given as °C/m, which is the same as K kg mol⁻¹ because a temperature difference in °C equals that in K.

Question 41. (i) What is de-icing agent ? How does it function ?
(ii) 19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid. (Pb. Board 2011)
Answer:
(i) Common salt is called de-icing agent because it lowers the freezing point of water to such an extent that it does not freeze to form ice. Hence, it is used to clear snow from roads.
(ii) Molar mass of CH₂FCOOH = 24 + 3 + 19 + 32 = 78 g mol⁻¹
Molality, \( m = \frac{19.5/78}{500} \times 1000 \) = 0.5 m
Calculated ΔT_f = K_f m = 1.86 × 0.5 = 0.93°
\( i = \frac{\text{Observed } \Delta T_f}{\text{Calculated } \Delta T_f} = \frac{1.0}{0.93} \) = 1.0753
CH₂FCOOH ⇄ CH₂FCOO⁻ + H⁺; initially C, at equilibrium C(1 − α), Cα, Cα
i = 1 + α ∴ α = 0.0753
\( K_a = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} = \frac{0.5 \times (0.0753)^2}{1 - 0.0753} \) = 3.07 × 10⁻³

Teacher's Note:
1. For a weak monobasic acid, i = 1 + α; molality is taken as molarity for this dilute aqueous solution.

Question 42. (i) Why elevation in boiling point is a colligative property ?
(ii) Calculate the osmotic pressure in Pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 1,85,000 in 450 mL of water at 37°C. (Pb. Board 2011)
Answer:
(i) For a dilute solution, ΔT_b = K_b m, i.e., elevation in boiling point is directly proportional to the molality of the solution. It depends only on the number of moles (particles) of solute dissolved in a given mass of solvent and not on the nature of the solute. Hence, it is a colligative property.
(ii) π = \( \frac{n}{V} \)RT; n = \( \frac{1.0}{185000} \) mol = 5.405 × 10⁻⁶ mol, V = 450 mL = 0.45 L = 0.45 × 10⁻³ m³, T = 37 + 273 = 310 K, R = 8.314 J K⁻¹ mol⁻¹
\( \pi = \frac{5.405 \times 10^{-6} \times 8.314 \times 310}{0.45 \times 10^{-3}} \) = 30.96 Pa ≈ 31 Pa

Teacher's Note:
1. To get pascals directly, use R = 8.314 J K⁻¹ mol⁻¹ and volume in m³.