CBSE Class 12 Chemistry Solutions MCQs Set 15

Question 1. Which of the following is the correct example of solid solution in which the solute is in gas phase?
(a) Copper dissolved in gold
(b) Camphor in nitrogen gas
(c) Hydrogen in palladium
(d) All of the options
Answer: (c) Hydrogen in palladium
In simple words: Hydrogen gas dissolved in solid palladium is a solid solution where the solute is a gas. It represents a rare scenario where the solute can occupy a larger volume than the solvent.

Exam Tip: In solid solutions, the solvent is always solid, while the solute can be gas, liquid, or solid. Hydrogen in palladium is a standard example of a gas-in-solid solution.

Question 2. Consider the following table.

'X' and 'Y' in the above table are
(a) X = Dust or smoke in air, Y = Solid and solid
(b) X = Humidity in air, Y = Liquid and solid
(c) X = Camphor in nitrogen gas, Y = Gas and solid
(d) X = Chloroform mixed with nitrogen gas, Y = Solid and solid
Answer: (d) X = Chloroform mixed with nitrogen gas, Y = Solid and solid
In simple words: A mixture of chloroform and nitrogen gas is a liquid solute in a gas solvent. Bronze and other alloys consist of solid solutes dissolved in solid solvents.

Exam Tip: Keep in mind that alloys like bronze are solid-in-solid mixtures, whereas chloroform in nitrogen gas represents a liquid solute in a gaseous solvent.

Question 3. 184 g ethyl alcohol is mixed with 72 g of water. The ratio of mole fraction of alcohol to water is
(a) 3 : 4
(b) 1 : 2
(c) 1 : 4
(d) 1 : 1
Answer: (d) 1 : 1
In simple words: Since both ethyl alcohol and water have 4 moles each in this mixture, they have equal mole fractions. This yields a 1 to 1 mole fraction ratio.

Exam Tip: When equal moles of two substances are mixed, their mole fractions are identical, which always results in a 1 : 1 ratio.

Question 4. 4 L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molarity of the resultant solution is ....
(a) 0.004
(b) 0.008
(c) 0.012
(d) 0.016
Answer: (d) 0.016
In simple words: Adding 1 liter of water increases the total volume of the mixture to 5 liters. Applying the dilution formula shows the concentration drops to 0.016 M.

Exam Tip: Use the dilution formula \( M_1 V_1 = M_2 V_2 \), ensuring that the final volume \( V_2 \) represents the sum of the initial volume and the added water.

Question 5. In a chemistry laboratory, Richa took 5 g of a solute from an unknown box and prepared a 0.25 M basic solution. The volume of the solution was 500 mL. Based on the above data, which of the following is likely to be the unknown substance used by Richa?
(Approx. Atomic masses of Ca = 40 u; Na = 23 u; Li = 7 u; Cs = 133 u; O = 16 u; H = 1 u)
(a) Ca(OH)₂
(b) NaOH
(c) LiOH
(d) CsOH
Answer: (b) NaOH
In simple words: A 500 mL solution with a strength of 0.25 M contains 0.125 moles of solute. Since 5 grams represents 0.125 moles, the molar mass is 40 g/mol, which matches NaOH.

Exam Tip: Determine the number of moles from molarity and volume first, then compute the molar mass to easily identify the unknown compound from the given atomic weights.

Question 6. A glycerine solution, at 293 K, has a molality of 3.89 molal and molarity of 5.33 M. Which of these would be correct for molarity and molality of the same glycerine solution at 450 K?
(a) Molarity < 5.33 M; Molality = 3.89 molal
(b) Molarity < 5.33 M; Molality < 3.89 molal
(c) Molarity > 5.33 M; Molality = 3.89 molal
(d) Molarity = 5.33 M; Molality = 3.89 molal
Answer: (a) Molarity < 5.33 M; Molality = 3.89 molal
In simple words: Heating a solution causes its volume to expand, which decreases its molarity. Molality is based on mass, so it is unaffected by temperature changes.

Exam Tip: Volume is temperature-dependent while mass is not; hence, volume-based concentrations (molarity) change with temperature, but mass-based concentrations (molality) stay constant.

Question 7. A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is .....
(a) saturated
(b) supersaturated
(c) unsaturated
(d) concentrated
Answer: (b) supersaturated
In simple words: A supersaturated solution contains more dissolved solute than it can normally hold. Adding even a tiny crystal triggers rapid precipitation of the excess dissolved substance.

Exam Tip: Adding solute to an unsaturated solution causes dissolution, to a saturated solution causes no change, and to a supersaturated solution triggers precipitation.

Question 8. Pressure does not have any significant effect on solubility of solids in liquids because
(a) solids are highly compressible.
(b) liquids are highly compressible.
(c) solubility of solid in liquid is directly proportional to partial pressure.
(d) solids and liquids are highly incompressible.
Answer: (d) solids and liquids are highly incompressible.
In simple words: Both solids and liquids cannot be compressed easily. Because pressure has almost no effect on their volume, it does not change how much solid can dissolve in a liquid.

Exam Tip: Note that pressure only significantly influences the solubility of gases in liquids (as stated by Henry's law) because gases are highly compressible, unlike solids and liquids.

Question 9. Solubility of gases in liquids decreases with rise in temperature because dissolution is an
(a) endothermic and reversible process
(b) exothermic and reversible process
(c) endothermic and irreversible process
(d) exothermic and irreversible process
Answer: (b) exothermic and reversible process
In simple words: Dissolving gas in a liquid releases heat, making it an exothermic reaction. According to Le Chatelier's principle, raising the temperature shifts this reversible process backward, releasing gas.

Exam Tip: Gas dissolution is exothermic (\( \Delta H < 0 \)). Thus, applying heat shifts the equilibrium to favor the undissolved gas state, reducing solubility.

Question 10. Value of Henry's constant \( K_H \),
(a) increases with decrease in temperature.
(b) decreases with increase in temperature.
(c) increases with increase in temperature.
(d) remains constant.
Answer: (c) increases with increase in temperature.
In simple words: Henry's constant rises when the temperature goes up. This explains why gases become less soluble in liquids at higher temperatures.

Exam Tip: Since solubility of a gas is inversely proportional to \( K_H \) (\( p = K_H \chi \)), a higher temperature increases \( K_H \), which directly explains the drop in gas solubility.

Question 11. Which of the following statements is not true?
(a) A temperature increase will cause an increase in vapour pressure.
(b) The total vapour pressure of a mixture of liquids is equal to the sum of the vapour pressure of each component in the mixture.
(c) At a given temperature, all materials have the same vapour pressure.
(d) The vapour pressure of a liquid in a mixture of liquids is less than the vapour pressure of that pure liquid.
Answer: (c) At a given temperature, all materials have the same vapour pressure.
In simple words: Different liquids have unique strengths of attractive forces between their molecules. Therefore, even at the exact same temperature, they will have different vapor pressures.

Exam Tip: Vapor pressure depends on the nature of the liquid (intermolecular forces) and temperature. Weaker intermolecular forces result in higher vapor pressure at a given temperature.

Question 12. Which one of the following pairs will not form an ideal solution?
(a) Benzene and toluene
(b) n-hexane and n-heptane
(c) Ethanol and acetone
(d) Bromoethane and chloroethane
Answer: (c) Ethanol and acetone
In simple words: Mixing ethanol and acetone disrupts the strong hydrogen bonding between ethanol molecules, causing a positive deviation from Raoult's law instead of behaving ideally.

Exam Tip: Ideal solutions are formed by compounds with very similar structures and intermolecular forces (like benzene/toluene). Dissimilar interactions lead to non-ideal behavior.

Question 13. When 1 mole of benzene is mixed with 1 mole of toluene. The vapours will contain (Given: vapour pressure of benzene = 12.8 kPa and vapour pressure of toluene = 3.85 kPa).
(a) equal amount of benzene and toluene as it forms an ideal solution.
(b) unequal amount of benzene and toluene as it forms a non-ideal solution.
(c) higher percentage of benzene.
(d) higher percentage of toluene.
Answer: (c) higher percentage of benzene.
In simple words: Because benzene is more volatile and has a much higher vapor pressure than toluene, it evaporates more readily, leading to a richer concentration of benzene in the vapor phase.

Exam Tip: According to Dalton's law and Raoult's law, the vapor phase is always richer in the component that has a higher pure vapor pressure (the more volatile component).

Question 14. 1 mole of liquid A and 2 moles of liquid B make a solution having a total vapour pressure 40 torr. The vapour pressure of pure A and pure B are 45 torr and 30 torr respectively. The above solution
(a) is an ideal solution.
(b) shows positive deviation.
(c) shows negative deviation.
(d) is a maximum boiling azeotrope.
Answer: (b) shows positive deviation.
In simple words: As computed using Raoult's law, the expected vapor pressure is 35 Torr. Since the actual measured pressure of 40 Torr is higher than 35 Torr, the mixture exhibits a positive deviation.

Exam Tip: Perform the calculation \( P_{\text{total}} = p_A^o \chi_A + p_B^o \chi_B \) to find the expected pressure. A measured value larger than this calculation indicates positive deviation.

Question 15. An azeotropic mixture of two liquids has a boiling point higher than either of the two liquids when it
(a) shows large negative deviation from Raoult’s law.
(b) shows no deviation from Raoult’s law.
(c) shows large positive deviation from Raoult’s law.
(d) obeys Raoult’s law.
Answer: (a) shows large negative deviation from Raoult’s law.
In simple words: When the attractive forces between different molecules are very strong, the vapor pressure drops. This lower vapor pressure means you must heat the mixture to a higher temperature to make it boil.

Exam Tip: A maximum boiling azeotrope corresponds to solutions showing large negative deviations from Raoult's law because a lower vapor pressure requires more heat to reach the boiling point.

Question 16. Which of the following azeotropic solutions has the boiling point more than the boiling point of its constituents molecules?
(a) \( \text{CHCl}_3 \) and \( \text{CH}_3\text{COCH}_3 \)
(b) \( \text{CS}_2 \) and \( \text{CH}_3\text{COCH}_3 \)
(c) \( \text{CH}_3\text{CH}_2\text{OH} \) and \( \text{CH}_3\text{COCH}_3 \)
(d) \( \text{C}_6\text{H}_6 \) and \( \text{CH}_3\text{COCH}_3 \)
Answer: (a) \( \text{CHCl}_3 \) and \( \text{CH}_3\text{COCH}_3 \)
In simple words: When chloroform and acetone are mixed, they form strong hydrogen bonds with each other. This reduces the vapor pressure and increases the boiling point above that of either pure liquid.

Exam Tip: A maximum boiling azeotrope is formed by mixtures showing a negative deviation from Raoult's law, such as chloroform and acetone due to hydrogen bonding between the different components.

Question 17. If molality of a dilute solution is doubled, the value of the molal elevation constant (\( K_b \)) will be
(a) halved
(b) doubled
(c) tripled
(d) unchanged
Answer: (d) unchanged
In simple words: The molal elevation constant is a special property that depends only on the choice of the solvent. Changing how much solute is dissolved does not affect this constant at all.

Exam Tip: Keep in mind that \( K_b \) (ebullioscopic constant) and \( K_f \) (cryoscopic constant) are characteristic of the solvent alone and do not vary with solute concentration or molality.

Question 18. How much ethyl alcohol must be added to 1 litre of water so that the solution will freeze at -14°C? (\( K_f \) for water = 1.86°C kg/mol)
(a) 7.5 mol
(b) 8.5 mol
(c) 9.5 mol
(d) 10.5 mol
Answer: (a) 7.5 mol
In simple words: To lower the freezing point of 1 kg of water by 14 degrees, we use the freezing point depression formula. Doing the math shows we need to add exactly 7.5 moles of ethyl alcohol.

Exam Tip: Use the formula \( \Delta T_f = K_f \times m \). Since 1 liter of water has a mass of 1 kg, the molality is simply equal to the number of moles of solute added.

Question 19. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because .........
(a) it gains water due to osmosis
(b) it loses water due to reverse osmosis
(c) it gains water due to reverse osmosis
(d) it loses water due to osmosis
Answer: (d) it loses water due to osmosis
In simple words: Water flows out from the mango cells into the highly concentrated salt water outside. This movement from a low-strength to a high-strength solution is osmosis, causing the mango to shrink.

Exam Tip: During osmosis, solvent molecules move through a semi-permeable membrane from a dilute solution (high solvent chemical potential) to a concentrated solution (low solvent chemical potential).

Question 20. At a given temperature, osmotic pressure of a concentrated solution of a substance .........
(a) is higher than that of a dilute solution.
(b) is lower than that of a dilute solution.
(c) is same as that of a dilute solution.
(d) cannot be compared with osmotic pressure of dilute solution.
Answer: (a) is higher than that of a dilute solution.
In simple words: Osmotic pressure is directly linked to the concentration of dissolved particles. Since a concentrated solution has more solute particles, it exerts a stronger osmotic pressure.

Exam Tip: Remember the formula \( \pi = CRT \). Since temperature (\( T \)) and gas constant (\( R \)) are constant, the osmotic pressure (\( \pi \)) is directly proportional to the molarity (\( C \)) of the solution.

Question 21. In case of association, abnormal molar mass of solute will
(a) increase
(b) decrease
(c) remain same
(d) first increase and then decrease
Answer: (a) increase
In simple words: When particles join together in a solution, they act like larger, heavier single particles. This grouping makes the measured molar mass of the solute appear higher than normal.

Exam Tip: Association of particles reduces the total number of particles (\( i < 1 \)). Since colligative properties are inversely proportional to molar mass, abnormal molar mass increases during association.

Question 22. We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1 M, 0.01 M and 0.001 M, respectively. The value of van't Hoff factor for these solutions will be in the order .........
(a) \( i_A < i_B < i_C \)
(b) \( i_A > i_B > i_C \)
(c) \( i_A = i_B = i_C \)
(d) \( i_A < i_B > i_C \)
Answer: (c) \( i_A = i_B = i_C \)
In simple words: Under ideal conditions where NaCl splits completely, each molecule always breaks into 2 ions regardless of the concentration. Therefore, the van't Hoff factor is the same for all three.

Exam Tip: For strong electrolytes undergoing complete ionization in dilute solutions, the van't Hoff factor (\( i \)) is equal to the total number of ions produced per formula unit and remains constant.

Question 23. Van’t Hoff factor for \( \text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} \) solution, assuming complete ionisation is
(a) 1
(b) 3
(c) 13
(d) 2
Answer: (b) 3
In simple words: When sodium sulfate crystals dissolve, they break apart into two sodium ions and one sulfate ion. This produces a total of three active ions, so the van't Hoff factor is 3.

Exam Tip: Do not count water of crystallization molecules in the van't Hoff factor; only the ions produced from the dissociation of the ionic solute determine the value of \( i \).

Question 24. Match the Column I with Column II.

Codes:
(a) A → (ii), B → (i), C → (iv), D → (iii)
(b) A → (iv), B → (i), C → (ii), D → (iii)
(c) A → (i), B → (ii), C → (iii), D → (iv)
(d) A → (iii), B → (ii), C → (i), D → (iv)
Answer: (a) A → (ii), B → (i), C → (iv), D → (iii)
In simple words: Matching is based on the ratio of the van't Hoff factors of the substances. For example, urea and glucose both have a factor of 1, giving a 1:1 ratio.

Exam Tip: Calculate the van't Hoff factor (\( i \)) for each solute in Column I assuming complete ionization, and then find their simple ratios to match with Column II.

Question 25. Which of the following statement is/are true?
I. The freezing point of 0.1 M KCl is higher than that of 0.1 M \( \text{C}_2\text{H}_5\text{OH} \).
II. The freezing point of a 4% aqueous solution of X having molecular weight as m is equal to the freezing point of 12% aqueous solution of Y having molecular weight 3m (assume that i = 1 for both X and Y).
III. The boiling point of pure water at sea level is greater than at mount Everest.
(a) I and II
(b) I, II and III
(c) II and III
(d) I and III
Answer: (c) II and III
In simple words: Since KCl dissociates into more particles than ethanol, it lowers the freezing point more, making statement I false. Statements II and III are mathematically and physically correct.

Exam Tip: Freezing point depression depends on the number of solute particles (\( \Delta T_f = i K_f m \)), so electrolytes like KCl cause a greater drop in freezing point than non-electrolytes.

Question 26. If van't Hoff factor of aqueous solutions of X, Y and Z are 1.9, 0.9 and 2.6. Then, which of the following relation is correct?
(a) Boiling point: Z < X < Y
(b) Freezing point: Z < X < Y
(c) Osmotic pressure: X = Y = Z
(d) Vapour pressure: X < Y = Z
Answer: (b) Freezing point: Z < X < Y
In simple words: A higher van't Hoff factor causes a greater depression in freezing point. Since Z has the highest factor, its freezing point drops the most, followed by X, then Y.

Exam Tip: Remember that freezing point is inversely related to the van't Hoff factor because a larger \( i \) value causes a greater depression in freezing point (\( \Delta T_f \)).

Question 27. Assertion (A): One molar aqueous solution has always higher concentration than one molal.
Reason (R): The molality of a solution depends upon the density of the solution whereas molarity does not.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: A 1 M aqueous solution is more concentrated because its volume contains less than 1000 g of water, meaning more solute particles per unit volume.

Exam Tip: For aqueous solutions, 1 M is always more concentrated than 1 m because some volume is occupied by the solute, leaving less than 1000 g of solvent.

Question 28. Assertion (A): Aquatic species are more comfortable in cold water rather than in warm water.
Reason (R): Different gases have different \( K_H \) values at the same temperature.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: Oxygen is more soluble in cold water, which helps aquatic life breathe much easier. While the reason about differing \( K_H \) values is true, it does not explain this direct comfort.

Exam Tip: Gas solubility in water decreases as temperature rises, so cold water contains a higher concentration of dissolved oxygen, benefiting aquatic organisms.

Question 29. Assertion (A): Ethanol and acetone shows positive deviation from Raoult's law.
Reason (R): Pure ethanol molecules exhibit hydrogen bond and on adding acetone hydrogen bond between ethanol molecules breaks.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Pure ethanol has strong hydrogen bonds between its molecules. When acetone is added, it sits between ethanol molecules and breaks these bonds, letting the molecules escape more easily into the vapor phase.

Exam Tip: Positive deviation occurs when the solute-solvent interactions (ethanol-acetone) are weaker than the solute-solute and solvent-solvent interactions, leading to a higher vapor pressure.

Question 30. Assertion (A): Chloroform and acetone mixture shows negative deviation from Raoult's law.
Reason (R): In chloroform and acetone mixture, A—A or B—B type intermolecular interactions are weaker than A—B type intermolecular interactions.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Chloroform and acetone form strong hydrogen bonds with each other. Because these new A-B bonds are stronger than the bonds in the separate liquids, fewer molecules escape, which lowers the vapor pressure.

Exam Tip: Stronger solute-solvent interactions (\( A-B \)) compared to original interactions (\( A-A \) or \( B-B \)) lead to negative deviation from Raoult's law and lower vapor pressure.

Question 31. Assertion (A): Nitric acid and water form maximum boiling azeotrope.
Reason (R): Azeotropes are binary mixture having the same composition in liquid and vapour phase.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: Nitric acid and water form a maximum boiling azeotrope due to a negative deviation from Raoult's law. While the definition of an azeotrope having identical liquid and vapor compositions is true, it is not the explanation for why it is maximum boiling.

Exam Tip: Azeotropes boil at a constant temperature because their composition does not change during boiling. Maximum boiling ones show large negative deviations from Raoult's law.

Question 32. Assertion (A): The vapour pressure of 0.1 M sugar solution is less than that of 0.1 M potassium chloride solution.
Reason (R): Lowering of vapour pressure is directly proportional to the number of species present in the solution.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: Since KCl splits into twice as many particles as sugar, it causes a much larger drop in vapor pressure. Consequently, the actual vapor pressure of the sugar solution remains higher than that of the KCl solution.

Exam Tip: Remember that lowering of vapor pressure is a colligative property. Higher particle concentration (due to dissociation of KCl) results in a lower vapor pressure for the electrolyte solution.

Question 33. Assertion (A): When NaCl is added to water, a depression in freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes depression in the freezing point.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: Adding NaCl to water lowers its vapor pressure. Because the vapor pressure is lower, the solution must be cooled to a lower temperature to freeze, causing freezing point depression.

Exam Tip: Freezing occurs when the vapor pressure of the liquid solvent equals the vapor pressure of the solid solvent. A lower liquid vapor pressure thus decreases the freezing point.

Question 34. Assertion (A): Cryoscopic constant depends on nature of solvent.
Reason (R): Cryoscopic constant is a universal constant.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: The cryoscopic constant is unique to each solvent, depending on its specific molecular properties and heat of fusion. Therefore, it is not a universal constant.

Exam Tip: Cryoscopic constant (\( K_f \)) depends on the molar mass, gas constant, and enthalpy of fusion of the solvent, making it solvent-specific.

Question 35. Assertion (A): Osmotic pressure is a colligative property.
Reason (R): Osmotic pressure is proportional to the molality.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: Osmotic pressure is a colligative property because it depends on solute particle concentration. However, it is mathematically defined as proportional to molarity rather than molality.

Exam Tip: Be careful with concentration terms: osmotic pressure is defined by the equation \( \pi = CRT \), where \( C \) is molarity (moles per liter), not molality.

Question 36. Assertion (A): NaCl in water and organic acids in benzene show abnormal molecular mass.
Reason (R): Abnormal molecular mass is obtained when the substance in the solution undergoes dissociation or association.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In simple words: NaCl splits into ions in water, while organic acids form pairs in benzene. Because these reactions alter the total count of particles, the measured molar masses deviate from normal.

Exam Tip: Electrolytes that dissociate (like NaCl) or molecules that associate (like benzoic acid in benzene) yield abnormal molecular masses due to changes in particle count.

Case Study 37. Henry’s law and Raoult’s law are key principles in understanding solution equilibria, particularly concerning gas-liquid and liquid-liquid interactions. Henry’s law is particularly relevant in environmental systems, such as gas exchange between the atmosphere and bodies of water, where factors like temperature, pressure and salinity play crucial roles. Conversely, Raoult’s law applies to ideal solutions, asserting that the vapour pressure of each component is proportional to its mole fraction. However, real solutions often deviate from ideal behaviour. Positive deviations occur when solute-solvent interactions are weaker than in the pure components, leading to higher than expected vapour pressures. Negative deviations, on the other hand, result from stronger solute-solvent interactions, causing lower vapour pressure. Both laws, while idealised provide valuable insights into molecular interactions, with deviations offering further understanding of non-ideal solutions in field like chemical engineering and environmental chemistry.

Question (i) What role does Henry’s law play in explaining gas exchange between the atmosphere and water bodies?
Answer: Henry’s law explains that the solubility of gases, like oxygen and nitrogen in water, is directly proportional to their partial pressure in the atmosphere. This rule is vital for understanding how gases dissolve in aquatic systems. For instance, a sudden decrease in external pressure causes dissolved gas to release quickly, which explains how oxygen dissolves from air into lakes and oceans.
In simple words: Henry's law states that the amount of gas dissolved in water is proportional to its pressure in the air, explaining how lakes absorb oxygen.

Exam Tip: Mention the direct proportional relationship between solubility of a gas and its partial pressure above the liquid surface to score full marks.

Question (ii) When the temperature increases, then Henry’s constant
(a) increases
(b) decreases
(c) remains same
(d) can’t predict
Answer: (a) increases
In simple words: As the temperature goes up, the value of Henry's constant increases, which leads to a decrease in the solubility of gases in liquids.

Exam Tip: Remember that Henry's constant (\( K_H \)) is temperature-dependent and increases with rising temperature, which makes gases less soluble in warm liquids.

Question (iii) Real solutions deviate from Raoult’s law
(a) due to difference in solute-solute interactions.
(b) due to difference in solvent-solvent interactions.
(c) due to difference in solute-solvent interactions.
(d) None of the options
Answer: (c) due to difference in solute-solvent interactions.
In simple words: Real solutions deviate from Raoult's law because the forces of attraction between solute and solvent molecules differ from those in the pure liquids.

Exam Tip: Deviations from ideal behavior are caused by changes in intermolecular forces upon mixing; positive deviation means weaker interactions, and negative deviation means stronger interactions.

Question (iii) Or Raoult’s law is primarily used for in solution chemistry
(a) to predict the mole fraction.
(b) to predict the vapour pressure.
(c) to predict the molality.
(d) to predict the molarity.
Answer: (b) to predict the vapour pressure.
In simple words: Raoult's law helps to determine the vapor pressure of a solution by multiplying the vapor pressure of the pure liquid by its mole fraction.

Exam Tip: Raoult's law states that the partial vapor pressure of any volatile component of a solution is directly proportional to its mole fraction in the solution.

Case Study 38. According to many researches the decrease in freezing point is directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution’s identity. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapour pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity.

Question (i) Why is osmotic pressure preferable to other colligative properties?
Answer: The osmotic pressure technique is favored because the measurement is carried out at room temperature, and it uses molarity rather than molality. Furthermore, its magnitude is notably larger than other colligative properties, even for extremely dilute solutions.
In simple words: Osmotic pressure is easier to measure at room temperature and gives much larger, more readable values for highly diluted solutions.

Exam Tip: This method is highly preferred for biomolecules like proteins and polymers because they are unstable at high temperatures and have low solubility.

Question (ii) What are the factors on which colligative properties depend?
Answer: Colligative properties depend on the identity of the solvent and the concentration of the solute particles, rather than the chemical nature of the solute.
In simple words: These properties depend only on how many solute particles are dissolved, not on what kind of solute is used.

Exam Tip: Clearly state that colligative properties depend strictly on the number (or concentration) of solute particles and not their identity.

Question (iii) Why does water boils above 100°C when non-volatile solid is added to pure water?
Answer: Adding a non-volatile solid lowers the vapor pressure of the water. As a result, a higher temperature is needed to bring its vapor pressure back up to atmospheric pressure, elevating the boiling point above 100°C.
In simple words: Solute particles block water molecules from evaporating, lowering the vapor pressure. We must heat it past 100°C to make it boil.

Exam Tip: Explain that vapor pressure lowering directly causes an elevation in boiling point since the liquid must reach a higher temperature to equal atmospheric pressure.

Question (iv) Or Why does osmotic pressure depend on temperature?
Answer: Osmotic pressure is the excess pressure required to halt osmosis. According to the van't Hoff equation \( \pi = iCRT \), osmotic pressure is directly proportional to temperature as heat increases the kinetic energy and movement of the solvent particles.
In simple words: Heating a solution makes the molecules move faster and hit the membrane harder, which increases the pressure needed to stop them.

Exam Tip: Write the equation \( \pi = iCRT \) to mathematically demonstrate the direct relationship between osmotic pressure and absolute temperature.