C.B.S.E. Chemistry XII – Solutions(NCERT) (76 MCQs with Explanations)

1. A solution is best defined as a:

1. Heterogeneous mixture of two or more components.
2. Homogeneous mixture of two or more components.
3. Pure substance with a fixed composition.
4. Mixture where components are chemically bonded.

Answer: (b) Homogeneous mixture of two or more components.

Explanation: The sources define solutions as homogeneous mixtures of two or more components, meaning their composition and properties are uniform throughout the mixture.

2. In a binary solution, the component present in the largest quantity is generally called the:

1. Solute
2. Solvent
3. Dispersed phase
4. Dispersing medium

Answer: (b) Solvent

Explanation: The component present in the largest quantity is known as the solvent, and it determines the physical state in which the solution exists. The other components are called solutes.

3. Which of the following is an example of a solid solution?

1. Oxygen dissolved in water
2. Ethanol dissolved in water
3. Camphor in nitrogen gas
4. Amalgam of mercury with sodium

Answer: (d) Amalgam of mercury with sodium

Explanation: In a solid solution, the solvent is a solid. An amalgam of mercury with sodium is an example where the solute is a liquid (mercury) and the solvent is a solid (sodium).

4. A solution described as “10% glucose in water by mass” means:

1. 10 g of glucose is dissolved in 100 g of water.
2. 10 g of glucose is dissolved in 90 g of water.
3. 10 mL of glucose is dissolved in 90 mL of water.
4. 10 g of glucose is dissolved to make 100 mL of solution.

Answer: (b) 10 g of glucose is dissolved in 90 g of water.

Explanation: Mass percentage (w/w) means that for a 10% solution, 10 g of the solute (glucose) is dissolved in 90 g of the solvent (water), resulting in a total of 100 g of solution.

5. Which of the following concentration units is most convenient for a solute present in trace quantities, such as pollutants in water?

1. Molarity
2. Molality
3. Parts per million (ppm)
4. Mass percentage

Answer: (c) Parts per million (ppm)

Explanation: When a solute is present in very small (trace) quantities, it is convenient to express its concentration in parts per million (ppm).

6. Which of the following concentration terms is independent of temperature?

1. Molarity
2. Molality
3. Volume percentage (V/V)
4. Mass by volume percentage (w/V)

Answer: (b) Molality

Explanation: Molality (m) is defined using the mass of the solvent, which does not change with temperature. Molarity and other volume-based units are temperature-dependent because volume changes with temperature.

7. Mole fraction is a useful unit for relating which physical property to the concentration of a solution?

1. Density
2. Viscosity
3. Vapour pressure
4. Surface tension

Answer: (c) Vapour pressure

Explanation: The mole fraction unit is very useful in relating physical properties like vapour pressure with the concentration of the solution and is also useful in calculations involving gas mixtures.

8. Calculate the molarity of a solution containing 5 g of NaOH in 450 mL of solution.

1. (A) 0.125 M
2. (B) 0.278 M
3.  (C) 0.556 M
4. (D) 1.111 M

Answer: (b) 0.278 M

Explanation: First, find the moles of NaOH: Moles = 5 g / 40 g mol⁻¹ = 0.125 mol. Then, calculate molarity: Molarity = Moles / Volume (L) = 0.125 mol / 0.450 L = 0.278 M.

9. The principle “like dissolves like” suggests that:

1. All solutes dissolve in all solvents.
2. Polar solutes dissolve in non-polar solvents.
3. Polar solutes dissolve in polar solvents, and non-polar solutes dissolve in non-polar solvents.
4. Non-polar solutes dissolve in polar solvents.

Answer: (c) Polar solutes dissolve in polar solvents, and non-polar solutes dissolve in non-polar solvents.

Explanation: A solute dissolves in a solvent if the intermolecular interactions are similar in both. This generally means polar solutes dissolve in polar solvents and non-polar solutes in non-polar ones.

10. In a nearly saturated solution, if the dissolution process is endothermic (Δsol H > 0), how does solubility change with an increase in temperature?

1.  It decreases.
2.  It increases.
3.  It remains the same.
4. It depends on the pressure.

Answer: (b) It increases.

Explanation: According to Le Chatelier’s Principle, if the dissolution process is endothermic (absorbs heat), an increase in temperature will shift the equilibrium towards the products, thus increasing the solubility.

11. Henry’s Law states that the partial pressure of a gas in the vapour phase (p) is proportional to the:

1. Temperature of the solution.
2. Mole fraction of the gas (x) in the solution.
3. Volume of the solvent.
4. Total pressure of the system.

Answer: (b) Mole fraction of the gas (x) in the solution.

Explanation: The most common form of Henry’s Law is expressed as p = KH * x, where p is the partial pressure of the gas and x is its mole fraction in the solution.

12. A higher value of Henry’s law constant (KH) at a given pressure indicates:

1. Higher solubility of the gas.
2. Lower solubility of the gas.
3. The gas is highly reactive.
4. The gas is ideal.

Answer: (b) Lower solubility of the gas.

Explanation: From the equation p = KH * x, it is clear that for a given pressure (p), if KH is higher, the mole fraction (x) must be lower. Therefore, a higher KH value corresponds to lower solubility.

13. Aquatic species are more comfortable in cold water than in warm water because:

1. Cold water has a higher density.
2. The solubility of dissolved oxygen is higher in cold water.
3. Cold water has lower pressure.
4. Warm water contains more pollutants.

Answer: (b) The solubility of dissolved oxygen is higher in cold water. Explanation: The KH values for gases like O₂ and N₂ increase with temperature, indicating that their solubility decreases. Therefore, cold water has a higher concentration of dissolved oxygen, making it more suitable for aquatic life.

14. The medical condition known as “bends,” experienced by scuba divers, is caused by:

1. Low concentration of oxygen in the blood.
2. The formation of nitrogen bubbles in the blood upon rapid ascent.
3. High concentration of carbon dioxide in the blood.
4. The toxic effects of helium.

Answer: (b) The formation of nitrogen bubbles in the blood upon rapid ascent. Explanation: Increased pressure underwater increases the solubility of nitrogen in the blood. When divers ascend, the pressure decreases, releasing dissolved gases and forming bubbles of nitrogen in the blood, which causes the “bends”.

15. Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component is:

1. Equal to the vapour pressure of the pure component.
2. Inversely proportional to its mole fraction.
3. Directly proportional to its mole fraction in the solution.
4. Directly proportional to the total pressure.

Answer: (c) Directly proportional to its mole fraction in the solution. Explanation: Raoult’s law states that the partial vapour pressure of each component (p₁) is directly proportional to its mole fraction (x₁) in the solution, given by the equation p₁ = p₁⁰ * x₁.

16. When a non-volatile solute is added to a pure solvent, the vapour pressure of the solution:

1. increases
2. decreases
3. Remains unchanged.
4. Becomes equal to the atmospheric pressure.

Answer: (b) Decreases. Explanation: In the solution, the surface has both solute and solvent molecules. This reduces the fraction of the surface covered by solvent molecules, which in turn reduces the number of solvent molecules escaping from the surface, thus lowering the vapour pressure.

17. Which of the following is NOT a characteristic of an ideal solution?

1. ΔmixH = 0
2. ΔmixV = 0
3. It obeys Raoult’s law over the entire range of concentration.
4. The intermolecular attractive forces between A-B are much stronger than A-A and B-B forces.

Answer: (d) The intermolecular attractive forces between A-B are much stronger than A-A and B-B forces.

Explanation: In an ideal solution, the intermolecular attractive forces between solute-solvent (A-B) are nearly equal to those between solute-solute (A-A) and solvent-solvent (B-B) molecules. Stronger A-B forces lead to negative deviation.

18. A mixture of ethanol and acetone shows positive deviation from Raoult’s law. This is because:

1. Ethanol and acetone form strong hydrogen bonds with each other.
2. Acetone molecules break some of the hydrogen bonds between ethanol molecules, weakening interactions.
3. The mixing process is highly exothermic (ΔmixH < 0).
4. The volume of the solution decreases upon mixing.

Answer: (b) Acetone molecules break some of the hydrogen bonds between ethanol molecules, weakening interactions.

Explanation: In pure ethanol, molecules are hydrogen-bonded. Adding acetone disrupts these bonds, making the new solute-solvent interactions weaker. This increases the escaping tendency of molecules, leading to a higher vapour pressure and positive deviation.

19. A mixture of chloroform and acetone exhibits negative deviation from Raoult’s law due to:

1. Weaker interactions between chloroform and acetone molecules.
2. The formation of a hydrogen bond between chloroform and acetone molecules.
3. An increase in volume upon mixing (ΔmixV > 0).
4. The mixture being nearly ideal.

Answer: (b) The formation of a hydrogen bond between chloroform and acetone molecules.

Explanation: A chloroform molecule can form a hydrogen bond with an acetone molecule, creating solute-solvent interactions that are stronger than the original interactions in the pure components. This reduces the escaping tendency of molecules, lowering the vapour pressure.

20. A binary mixture that boils at a constant temperature and has the same composition in both liquid and vapour phases is called:

1. An ideal solution
2. A saturated solution
3. An azeotrope
4. A colligative solution

Answer: (c) An azeotrope

Explanation: Azeotropes are binary mixtures with the same composition in liquid and vapour phases, which boil at a constant temperature, making them inseparable by fractional distillation.

21. Solutions that show a large positive deviation from Raoult’s law form:

1. Maximum boiling azeotropes.
2. Minimum boiling azeotropes.
3. Ideal solutions.
4. No azeotropes.

Answer: (b) Minimum boiling azeotropes.

Explanation: A large positive deviation means the mixture’s vapour pressure is significantly higher than predicted, which corresponds to a lower boiling point. Such solutions form a minimum boiling azeotrope at a specific composition. An example is the ethanol-water mixture.

22. Which of the following is NOT a colligative property?

1. Relative lowering of vapour pressure
2. Elevation of boiling point
3. Vapour pressure of the solution
4. Osmotic pressure

Answer: (c) Vapour pressure of the solution.

Explanation: Colligative properties depend only on the number of solute particles, not their nature. The four main colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. The absolute vapour pressure is not a colligative property, but its relative lowering is.

23. According to Raoult’s law, the relative lowering of vapour pressure for a solution containing a non-volatile solute is equal to the:

1. Mole fraction of the solvent.
2. Mole fraction of the solute.
3. Molality of the solution.
4. Molarity of the solution.

Answer: (b) Mole fraction of the solute. Explanation: The expression for relative lowering of vapour pressure, (p₁⁰ – p₁)/p₁⁰, is equal to the mole fraction of the solute (x₂).

24. The elevation of boiling point (ΔTb) for a dilute solution is directly proportional to the:

1. Molarity of the solution.
2. Mole fraction of the solute.
3. Molal concentration (molality) of the solution.
4. Mass percentage of the solute.

Answer: (c) Molal concentration (molality) of the solution.

Explanation: For dilute solutions, the elevation of boiling point (ΔTb) is directly proportional to the molality (m) of the solute, expressed as ΔTb = Kb * m.

25. The freezing point of a solution is lower than that of the pure solvent because:

1. The vapour pressure of the solution is higher than the pure solvent.
2. The solution’s vapour pressure becomes equal to the solid solvent’s vapour pressure at a lower temperature.
3. The solute particles have zero kinetic energy at the freezing point.
4. The solute increases the enthalpy of fusion of the solvent.

Answer: (B) The solution’s vapour pressure becomes equal to the solid solvent’s vapour pressure at a lower temperature.

Explanation: Adding a non-volatile solute lowers the solvent’s vapour pressure. A liquid freezes when its vapour pressure equals that of its solid phase. Since the solution’s vapour pressure is already lower, it needs to be cooled to a lower temperature to reach this point.

26. The spontaneous flow of solvent molecules from a pure solvent to a solution through a semipermeable membrane is called:

1. Reverse osmosis
2. Diffusion
3. Osmosis
4. Effusion

Answer: (c) Osmosis

Explanation: Osmosis is the process where solvent molecules flow through a semipermeable membrane from a region of pure solvent (or lower solute concentration) to a region of solution (or higher solute concentration).

27. A 0.9% (mass/volume) sodium chloride solution is said to be isotonic with the fluid inside blood cells. If blood cells are placed in a solution with a salt concentration of 0.5%, the solution is considered:

1. Isotonic
2. Hypertonic
3. Hypotonic
4. Saturated

Answer: (c) Hypotonic

Explanation: A solution with a salt concentration less than 0.9% (mass/volume) is called hypotonic. In this case, water will flow into the blood cells, causing them to swell.

28. The process of reverse osmosis, used for desalination of seawater, occurs when:

1. The temperature of the seawater is increased.
2. The semipermeable membrane is removed.
3. A pressure larger than the osmotic pressure is applied to the solution side. (D) A pressure lower than the osmotic pressure is applied to the solution side.
4. none of these

Answer: (c) A pressure larger than the osmotic pressure is applied to the solution side.

Explanation: The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution. This forces the pure solvent to flow out of the solution through the semipermeable membrane.

29. When a solute like KCl dissociates in water, the experimentally determined molar mass is:

1. Higher than the true value.
2. Lower than the true value.
3. The same as the true value.
4. Equal to zero.

Answer: (b) Lower than the true value.

Explanation: Dissociation increases the total number of particles in the solution. Since colligative properties depend on the number of particles, the observed effect is larger than expected, which leads to an experimentally determined molar mass that is always lower than the true value.

30. When molecules of ethanoic acid associate (dimerise) in benzene, the van’t Hoff factor (i) is:

1. Greater than 1
2. Less than 1
3. Equal to 1
4. Equal to 2

Answer: (b) Less than 1.

Explanation: Association reduces the total number of moles of particles in the solution. Since the van’t Hoff factor (i) is the ratio of moles of particles after association to moles before association, its value for association is less than unity.

31. What is the theoretical van’t Hoff factor (i) for a completely dissociated K₂SO₄ solution?

1. 1
2. 2
3. 3
4. 4

Answer: (c) 3

Explanation: K₂SO₄ dissociates into three ions: two K⁺ ions and one SO₄²⁻ ion. Therefore, one mole of the solute produces three moles of particles, and the theoretical van’t Hoff factor i approaches 3 in a very dilute solution.

32. The modified equation for osmotic pressure (Π) that accounts for association or dissociation is:

1. Π = n₂RT / V
2. Π = mRT / i
3. Π = i (n₂/V)RT
4. Π = i * K * m

Answer: (c) Π = i (n₂/V)RT

Explanation: The inclusion of the van’t Hoff factor (i) modifies the standard colligative property equations. The correct modified equation for osmotic pressure is Π = i n₂RT / V.

33. Calculate the mass percentage of benzene (C₆H₆) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride (CCl₄).

1. 03%
2. 28%
3. 72%
4. 50%

Answer: (b) 15.28%

Explanation: The total mass of the solution is the sum of the masses of the components: 22 g (benzene) + 122 g (CCl₄) = 144 g. The mass percentage of benzene is calculated using the formula: Mass % of component = (Mass of the component / Total mass of the solution) × 100. Mass % of benzene = (22 g / 144 g) × 100 = 15.28%.

34. Why is molality (m) preferred over molarity (M) for expressing concentration in experiments involving temperature changes?

1. Molality is easier to calculate.
2. Molarity depends on the mass of the solvent, which changes with temperature.
3. Molality is independent of temperature because it is based on mass, while molarity is based on volume, which is temperature-dependent.
4. Molarity applies only to aqueous solutions.

Answer: (c) Molality is independent of temperature because it is based on mass, while molarity is based on volume, which is temperature-dependent. Explanation: The source states that molality is independent of temperature, whereas molarity is a function of temperature. This is because volume depends on temperature, but mass does not. Therefore, molality is a more robust unit for concentration when temperature is a variable.

35. A solution of glucose (molar mass = 180 g mol⁻¹) in water is labelled as 10% w/w. What is the molality of this solution?

1. 1 m
2. 556 m
3. 617 m
4. 0 m

Answer: (c) 0.617 m

Explanation: A 10% w/w solution means 10 g of glucose is dissolved in 90 g of water. Moles of glucose = Mass / Molar mass = 10 g / 180 g mol⁻¹ = 0.0555 mol. Mass of solvent (water) in kg = 90 g / 1000 = 0.090 kg. Molality (m) = Moles of solute / Mass of solvent in kg = 0.0555 mol / 0.090 kg = 0.617 mol kg⁻¹.

36. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂, molar mass = 62 g mol⁻¹) and 200 g of water. What is the molality of the solution?

1. 95 m
2. 11 m
3. 056 m
4. 79 m

Answer: (a) 17.95 m

Explanation: First, calculate the moles of the solute (ethylene glycol): Moles = 222.6 g / 62 g mol⁻¹ = 3.59 mol. Next, convert the mass of the solvent (water) to kg: 200 g = 0.2 kg. Molality (m) = Moles of solute / Mass of solvent in kg = 3.59 mol / 0.2 kg = 17.95 m.

37. Why do climbers at high altitudes experience a condition known as anoxia?

1. The temperature is too low.
2. The partial pressure of oxygen is high, increasing its solubility in blood.
3. The partial pressure of oxygen is low, leading to low concentrations of oxygen in the blood and tissues.
4. The air contains toxic gases at high altitudes.

Answer: (c) The partial pressure of oxygen is low, leading to low concentrations of oxygen in the blood and tissues.

Explanation: The source explains that at high altitudes, the partial pressure of oxygen is less than at ground level. According to Henry’s law, this leads to low concentrations of oxygen in the blood and tissues, causing weakness and an inability to think clearly, a condition known as anoxia.

38. The Henry’s law constant (Kʜ) for N₂ gas in water at 293 K is 76.48 kbar. If N₂ exerts a partial pressure of 0.987 bar, what is the mole fraction of N₂ in the solution?

1. 29 × 10⁻³
2. 29 × 10⁻⁵
3. 74 × 10⁴
4. 0129

Answer: (b) 1.29 × 10⁻⁵

Explanation: Henry’s law is given by p = Kʜ * x. To find the mole fraction (x), we rearrange the formula to x = p / Kʜ. Given: p = 0.987 bar, Kʜ = 76.48 kbar = 76480 bar. x = 0.987 bar / 76480 bar = 1.29 × 10⁻⁵.

39. Why do gases always tend to be less soluble in liquids as the temperature is raised?

1. The dissolution process for gases is endothermic.
2. The kinetic energy of gas molecules decreases with temperature.
3. The dissolution process for gases is an exothermic process.
4. The partial pressure of the gas increases with temperature.

Answer: (c) The dissolution process for gases is an exothermic process. Explanation: The dissolution of a gas in a liquid can be considered similar to condensation, and heat is evolved in this process (exothermic). According to Le Chatelier’s Principle, for an exothermic process, solubility should decrease with an increase in temperature. This is also supported by data showing that Kʜ values increase with temperature, indicating lower solubility.

40. Raoult’s law becomes a special case of Henry’s law when:

1. The solute is non-volatile.
2. The solution is ideal.
3. The Henry’s law constant (Kʜ) becomes equal to the vapour pressure of the pure component (p₁⁰).
4. The temperature is constant.

Answer: (c) The Henry’s law constant (Kʜ) becomes equal to the vapour pressure of the pure component (p₁⁰).

Explanation: Both laws state that the partial pressure of a volatile component is directly proportional to its mole fraction. The proportionality constants are Kʜ for Henry’s law and p₁⁰ for Raoult’s law. Thus, Raoult’s law can be seen as a special case of Henry’s law where Kʜ becomes equal to p₁⁰.

41. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of pure heptane and octane are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane (molar mass = 100 g/mol) and 35.0 g of octane (molar mass = 114 g/mol)?

1. 5 kPa
2. 0 kPa
3. 0 kPa
4. 4 kPa

Answer: (A) 73.5 kPa

Explanation: Moles of heptane (n₁) = 26.0 g / 100 g/mol = 0.26 mol. Moles of octane (n₂) = 35.0 g / 114 g/mol = 0.307 mol. Total moles = 0.26 + 0.307 = 0.567 mol. Mole fraction of heptane (x₁) = 0.26 / 0.567 = 0.458. Mole fraction of octane (x₂) = 0.307 / 0.567 = 0.541. According to Raoult’s law, P_total = p₁⁰x₁ + p₂⁰x₂. P_total = (105.2 kPa * 0.458) + (46.8 kPa * 0.541) = 48.18 kPa + 25.32 kPa = 73.5 kPa.

42. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile solid weighing 0.5 g is added to 39.0 g of benzene (molar mass = 78 g/mol). The vapour pressure of the solution is 0.845 bar. What is the molar mass of the solid?

1. 170 g/mol
2. 58 g/mol
3. 180 g/mol
4. 85 g/mol

Answer: (a) 170 g/mol

Explanation: Using the formula for relative lowering of vapour pressure for dilute solutions: (p₁⁰ – p₁) / p₁⁰ = (w₂ × M₁) / (M₂ × w₁). (0.850 – 0.845) / 0.850 = (0.5 g × 78 g/mol) / (M₂ × 39.0 g). 0.005 / 0.850 = 39 / (M₂ × 39). 0.00588 = 1 / M₂. M₂ = 1 / 0.00588 = 170 g/mol.

43. 18 g of glucose (C₆H₁₂O₆, molar mass = 180 g/mol) is dissolved in 1 kg of water. At what temperature will this solution boil if Kb for water is 0.52 K kg mol⁻¹ and pure water boils at 373.15 K?

1. 15 K
2. 202 K
3. 098 K
4. 52 K

Answer: (b) 373.202 K

Explanation: Moles of glucose = 18 g / 180 g/mol = 0.1 mol. Molality (m) = Moles of solute / Mass of solvent in kg = 0.1 mol / 1 kg = 0.1 m. Elevation in boiling point (ΔTb) = Kb × m = 0.52 K kg mol⁻¹ × 0.1 m = 0.052 K. Boiling point of solution = Boiling point of pure water + ΔTb = 373.15 K + 0.052 K = 373.202 K.

44. 45 g of ethylene glycol (C₂H₆O₂, molar mass = 62 g/mol) is mixed with 600 g of water. What is the freezing point of the solution? (Kf for water = 1.86 K kg mol⁻¹, Freezing point of pure water = 273.15 K)

1. 95 K
2. 35 K
3. 15 K
4. 83 K

Answer: (a) 270.95 K

Explanation: Moles of ethylene glycol = 45 g / 62 g/mol = 0.73 mol. Molality (m) = 0.73 mol / 0.6 kg = 1.2 mol kg⁻¹. Depression in freezing point (ΔTf) = Kf × m = 1.86 K kg mol⁻¹ × 1.2 m = 2.2 K. Freezing point of solution = Freezing point of pure water – ΔTf = 273.15 K – 2.2 K = 270.95 K.

45. What is the molar mass of a protein if 1.26 g of it in 200 cm³ of an aqueous solution at 300 K has an osmotic pressure of 2.57 × 10⁻³ bar? (R = 0.083 L bar mol⁻¹ K⁻¹)

1. 30,511 g/mol
2. 122,044 g/mol
3. 61,022 g/mol
4. 6,102 g/mol

Answer: (c) 61,022 g/mol

Explanation: Using the osmotic pressure equation to find molar mass: M₂ = (w₂RT) / (ΠV). Given: w₂ = 1.26 g, R = 0.083 L bar mol⁻¹ K⁻¹, T = 300 K, Π = 2.57 × 10⁻³ bar, V = 200 cm³ = 0.200 L. M₂ = (1.26 g × 0.083 L bar K⁻¹ mol⁻¹ × 300 K) / (2.57 × 10⁻³ bar × 0.200 L) = 31.374 / 0.000514 = 61,022 g/mol.

46. When 2 g of benzoic acid (C₆H₅COOH) is dissolved in 25 g of benzene, it shows a depression in freezing point of 1.62 K. If the molar mass calculated from this data is 241.98 g/mol, what is the van’t Hoff factor (i)? (Normal molar mass of benzoic acid = 122 g/mol)

1. 0
2. 504
3. 0
4. 992

Answer: (b) 0.504

Explanation: The van’t Hoff factor (i) is defined as the ratio of the normal molar mass to the abnormal (experimentally determined) molar mass. i = Normal molar mass / Abnormal molar mass = 122 g mol⁻¹ / 241.98 g mol⁻¹ = 0.504. The value is less than 1, indicating association of the solute molecules.

47. Which of the following aqueous solutions would have the highest boiling point, assuming equal molal concentrations?

1. Glucose (C₆H₁₂O₆)
2. Sodium Chloride (NaCl)
3. Potassium Sulphate (K₂SO₄)
4. Acetic Acid (CH₃COOH)

 Answer: (c) Potassium Sulphate (K₂SO₄)

Explanation: Elevation of boiling point is a colligative property modified by the van’t Hoff factor (ΔTb = i * Kb * m). A higher value of ‘i’ leads to a greater elevation in boiling point.

◦ Glucose does not dissociate (i=1).

◦ NaCl dissociates into 2 ions (i ≈ 2).

◦ K₂SO₄ dissociates into 3 ions (2K⁺ and SO₄²⁻), giving an ideal i-value of 3.

◦ Acetic acid is a weak electrolyte and only partially dissociates (1 < i < 2). Therefore, K₂SO₄ will produce the most particles and have the highest boiling point.

49. If the degree of dissociation (x) of a weak electrolyte is 0.041, what is the van’t Hoff factor (i) for the reaction CH₃COOH ⇌ CH₃COO⁻ + H⁺?

1. 959
2. 0
3. 041
4. 0

Answer: (c) 1.041

Explanation: For a solute that dissociates, the total number of moles of particles at equilibrium is n(1 – x + x + x) = n(1 + x). The van’t Hoff factor ‘i’ is the ratio of total moles of particles at equilibrium to the initial moles, so i = n(1+x)/n = 1+x. Given x = 0.041, then i = 1 + 0.041 = 1.041.

51. A raw mango shrivels when pickled in concentrated brine (salt water). This phenomenon is an example of:

1. Reverse Osmosis
2. Osmosis
3. Diffusion
4. Dissolution

Answer: (b) Osmosis

Explanation: The source explains that a raw mango placed in a concentrated salt solution loses water via osmosis and shrivels. This is because the water concentration inside the mango is higher than in the concentrated salt solution (a hypertonic solution), causing water to move out of the mango through its semipermeable membrane.

52. The preservation of meat by salting protects against bacterial action because:

1. Salt is toxic to bacteria.
2. The salt raises the boiling point of water in the meat.
3. Through osmosis, a bacterium on the salted meat loses water, shrivels, and dies.
4. The salt forms a protective layer on the meat.

Answer: (c) Through osmosis, a bacterium on the salted meat loses water, shrivels, and dies.

Explanation: The source explicitly states that the preservation of meat by salting works because, “Through the process of osmosis, a bacterium on salted meat…loses water, shrivels and dies“.

53. A mixture of nitric acid and water forms a maximum boiling azeotrope. This indicates that the solution exhibits:

1. Positive deviation from Raoult’s law
2. Negative deviation from Raoult’s law
3. Ideal behavior
4. No deviation from Raoult’s law

Answer: (b) Negative deviation from Raoult’s law

Explanation: The source states that solutions that show a large negative deviation from Raoult’s law form maximum boiling azeotropes at a specific composition. This occurs because the intermolecular forces between the solute and solvent are stronger than in the pure components, which lowers the vapour pressure and raises the boiling point.

54. In a solution showing positive deviation from Raoult’s law, the enthalpy of mixing (ΔmixH) is:

1. Zero (ΔmixH = 0)
2. Negative (ΔmixH < 0, exothermic)
3. Positive (ΔmixH > 0, endothermic)
4. Always equal to the volume of mixing (ΔmixV)

Answer: (c) Positive (ΔmixH > 0, endothermic)

Explanation: A positive deviation occurs when the solute-solvent interactions (A-B) are weaker than the solute-solute (A-A) and solvent-solvent (B-B) interactions. To form the solution, stronger intermolecular forces must be broken, and weaker ones are formed. This requires an input of energy, making the process endothermic (ΔmixH > 0). For ideal solutions, ΔmixH = 0.

55. Why is it impossible to separate the components of an azeotropic mixture by fractional distillation?

1. The components are chemically bonded to each other.
2. The mixture has the same composition in the liquid and vapour phases.
3. The components have the same boiling point in their pure state.
4. The mixture forms a heterogeneous system.

Answer: (b) The mixture has the same composition in the liquid and vapour phases.

Explanation: The source defines azeotropes as binary mixtures that have the same composition in the liquid and vapour phase and boil at a constant temperature. Because the vapour has the same composition as the liquid, no separation can be achieved by distilling it.

56. Why is the osmotic pressure method preferred over other colligative properties for determining the molar masses of macromolecules like proteins?

1. It involves measuring a very small change in temperature.
2. Its magnitude is large even for very dilute solutions, and measurements can be done at room temperature.
3. Biomolecules are most stable at their boiling points.
4. It does not require a semipermeable membrane.

Answer: (b) Its magnitude is large even for very dilute solutions, and measurements can be done at room temperature.

Explanation: The source highlights several advantages of the osmotic pressure method for biomolecules: its magnitude is large even for very dilute solutions, making it easier to measure accurately; measurements are done around room temperature, at which biomolecules are generally stable; and it uses molarity instead of molality.

57. According to Raoult’s law, the freezing point of a solvent is depressed upon the addition of a non-volatile solute because:

1. The solute particles physically block the formation of the solid solvent crystal lattice.
2. The vapour pressure of the solution is lowered, and it becomes equal to the vapour pressure of the solid solvent at a lower temperature.
3. The solute increases the kinetic energy of the solvent molecules, preventing them from freezing.
4. The solute has a very low freezing point itself.

Answer: (b) The vapour pressure of the solution is lowered, and it becomes equal to the vapour pressure of the solid solvent at a lower temperature. Explanation: The source explains that a substance freezes when the vapour pressure of its liquid phase equals the vapour pressure of its solid phase. When a solute is added, the solution’s vapour pressure decreases. To reach the point where the solution’s vapour pressure equals that of the solid solvent, the temperature must be lowered. Thus, the freezing point decreases.

1. A raw mango shrivels when pickled in concentrated salt water. This phenomenon is an example of:
2. Reverse Osmosis
3.  Osmosis
4. Diffusion
5. Dissolution

Answer: (b) Osmosis

Explanation: The source states that a raw mango placed in a concentrated salt solution loses water via osmosis and shrivels. This is because the salt solution is hypertonic (has a lower water concentration) compared to the fluid inside the mango, causing water to move out of the mango.

59. The preservation of meat by salting and fruits by adding sugar protects against bacterial action because:

1. Salt and sugar are directly toxic to bacteria.
2. The salt and sugar raise the boiling point of water inside the bacteria.
3. Through osmosis, a bacterium on the salted meat or sugared fruit loses water, shrivels, and dies.
4. The salt and sugar form an impermeable layer that prevents bacteria from reaching the food.

Answer: (c) Through osmosis, a bacterium on the salted meat or sugared fruit loses water, shrivels, and dies.

Explanation: This is explained in the source: “Through the process of osmosis, a bacterium on salted meat or candid fruit loses water, shrivels and dies“.

60. A solution of 1 mol of sucrose in 1000 g of water boils at 373.52 K. What does this demonstrate?

1. Depression of freezing point
2. Relative lowering of vapour pressure
3. Osmotic pressure
4. Elevation of boiling point

Answer: (d) Elevation of boiling point

Explanation: Pure water boils at 373.15 K. The solution boils at a higher temperature (373.52 K), which is an example of the elevation of boiling point, a colligative property that occurs when a non-volatile solute is added to a solvent.

61. To calculate the molarity of a solution that is 20% (mass/mass) aqueous KI, what additional piece of information is essential?

1. The molar mass of KI.
2. The temperature of the solution.
3. The density of the solution.
4. The vapour pressure of the solution.

Answer: (c) The density of the solution.

Explanation: Molarity is defined as moles of solute per volume of solution in litres. A 20% (mass/mass) solution gives the mass of solute and solvent, allowing you to find the total mass of the solution. To convert this mass into the required volume of the solution, the density of the solution is needed (Volume = Mass/Density).

62. For a solute that undergoes association (e.g., dimerization) in a solvent, the van’t Hoff factor (i) will be:

1. i > 1
2. i < 1
3. i = 1
4. i = 0

Answer: (b) i < 1

Explanation: Association, like the dimerization of ethanoic acid, reduces the total number of moles of particles in the solution compared to the number of moles of solute initially dissolved. Since ‘i’ is the ratio of moles of particles after association to the moles before association, its value is less than unity.

63. Which of the following aqueous solutions is expected to have the highest van’t Hoff factor (i), assuming complete dissociation?

1. NaCl
2. MgSO₄
3. K₂SO₄
4. Glucose (C₆H₁₂O₆)

Answer: (c) K₂SO₄

Explanation: The van’t Hoff factor reflects the number of particles a solute dissociates into.

◦ NaCl → Na⁺ + Cl⁻ (2 particles, i=2)

◦ MgSO₄ → Mg²⁺ + SO₄²⁻ (2 particles, i=2)

◦ K₂SO₄ → 2K⁺ + SO₄²⁻ (3 particles, i=3)

◦ Glucose does not dissociate (1 particle, i=1). Therefore, K₂SO₄, which dissociates into three ions, is expected to have the highest ‘i’ value.

64. If a solute dissociates in a solution, the experimentally determined molar mass (abnormal molar mass) will be:

1. Higher than the true molar mass.
2. Lower than the true molar mass.
3. The same as the true molar mass.
4. Dependent on the solvent used.

Answer: (b) Lower than the true molar mass.

Explanation: Dissociation increases the number of particles in the solution, which leads to a larger-than-expected colligative effect (e.g., a larger ΔTf). Since molar mass is calculated as being inversely proportional to the colligative property, a larger effect results in a calculated molar mass that is lower than the true value.

65. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point was raised to 354.11 K from 353.23 K. What is the molar mass of the solute? (Kb for benzene = 2.53 K kg mol⁻¹)

1. 3 g/mol
2. 58 g/mol
3. 180 g/mol
4. 90 g/mol

Answer: (b) 58 g/mol

Explanation: This is a direct calculation from Example 1.8 in the source. First, find the elevation in boiling point: ΔTb = 354.11 K – 353.23 K = 0.88 K. Use the formula M₂ = (Kb × w₂ × 1000) / (ΔTb × w₁). M₂ = (2.53 K kg mol⁻¹ × 1.80 g × 1000) / (0.88 K × 90 g) = 4554 / 79.2 = 58 g mol⁻¹.

66. The vapour pressures of pure liquids A and B at 350 K are 450 mm Hg and 700 mm Hg, respectively. If the total vapour pressure of the mixture is 600 mm Hg, what is the mole fraction of component A in the liquid mixture?

1. 60
2. 50
3. 40
4. 30

Answer: (c) 0.40

Explanation: Let the mole fraction of A be xA. Then the mole fraction of B is (1 – xA). According to Raoult’s law for ideal solutions: P_total = pA⁰xA + pB⁰(1 – xA). 600 = 450(xA) + 700(1 – xA) 600 = 450xA + 700 – 700xA -100 = -250xA xA = 100 / 250 = 0.40.

67. Using the data from the previous question (P_total = 600 mm Hg, xA = 0.40), what is the mole fraction of component A in the vapour phase?

1. 30
2. 40
3. 50
4. 70

Answer: (a) 0.30

Explanation: The mole fraction in the vapour phase (yA) is determined by Dalton’s law: yA = pA / P_total. First, find the partial pressure of A: pA = pA⁰xA = 450 mm Hg × 0.40 = 180 mm Hg. Then, calculate the mole fraction in the vapour phase: yA = 180 mm Hg / 600 mm Hg = 0.30.

68. A solute has a normal molar mass of 122 g/mol. When dissolved in benzene, its experimentally determined molar mass is 241.98 g/mol. What is the van’t Hoff factor (i) for the solute in this solvent?

1. 0
2. 0
3. 504
4. 992

Answer: (c) 0.504

Explanation: The van’t Hoff factor is defined as the ratio of the normal molar mass to the abnormal (experimental) molar mass. i = Normal Molar Mass / Abnormal Molar Mass = 122 g/mol / 241.98 g/mol = 0.504. The value is less than 1, indicating association of the solute molecules.

69. Calculate the molality of 2.5 g of ethanoic acid (CH₃COOH, Molar Mass = 60 g/mol) in 75 g of benzene.

1. 0417 m
2. 556 m
3. 333 m
4. 80 m

Answer: (b) 0.556 m

Explanation: This calculation is shown in Example 1.3. Moles of solute = 2.5 g / 60 g/mol = 0.0417 mol. Mass of solvent in kg = 75 g / 1000 = 0.075 kg. Molality (m) = Moles of solute / Mass of solvent in kg = 0.0417 mol / 0.075 kg = 0.556 mol kg⁻¹.

70. The Henry’s law constant (Kʜ) for O₂ gas in water at 303 K is 46.82 kbar, while for N₂ it is 88.84 kbar. This indicates that at the same pressure:

1. O₂ is more soluble than N₂.
2. N₂ is more soluble than O₂.
3. O₂ and N₂ have the same solubility.
4. Solubility cannot be compared using Kʜ values.

Answer: (a) O₂ is more soluble than N₂.

Explanation: The source states that higher the value of Kʜ at a given pressure, the lower is the solubility of the gas in the liquid. Since O₂ has a lower Kʜ value (46.82 kbar) than N₂ (88.84 kbar), it is more soluble in water at this temperature.

71. Which of the following pairs will form a nearly ideal solution?

1. Phenol and aniline
2. Chloroform and acetone
3. Ethanol and acetone
4. n-hexane and n-heptane

Answer: (d) n-hexane and n-heptane

Explanation: An ideal solution is formed when the components have similar structures and intermolecular forces. n-hexane and n-heptane are both non-polar alkanes of similar size, so their interactions are very similar. The source lists this pair as an example of a nearly ideal solution. The other pairs are given as examples of non-ideal solutions showing negative (A, B) or positive (C) deviations.

72. Which statement correctly describes a solution that shows negative deviation from Raoult’s law?

1. A-B interactions are weaker than A-A and B-B interactions.
2. The total vapour pressure of the solution is higher than predicted.
3. The mixing process is endothermic (ΔmixH > 0).
4. The intermolecular attractive forces between solute (A) and solvent (B) are stronger than A-A and B-B interactions.

Answer: (d) The intermolecular attractive forces between solute (A) and solvent (B) are stronger than A-A and B-B interactions.

Explanation: The source explains that in the case of negative deviations, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B. This leads to a decrease in vapour pressure.

73. What is the mole fraction of water in a solution containing 20% of ethylene glycol (C₂H₆O₂) by mass?

1. 068
2. 322
3. 500
4. 932

Answer: (d) 0.932

Explanation: This is from Example 1.1 in the source. If the mole fraction of ethylene glycol is calculated to be 0.068, then the mole fraction of water is simply 1 minus the mole fraction of ethylene glycol, because the sum of mole fractions must be unity. Mole fraction of water = 1 – 0.068 = 0.932.

74. A 35% (v/v) solution of ethylene glycol is used as an antifreeze in cars. This concentration means that:

1. 35 g of ethylene glycol is dissolved in 65 g of water.
2. 35 mL of ethylene glycol is dissolved in water to make a total solution volume of 100 mL.
3. 35 mL of ethylene glycol is dissolved in 100 mL of water.
4. 35 g of ethylene glycol is dissolved in 100 mL of solution.

Answer: (b) 35 mL of ethylene glycol is dissolved in water to make a total solution volume of 100 mL.

Explanation: Volume percentage (V/V) is defined as the volume of the component divided by the total volume of the solution, multiplied by 100. Therefore, a 35% solution means 35 mL of the component is present in a total volume of 100 mL of solution.

1. At high altitudes, climbers may suffer from a condition called anoxia. This is explained by:

1. Raoult’s Law
2. The concept of osmosis
3. Henry’s Law
4. The law of conservation of mass

Answer: (c) Henry’s Law

Explanation: The source explains anoxia as a biological phenomenon related to Henry’s law. At high altitudes, the partial pressure of oxygen is low, which decreases the solubility of oxygen in the blood, leading to low oxygen levels in tissues.

76. The colligative property equations are modified by including the van’t Hoff factor (i) to account for:

1. Changes in temperature and pressure.
2. The volatility of the solute.
3. The association or dissociation of the solute.
4. The formation of non-ideal solutions.

Answer: (c) The association or dissociation of the solute.

Explanation:  The van’t Hoff factor, ‘i’, specifically “to account for the extent of dissociation or association” of solute particles in a solution, which causes the observed colligative properties to deviate from the calculated values.

Dr.Dash Srimanta

Hi…! Currently, I am working as an  Professor at Department of Pharmaceutical Chemistry(H.O.D),The Pharmaceutical College, Barpali, Odisha. I have more than 19 years of teaching & research experience in the field of Chemistry & Pharmaceutical sciences.