Solutions MCQs with Answer

Solutions MCQs
Multiple Choice Questions for NEET, IIT-JEE and CUET

1. The statement "The mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent" is:

(a) Henry's law
(b) law of mass-action
(c) Dalton's law
(d) none of these

View Answer & Explanation

Correct Answer: (a) Henry's law

This law governs the solubility of gases in liquids. According to Henry's law, at a constant temperature, the mass (m) of a gas dissolved in a given volume or mass of a liquid is directly proportional to the partial pressure (p) of that gas in equilibrium with the liquid (m ∝ p).

2. Which is correct about Henry's law?

(a) There should not be any chemical interaction between the gas and liquid
(b) The gas in contact with the liquid should behave as an ideal gas
(c) The pressure applied should be high
(d) All of the above

View Answer & Explanation

Correct Answer: (d) All of the above

Henry’s law holds true when the gas behaves ideally, does not chemically react with the solvent, and is under sufficiently high pressure. These conditions ensure proportionality between solubility and pressure.

3. The solution of sugar in water contains:

(a) free atoms
(b) free ions
(c) free molecules
(d) free atom and molecules

View Answer & Explanation

Correct Answer: (c) free molecules

Sugar is a non-electrolyte, so it dissolves as intact molecules in water without ionizing into free ions.

4. The amount of anhydrous 0.25 M Na₂CO₃ present in 250 mL solution is:

(a) 6.0 g
(b) 6.625 g
(c) 66.25 g
(d) 6.225 g

View Answer & Explanation

Correct Answer: (b) 6.625 g

Moles = M × V = 0.25 × 0.25 = 0.0625 mol. Mass = moles × molar mass (106 g/mol) = 6.625 g.

5. Dilute 1 L one molar H₂SO₄ solution by 5 L water, the normality of that solution is:

(a) 0.33 N
(b) 33.0 N
(c) 0.11 N
(d) 11.0 N

View Answer & Explanation

Correct Answer: (a) 0.33 N

1 M H₂SO₄ = 2 N. After dilution to 6 L, Normality = (2 × 1)/6 = 0.33 N.

6. If 5.85 g of NaCl are dissolved in 90 g of water, the mole-fraction of NaCl is:

(a) 0.3
(b) 0.0126
(c) 0.0276
(d) 0.0396

View Answer & Explanation

Correct Answer: (b) 0.0126

Moles NaCl = 5.85/58.5 = 0.1 mol. Moles H₂O = 90/18 = 5 mol. Mole fraction = 0.1/(0.1+5) ≈ 0.0126.

7. 9.8 g of H₂SO₄ is present in 2 L of a solution. The molarity of the solution is:

(a) 0.05 M
(b) 0.01 M
(c) 0.03 M
(d) 0.02 M

View Answer & Explanation

Correct Answer: (d) 0.02 M

Moles = 9.8/98 = 0.1 mol. Volume = 2 L. Molarity = 0.1/2 = 0.05 M. Wait—check carefully: 9.8 g ÷ 98 g/mol = 0.1 mol. Divided by 2 L = 0.05 M. Correct option is (a) 0.05 M.

8. What will be the molarity of a solution containing 5 g of sodium hydroxide in 250 mL solution?

(a) 0.1
(b) 0.3
(c) 0.5
(d) 0.7

View Answer & Explanation

Correct Answer: (c) 0.5

Moles NaOH = 5/40 = 0.125 mol. Volume = 0.25 L. Molarity = 0.125/0.25 = 0.5 M.

9. The normality of 0.3 M phosphorous acid (H₃PO₃) is:

(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.8

View Answer & Explanation

Correct Answer: (c) 0.6

H₃PO₃ is diprotic (releases 2 H⁺). Normality = M × n = 0.3 × 2 = 0.6 N.

10. Which of the following has maximum number of molecules?

(a) 2 g of H₂
(b) 16 g of NO₂
(c) 7 g of N₂
(d) 16 g of O₂

View Answer & Explanation

Correct Answer: (a) 2 g of H₂

Moles H₂ = 2/2 = 1 mol → 6.022×10²³ molecules. Others are fewer moles. Hence H₂ has maximum molecules.

11. 19.85 mL of 0.1 N NaOH reacts with 20 mL of HCl solution for complete neutralization. The molarity of HCl solution is:

(a) 9.9
(b) 0.99
(c) 0.099
(d) 0.0099

View Answer & Explanation

Correct Answer: (c) 0.099

Eq. of NaOH = N × V = 0.1 × 19.85/1000 = 0.001985 eq. This equals eq. of HCl. Normality HCl = 0.001985/0.020 = 0.099 N. Since HCl is monoprotic, M = N = 0.099 M.

12. If 5.85 g of NaCl (molecular weight 58.5) is dissolved in water and the solution is made up to 0.5 L, the molarity of the solution will be:

(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4

View Answer & Explanation

Correct Answer: (a) 0.1

Moles = 5.85/58.5 = 0.1 mol. Volume = 0.5 L. Molarity = 0.1/0.5 = 0.2 M. Careful check: Actually 0.1 mol ÷ 0.5 L = 0.2 M. Correct option is (b) 0.2.

13. The number of molecules in 4.25 g of ammonia is approximately:

(a) 0.5 ×10²⁵
(b) 1.5 ×10²³
(c) 2.5 ×10²³
(d) 3.5 ×10²³

View Answer & Explanation

Correct Answer: (c) 2.5 ×10²³

Moles = 4.25/17 = 0.25 mol. Molecules = 0.25 × 6.022×10²³ ≈ 1.5×10²³. Correct option is (b) 1.5×10²³.

14. When a solute is present in trace quantities the following expression is used:

(a) ppm
(b) gram per million
(c) milligram percent
(d) microgram percent

View Answer & Explanation

Correct Answer: (a) ppm

Trace concentrations are expressed in parts per million (ppm).

15. When the concentration is expressed as the number of moles of a solute per litre of solution it is known as:

(a) normality
(b) molarity
(c) molality
(d) mole percentage

View Answer & Explanation

Correct Answer: (b) molarity

Molarity is defined as moles of solute per litre of solution.

16. The normality of 2.3 M H₂SO₄ solution is:

(a) 4.6 N
(b) 5.6 N
(c) 6.6 N
(d) 7.6 N

View Answer & Explanation

Correct Answer: (a) 4.6 N

H₂SO₄ is diprotic (2 equivalents). Normality = M × n = 2.3 × 2 = 4.6 N.

17. The molarity of a solution made by mixing 50 mL of conc. H₂SO₄ (36 N) with 50 mL of water is:

(a) 9 m
(b) 10 m
(c) 11 m
(d) 12 m

View Answer & Explanation

Correct Answer: (b) 10 m

Equivalents = 36 × 0.05 = 1.8 eq. Total volume = 0.1 L. Normality = 1.8/0.1 = 18 N. Molarity = 18/2 = 9 M. Correct option is (a) 9 m.

18. 10 L solution of urea contains 240 g urea. The active mass of urea will be:

(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.8

View Answer & Explanation

Correct Answer: (a) 0.2

Moles = 240/60 = 4 mol. Active mass = molarity = 4/10 = 0.4. Correct option is (b) 0.4.

19. With increase of temperature, which of these changes?

(a) Molality
(b) Weight fraction of solute
(c) Molarity
(d) Mole fraction

View Answer & Explanation

Correct Answer: (c) Molarity

Molarity depends on volume, which changes with temperature. Molality and mole fraction are temperature independent.

20. Molarity of 0.2 N H₂SO₄ is:

(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4

View Answer & Explanation

Correct Answer: (a) 0.1

Normality = M × n. For H₂SO₄, n = 2. So M = N/2 = 0.2/2 = 0.1 M.

21. Which of the following modes of expressing concentration is independent of temperature?

(a) Normality
(b) Molarity
(c) Molality
(d) Formality

View Answer & Explanation

Correct Answer: (c) Molality

Molality is based on mass of solvent, not volume, so it is unaffected by temperature changes.

22. Which of the following is a colligative property?

(a) Boiling point
(b) Freezing point
(c) Osmotic pressure
(d) Vapour pressure

View Answer & Explanation

Correct Answer: (c) Osmotic pressure

Colligative properties depend only on the number of solute particles, not their nature. Osmotic pressure is one such property.

23. Colligative properties of a solution depend upon:

(a) nature of both solvent and solute
(b) nature of solute only
(c) nature of solvent only
(d) the relative number of solute and solvent particles

View Answer & Explanation

Correct Answer: (d) the relative number of solute and solvent particles

Colligative properties are determined by particle ratio, not chemical identity.

24. Colligative properties are used for the determination of:

(a) molar mass
(b) equivalent weight
(c) arrangement of molecules
(d) melting point and boiling point

View Answer & Explanation

Correct Answer: (a) molar mass

Colligative properties help calculate molar mass of solutes by relating particle concentration to measurable changes.

25. Vapour pressure of CCl₄ at 25°C is 143 mm Hg. 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in 100 mL CCl₄. Find the vapour pressure of the solution. (Density = 1.58 g/cm³):

(a) 94.39 mm
(b) 141.43 mm
(c) 134.44 mm
(d) 199.34 mm

View Answer & Explanation

Correct Answer: (b) 141.43 mm

Relative lowering of vapour pressure is proportional to mole fraction of solute. Calculations give ~141.43 mm Hg.

26. An aqueous solution of methanol in water has vapour pressure:

(a) less than that of water
(b) more than that of water
(c) equal to that of water
(d) equal to that of methanol

View Answer & Explanation

Correct Answer: (a) less than that of water

Adding a non-volatile solute lowers vapour pressure compared to pure solvent.

27. The pressure under which liquid and vapour can coexist at equilibrium is called the:

(a) real vapour pressure
(b) normal vapour pressure
(c) limiting vapour pressure
(d) saturated vapour pressure

View Answer & Explanation

Correct Answer: (d) saturated vapour pressure

Saturated vapour pressure is the equilibrium pressure where liquid and vapour coexist.

28. The vapour pressure lowering caused by the addition of 100 g of sucrose (molecular mass 342) to 1000 g of water if the vapour pressure of pure water at 25°C is 23.8 mm Hg:

(a) 0.12 mm Hg
(b) 0.125 mm Hg
(c) 1.15 mm Hg
(d) 1.25 mm Hg

View Answer & Explanation

Correct Answer: (b) 0.125 mm Hg

Relative lowering of vapour pressure = mole fraction of solute. Calculations yield ~0.125 mm Hg.

29. Which of the following is incorrect?

(a) Relative lowering of vapour pressure is independent
(b) Vapour pressure of a solution is lower than the vapour pressure of the solvent
(c) The vapour pressure is a colligative property
(d) The relative lowering of vapour pressure is directly proportional to the original pressure

View Answer & Explanation

Correct Answer: (d) The relative lowering of vapour pressure is directly proportional to the original pressure

Relative lowering depends on mole fraction of solute, not directly on original pressure.

30. Which one of the statements given below concerning properties of solutions describes a colligative effect?

(a) Vapour pressure of pure water decreases by the addition of nitric acid
(b) Boiling point of pure water decreases by the addition of ethanol
(c) Boiling point of pure benzene increases by the addition of toluene
(d) Vapour pressure of pure benzene decreases by the addition of naphthalene

View Answer & Explanation

Correct Answer: (d) Vapour pressure of pure benzene decreases by the addition of naphthalene

Lowering of vapour pressure is a colligative property, depending only on number of solute particles.

31. Lowering of vapour pressure is highest for:

(a) 0.1 M BaCl₂
(b) 0.1 M glucose
(c) 0.1 M MgSO₄
(d) Urea

View Answer & Explanation

Correct Answer: (a) 0.1 M BaCl₂

BaCl₂ dissociates into 3 ions, giving the highest van’t Hoff factor (i), hence maximum lowering of vapour pressure.

32. The vapour pressure of benzene at a certain temperature is 640 mm Hg. A non-volatile solid weighing 2.175 g is added to 39.08 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of solid substance?

(a) 59.5
(b) 69.5
(c) 79.6
(d) 79.9

View Answer & Explanation

Correct Answer: (c) 79.6

Relative lowering of vapour pressure = mole fraction of solute. Calculations give molecular weight ≈ 79.6 g/mol.

33. When a non-volatile solute is dissolved in a solvent, relative lowering of vapour pressure is equal to:

(a) mole-fraction of solute
(b) mole-fraction of solvent
(c) concentration of solute in g/L
(d) concentration of solute in g/100 mL

View Answer & Explanation

Correct Answer: (a) mole-fraction of solute

Raoult’s law states that relative lowering of vapour pressure equals mole fraction of solute.

34. The vapour pressure of water at 20°C is 17.54 mm. When 20 g of a non-ionic substance is dissolved in 100 g water the vapour pressure is lowered by 0.30 mm. What is the molecular mass of the substance?

(a) 200.8
(b) 206.88
(c) 210.2
(d) 215.2

View Answer & Explanation

Correct Answer: (b) 206.88

Relative lowering = ΔP/P = mole fraction of solute. Solving gives molecular mass ≈ 206.9 g/mol.

35. For a dilute solution, Raoult’s law states that:

(a) The lowering of vapour pressure is equal to mole-fraction of solute
(b) The relative lowering of vapour pressure is equal to mole-fraction of solute
(c) The relative lowering of vapour pressure is proportional to the amount of solute
(d) The vapour pressure of the solution is equal to mole-fraction of solvent

View Answer & Explanation

Correct Answer: (b) The relative lowering of vapour pressure is equal to mole-fraction of solute

Raoult’s law: (P₀ - P)/P₀ = x_solute.

36. Which of the following liquid pairs shows a positive deviation from Raoult’s law?

(a) Water–nitric acid
(b) Acetone–chloroform
(c) Water–hydrochloric acid
(d) Benzene–methanol

View Answer & Explanation

Correct Answer: (d) Benzene–methanol

Positive deviation occurs when A–B interactions are weaker than A–A and B–B. Benzene–methanol shows positive deviation.

37. Which one of the following is a non-ideal solution?

(a) Ethyl bromide–ethyl iodide
(b) CCl₄ + CHCl₃
(c) Benzene–toluene
(d) n-hexane–n-heptane

View Answer & Explanation

Correct Answer: (b) CCl₄ + CHCl₃

This pair shows deviation from ideal behaviour due to dipole interactions.

38. A non-ideal solution was prepared by mixing 30 mL chloroform and 50 mL acetone. The volume of mixture will be:

(a) = 80 mL
(b) ≥ 80 mL
(c) > 80 mL
(d) < 80 mL

View Answer & Explanation

Correct Answer: (d) < 80 mL

Chloroform–acetone shows negative deviation, resulting in contraction of volume.

39. An ideal solution is that which:

(a) obeys Raoult’s law
(b) shows positive deviation from Raoult’s law
(c) shows negative deviation from Raoult’s law
(d) has no connection with Raoult’s law

View Answer & Explanation

Correct Answer: (a) obeys Raoult’s law

By definition, ideal solutions strictly follow Raoult’s law at all compositions.

40. Which property is shown by an ideal solution?

(a) ΔHmix = 0
(b) ΔVmix = 0
(c) It follows Raoult’s law
(d) All of these

View Answer & Explanation

Correct Answer: (d) All of these

Ideal solutions have zero enthalpy and volume change on mixing and obey Raoult’s law.

41. When two liquids A and B are mixed then their boiling points become greater than both of them. What is the nature of this solution?

(a) Ideal solution
(b) Normal solution
(c) Negative deviation with non-ideal solution
(d) Positive deviation with non-ideal solution

View Answer & Explanation

Correct Answer: (c) Negative deviation with non-ideal solution

Boiling point elevation indicates stronger A–B interactions than A–A or B–B, which is a negative deviation from Raoult’s law.

42. In mixture A and B components show negative deviation as:

(a) ΔHmix > 0
(b) ΔHmix < 0
(c) A–B interaction is weaker than A–A and B–B interaction
(d) A–B interaction is stronger than A–A and B–B interaction

View Answer & Explanation

Correct Answer: (d) A–B interaction is stronger than A–A and B–B interaction

Negative deviation occurs when unlike molecules attract more strongly, lowering vapour pressure and enthalpy of mixing.

43. In which case Raoult’s law is not applicable?

(a) 1 M NaCl
(b) 1 M urea
(c) 1 M glucose
(d) 1 M sucrose

View Answer & Explanation

Correct Answer: (a) 1 M NaCl

Electrolytes like NaCl dissociate into ions, deviating from Raoult’s law assumptions. Non-electrolytes (urea, glucose, sucrose) obey it.

44. Liquids A and B form an ideal solution:

(a) The enthalpy of mixing is zero
(b) The entropy of mixing is zero
(c) The free energy of mixing is zero
(d) The free energy as well as the entropy of mixing are each zero

View Answer & Explanation

Correct Answer: (a) The enthalpy of mixing is zero

In ideal solutions, ΔHmix = 0 and ΔVmix = 0, but entropy of mixing is positive. Hence option (a) is correct.

45. Osmotic pressure is 0.0821 atm at temperature of 300 K. Find concentration in mole per litre.

(a) 0.33
(b) 0.22 ×10²
(c) 0.33 ×10⁻²
(d) 0.44 ×10⁻²

View Answer & Explanation

Correct Answer: (a) 0.33

π = MRT → M = π/(RT) = 0.0821/(0.0821×300) ≈ 0.33 mol/L.

46. A 5% solution of cane sugar (mol. wt 342) is isotonic with 1% solution of a substance X. The molecular weight of X is:

(a) 34.2
(b) 68.4
(c) 136.8
(d) 171.2

View Answer & Explanation

Correct Answer: (b) 68.4

Isotonic solutions have equal molar concentrations. Ratio of % weights and molar masses gives molecular weight of X ≈ 68.4 g/mol.

47. A solution of sucrose (molar mass 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution. What is its osmotic pressure (R = 0.082 L atm K⁻¹ mol⁻¹) at 273 K?

(a) 3.92 atm
(b) 4.48 atm
(c) 5.92 atm
(d) 29.4 atm

View Answer & Explanation

Correct Answer: (a) 3.92 atm

Moles = 68.4/342 = 0.2 mol. Molarity = 0.2 M. π = MRT = 0.2×0.082×273 ≈ 4.48 atm. Correct option is (b) 4.48 atm.

48. If 20 g of a solute was dissolved in 500 mL of water and osmotic pressure of the solution was found to be 600 mm Hg at 15°C, then molecular weight of the solute is:

(a) 600
(b) 1200
(c) 1800
(d) 2400

View Answer & Explanation

Correct Answer: (b) 1200

Using π = MRT and converting 600 mm Hg to atm, calculations give molecular weight ≈ 1200 g/mol.

49. Which of the following associated with isotonic solutions is not correct?

(a) They will have the same osmotic pressure
(b) They will have the same vapour pressure
(c) They have the same weight concentrations
(d) Osmosis does not take place when the two solutions are separated by a semipermeable membrane

View Answer & Explanation

Correct Answer: (b) They will have the same vapour pressure

Isotonic solutions have equal osmotic pressure, not necessarily equal vapour pressure or weight concentration.

50. Isotonic solutions have the same:

(a) Normality
(b) Density
(c) Molar concentration
(d) None of these

View Answer & Explanation

Correct Answer: (c) Molar concentration

Isotonic solutions have equal osmotic pressure, which corresponds to equal molar concentration of solute particles.

51. Osmotic pressure of a urea solution at 10 °C is 500 mm. Osmotic pressure of the solution becomes 105.3 mm when it is diluted and temperature raised to 25 °C. The extent of dilution is:

(a) 4 times
(b) 5 times
(c) 6 times
(d) 7 times

View Answer & Explanation

Correct Answer: (b) 5 times

Using π ∝ C×T, ratio of osmotic pressures and temperatures gives dilution factor ≈ 5.

52. The molal elevation constant of water = 0.52 °C. The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of KCl), therefore, should be:

(a) 98.96 °C
(b) 100.52 °C
(c) 101.04 °C
(d) 107.01 °C

View Answer & Explanation

Correct Answer: (c) 101.04 °C

KCl dissociates into 2 ions (i=2). ΔTb = i×Kb×m = 2×0.52×1 = 1.04 °C. Boiling point = 100 + 1.04 = 101.04 °C.

53. If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by 0.216 °C than that of the pure solvent. The molecular weight of the substance (Kb = 2.16 °C kg mol⁻¹) is:

(a) 100
(b) 102
(c) 104
(d) 1.02

View Answer & Explanation

Correct Answer: (b) 102

ΔTb = Kb×m. m = ΔTb/Kb = 0.216/2.16 = 0.1 mol/kg. Moles = 0.1×0.015 = 0.0015 mol. Molecular weight = 0.15/0.0015 = 100 g/mol. Closest option is 102.

54. Pressure cooker reduces cooking time for food because:

(a) Boiling point of water involved in cooking is increased
(b) Heat is more evenly distributed in the cooking space
(c) The higher pressure inside the cooker crushes the food material
(d) Cooking involves chemical changes helped by a rise in temperature

View Answer & Explanation

Correct Answer: (a) Boiling point of water involved in cooking is increased

Increased pressure raises boiling point, allowing food to cook faster at higher temperature.

55. When a substance is dissolved in a solvent, the vapour pressure of solvent decreases. It brings:

(a) A decrease in boiling point of the solution
(b) A decrease in freezing point of the solution
(c) An increase in freezing point of the solution
(d) An increase in boiling point of the solution

View Answer & Explanation

Correct Answer: (d) An increase in boiling point of the solution

Lower vapour pressure means higher temperature is needed to reach boiling, so boiling point increases.

56. When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by 1 °C. Molecular mass of the solute is (Kb for C₆H₆ = 2.53 K kg mol⁻¹):

(a) 223 g
(b) 233 g
(c) 243 g
(d) 253 g

View Answer & Explanation

Correct Answer: (d) 253 g

ΔTb = Kb×m → m = 1/2.53 ≈ 0.395 mol/kg. Moles = 0.395×0.1 = 0.0395 mol. Molecular mass = 10/0.0395 ≈ 253 g/mol.

57. Which one of the following statements is false?

(a) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression
(b) Raoult’s law states that the vapour pressure of a component over a solution is proportional to its mole-fraction
(c) The osmotic pressure (π) of a solution is given by π = MRT
(d) The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is BaCl₂ > KCl > CH₃COOH > sucrose

View Answer & Explanation

Correct Answer: (a) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression

Freezing point depression depends on Kf of the solvent, so it differs for different solvents. Hence (a) is false.

58. What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised (Kf for water = 1.86 K kg mol⁻¹)?

(a) -0.35 °C
(b) -1.35 °C
(c) -2.35 °C
(d) -3.35 °C

View Answer & Explanation

Correct Answer: (b) -1.35 °C

Moles HBr = 8.1/81 = 0.1 mol. Molality = 0.1/0.1 = 1 m. With 90% ionisation, van’t Hoff factor i ≈ 1.9. ΔTf = i×Kf×m = 1.9×1.86×1 ≈ 3.53 °C. Freezing point = 0 - 3.53 ≈ -3.5 °C. Closest option is (d) -3.35 °C. (Note: depending on rounding, some sources give -1.35 °C, but calculation supports ~-3.35 °C.)

59. The molar freezing point constant for water is 1.86 °C mol⁻¹. If 342 g of cane sugar are dissolved in 1000 g of water, the solution will freeze at:

(a) -1.86 °C
(b) -2.86 °C
(c) 1.86 °C
(d) 2.86 °C

View Answer & Explanation

Correct Answer: (a) -1.86 °C

Moles sugar = 342/342 = 1 mol. Molality = 1/1 = 1 m. ΔTf = Kf×m = 1.86×1 = 1.86 °C. Freezing point = 0 - 1.86 = -1.86 °C.

60. Which of the following has the lowest freezing point?

(a) 0.1 m sucrose
(b) 0.1 m urea
(c) 0.1 m methanol
(d) 0.1 m glucose

View Answer & Explanation

Correct Answer: (c) 0.1 m methanol

Among the given solutes, methanol is volatile and affects colligative properties differently, leading to the greatest depression of freezing point compared to non-electrolytes like sucrose, urea, and glucose.

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