Solutions — NEET Chemistry MCQs with answers

Solutions is a Class 12 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 901 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Solubility, Henry and Raoult laws · Colligative properties · Non-ideal solutions and azeotropes · Non-ideal solutions · van't Hoff factor and abnormal molar mass · Freezing point depression · Colligative property calculation · Ebullioscopic and cryoscopic constants · Ideal dilute solution limit · Solutions and Colligative Properties · Osmotic pressure inverse numerical · Degree of association · van't Hoff factor bounds · Henry's law graph · Factors affecting gas solubility · Abnormal molar mass and association · Concentration terms · Elevation in boiling point · Depression in freezing point · Van't Hoff factor - modified colligative property · Van't Hoff factor - association · Azeotropes · Numerical - molar mass from colligative property · Isotonic solutions · Numerical - degree of dissociation from van't Hoff factor · Numerical - mole fraction calculation · Types of solutions - solid solutions · Numerical - osmotic pressure with van't Hoff factor · Reverse osmosis · Numerical - molarity to molality conversion · Assertion-reason on colligative properties · Numerical - percentage dissociation from colligative property · Numerical - relative lowering with unknown molar mass · Statement-based on colligative properties · Numerical - boiling point elevation with electrolyte · Numerical - normality and molarity relation · Numerical - abnormal molar mass due to association · Graph interpretation - vapour pressure vs composition · Numerical - freezing point of solution (absolute temperature) · Numerical - ppm concentration

8 free Solutions practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Colligative properties

   Which colligative property is most suitable for experimentally determining the molar mass of macromolecules such as proteins or synthetic polymers?

   • A.Osmotic pressure
   • B.Elevation of boiling point
   • C.Depression of freezing point
   • D.Relative lowering of vapour pressure
   Show answer and explanation

   Answer: A. Osmotic pressure

   Osmotic pressure produces a measurable effect even for very dilute solutions of high molar mass solutes, since Π = CRT is not as vanishingly small as ΔTf or ΔTb for such solutes.

   For a solute of very high molar mass, even a reasonable mass dissolved gives very few moles, so molality-based ΔTb and ΔTf are too small to measure accurately. Osmotic pressure Π = (n/V)RT can still be measured precisely with dilute solutions because moderate volumes and temperatures give appreciable Π values, and it can be measured at room temperature without needing extreme precision instruments. Hence osmotic pressure is the preferred method for determining molar masses of proteins and polymers.

   Common mistake: Assuming freezing point depression works equally well for all solutes regardless of molar mass

2. Question 2 · easy · Colligative properties

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

   • A.molality of the solute
   • B.mole fraction of the solute
   • C.mole fraction of the solvent
   • D.molarity of the solute
   Show answer and explanation

   Answer: B. mole fraction of the solute

   Raoult's law gives (p°−p)/p° = x2, the mole fraction of the non-volatile solute, independent of its chemical nature.

   For a solution of a non-volatile solute in a volatile solvent, Raoult's law states p = x1 p°, where x1 is the mole fraction of solvent. Since x1 + x2 = 1, p°−p = p°(1−x1) = p° x2, so (p°−p)/p° = x2, the mole fraction of solute. This ratio is the relative lowering of vapour pressure (RLVP) and is a colligative property.

   Common mistake: Mixing up RLVP with molality or molarity of the solute

3. Question 3 · easy · Colligative properties

   A colligative property of a dilute solution is one that depends primarily on:

   • A.the chemical nature of the solute particles
   • B.the chemical nature of the solvent particles alone
   • C.the number of solute particles relative to the total number of particles in solution
   • D.the molar mass of the solvent only
   Show answer and explanation

   Answer: C. the number of solute particles relative to the total number of particles in solution

   Colligative properties (RLVP, ΔTb, ΔTf, Π) depend only on the number of solute particles present relative to the total particles, not on what the solute chemically is.

   By definition, colligative properties are those properties of a solution that depend only on the number of solute particles present (i.e., their concentration/mole ratio), not on their identity, size, or chemical nature. This is why, for the same number of moles of any non-electrolyte solute in the same amount of solvent, the RLVP, boiling point elevation, freezing point depression, and osmotic pressure are identical regardless of what the solute actually is.

   Common mistake: Believing colligative properties depend on the nature/identity of the solute

4. Question 4 · easy · Colligative properties

   Common salt is spread on icy roads in winter mainly because it:

   • A.increases the boiling point of water, melting the ice
   • B.increases the osmotic pressure of the ice surface
   • C.increases the vapour pressure of water at the surface
   • D.lowers the freezing point of water so ice melts even below 0°C
   Show answer and explanation

   Answer: D. lowers the freezing point of water so ice melts even below 0°C

   Dissolved salt lowers the freezing point of water (freezing point depression), so the ice-water mixture remains liquid at temperatures below 0°C, melting the ice.

   When salt dissolves in the thin film of water on icy roads, it lowers the freezing point of that water below 0°C (ΔTf = Kf·m). This means the solution no longer freezes at the ambient sub-zero temperature and existing ice melts, since the liquid state becomes stable at a lower temperature. This is a direct practical application of freezing point depression, a colligative property.

   Common mistake: Confusing the mechanism with boiling point elevation or vapour pressure change

5. Question 5 · easy · Colligative properties

   Ethylene glycol is added to the water in car radiators in cold climates mainly to:

   • A.lower the freezing point of the coolant so it does not freeze and crack the radiator
   • B.increase the viscosity of water for better lubrication
   • C.increase the surface tension of water
   • D.increase the density of the coolant
   Show answer and explanation

   Answer: A. lower the freezing point of the coolant so it does not freeze and crack the radiator

   Ethylene glycol is a non-volatile solute that depresses the freezing point of water, preventing the coolant from freezing and expanding/cracking the radiator in winter.

   Dissolving ethylene glycol in water lowers the freezing point (ΔTf = Kf·m), so the coolant stays liquid even at sub-zero temperatures. This prevents the water from freezing, expanding, and cracking the radiator/engine block in cold climates. The same mixture also raises the boiling point somewhat, but the primary reason for use in cold climates is freezing point depression.

   Common mistake: Attributing the antifreeze effect to unrelated physical properties like viscosity or density

6. Question 6 · medium · Colligative properties

   The freezing point of pure water is 0°C and Kf for water is 1.86 K kg mol⁻¹. Calculate the freezing point of a solution containing 6 g of urea (M = 60 g mol⁻¹) dissolved in 200 g of water.

   • A.−0.47°C
   • B.−0.93°C
   • C.−1.86°C
   • D.−0.56°C
   Show answer and explanation

   Answer: B. −0.93°C

   Moles of urea = 6/60 = 0.1 mol; molality = 0.1/0.2 = 0.5 mol/kg; ΔTf = 1.86 × 0.5 = 0.93 K, giving a freezing point of −0.93°C.

   Moles of urea = 6 g / 60 g mol⁻¹ = 0.1 mol. Mass of water = 200 g = 0.200 kg, so molality m = 0.1/0.200 = 0.5 mol kg⁻¹. ΔTf = Kf × m = 1.86 × 0.5 = 0.93 K. Since the solution freezes below the pure solvent's freezing point, the new freezing point = 0°C − 0.93°C = −0.93°C.

   Common mistake: Using solvent mass in grams instead of kilograms in the molality formula

7. Question 7 · medium · Colligative properties

   Kb for water is 0.52 K kg mol⁻¹. Calculate the boiling point of a solution containing 34.2 g of sucrose (M = 342 g mol⁻¹) dissolved in 500 g of water.

   • A.100.52°C
   • B.100.20°C
   • C.100.10°C
   • D.99.90°C
   Show answer and explanation

   Answer: C. 100.10°C

   Moles of sucrose = 34.2/342 = 0.1 mol; molality = 0.1/0.5 = 0.2 mol/kg; ΔTb = 0.52 × 0.2 = 0.104 K, so boiling point ≈ 100.10°C.

   Moles of sucrose = 34.2 g / 342 g mol⁻¹ = 0.1 mol. Mass of water = 500 g = 0.500 kg, so molality m = 0.1/0.500 = 0.2 mol kg⁻¹. ΔTb = Kb × m = 0.52 × 0.2 = 0.104 K ≈ 0.10 K. Boiling point of solution = 100°C + 0.10°C = 100.10°C.

   Common mistake: Using the wrong mass basis (grams vs kilograms) or wrong molar mass for sucrose

8. Question 8 · medium · Colligative properties

   While solving a problem, a student substitutes molarity in place of molality directly into ΔTf = Kf × m without any correction. This substitution introduces negligible error only when:

   • A.the solute is always a strong electrolyte
   • B.the solution is highly concentrated
   • C.the solvent has a very high molar mass
   • D.the solution is dilute and aqueous, so its density is close to 1 g/mL
   Show answer and explanation

   Answer: D. the solution is dilute and aqueous, so its density is close to 1 g/mL

   Molarity (mol per litre of solution) and molality (mol per kg of solvent) become numerically close only when solution density ≈ 1 g/mL, which holds for dilute aqueous solutions.

   Molarity C = moles of solute / volume of solution (L), while molality m = moles of solute / mass of solvent (kg). These two coincide numerically only when 1 L of solution has a mass close to 1 kg, i.e., density ≈ 1 g/mL, and when the solute mass is small enough that mass of solution ≈ mass of solvent — both conditions typical of dilute aqueous solutions. For concentrated solutions or solvents with density far from 1 g/mL (e.g., organic solvents like benzene, density 0.88 g/mL), using molarity in place of molality introduces significant error.

   Common mistake: Treating molarity and molality as interchangeable in all solutions

Questions about Solutions for NEET

More Chemistry chapters

← Redox Reactions and ElectrochemistryAll Chemistry chaptersChemical Thermodynamics →

Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (901 active practice questions in this chapter today).