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Solutions and Colligative Properties: JEE Main 2026 PYQs

Every 2026 Session 2 question on this chapter, in one place.

Last updated: 26 October 2026

Direct answer

JEE Main 2026 asked 13 questions on Solutions and Colligative Properties across 9 shifts (Session 2).

Revise the chapter first: Solutions and Colligative Properties notes and formulas. Then solve these without looking at solutions.

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Solutions and Colligative Properties questions

Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.

2 April 2026 Shift 1, Q55 · Q55 · MCQ · Solutions · Medium

19.5 g of fluoro acetic acid (molar mass = 78 g mol$^{-1}$) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was 1°C. What is K$_a$ of fluoro acetic acid? (For water, K$_f$ = 1.86 K kg mol$^{-1}$). Assume molarity and molality to have same values.

  1. (1) $10^{-6}$
  2. (2) $4\times 10^{-4}$
  3. (3) $3\times 10^{-5}$
  4. (4) $3\times 10^{-3}$
Show answer and solution

Answer: Option (4)

Key idea: Colligative Properties and Weak Acid Dissociation

  1. Calculate moles of fluoroacetic acid: $n = \frac{19.5}{78} = 0.25$ mol
  2. Calculate molality: $m = \frac{0.25}{0.5} = 0.5$ mol/kg
  3. Find van't Hoff factor using $\Delta T_f = i \times K_f \times m$: $1 = i \times 1.86 \times 0.5$, so $i = 1.0753$
  4. For weak acid dissociation $HA \rightleftharpoons H^+ + A^-$, we have $i = 1 + \alpha$, so $\alpha = 0.0753$
  5. Calculate $K_a$ using $K_a = \frac{\alpha^2 C}{1-\alpha} = \frac{(0.0753)^2 \times 0.5}{1-0.0753} = 3 \times 10^{-3}$

2 April 2026 Shift 1, Q71 · Q71 · Numerical · Solutions · Medium

5.33 g of $\text{CrCl}_3 \cdot 6\text{H}_2\text{O}$, which is a 1 : 3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with $\text{AgNO}_3$ results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl formed is ___________ $\times 10^{-2}$. (Nearest integer) [Molar mass in g mol$^{-1}$ Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

Show answer and solution

Answer: 33

Key idea: Coordination compounds, electrolyte behavior, ion exchange, and stoichiometry

  1. Calculate molar mass of $\text{CrCl}_3 \cdot 6\text{H}_2\text{O} = 52 + 3(35.5) + 6(18) = 266.5$ g/mol
  2. Find moles of complex: $n_{\text{complex}} = \frac{5.33}{266.5} = 0.02$ mol
  3. Since it's a 1:3 electrolyte, the structure is $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ which gives 3 Cl$^-$ ions per complex
  4. Calculate moles of AgCl formed: Molar mass of AgCl $= 143.5$ g/mol, so $n_{\text{AgCl}} = \frac{8.61}{143.5} = 0.06$ mol
  5. Find the ratio: $\frac{n_{\text{complex}}}{n_{\text{AgCl}}} = \frac{0.02}{0.06} = \frac{1}{3}$
  6. Convert to required format: $\frac{1}{3} \times 10^2 = 33.33 \approx 33$

2 April 2026 Shift 2, Q55 · Q55 · MCQ · Solutions · Medium

Solution A is prepared by dissolving 1 g of a protein (molar mass = 50000 g mol$^{-1}$) in 0.5 L of water at 300 K. Its osmotic pressure is $x$ bar. Solution B is made by dissolving 2 g of same protein in 1 L of water at 300 K. Osmotic pressure of solution B is $y$ bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is $z$ bar. $x$, $y$ and $z$ respectively are : (R = 0.083 L bar mol$^{-1}$ K$^{-1}$)

  1. (1) $9.96 \times 10^{-4}$; $9.96 \times 10^{-4}$; $9.96 \times 10^{-4}$
  2. (2)
  3. (3)
  4. (4)
Show answer and solution

Answer: Option (1)

Key idea: Osmotic Pressure and Colligative Properties

  1. Use osmotic pressure formula: $\pi = \frac{n}{V}RT$ where $n$ is moles, $V$ is volume in L
  2. For Solution A: $n_A = \frac{1}{50000} = 2 \times 10^{-5}$ mol, so $x = \frac{2 \times 10^{-5}}{0.5} \times 0.083 \times 300 = 9.96 \times 10^{-4}$ bar
  3. For Solution B: $n_B = \frac{2}{50000} = 4 \times 10^{-5}$ mol, so $y = \frac{4 \times 10^{-5}}{1} \times 0.083 \times 300 = 9.96 \times 10^{-4}$ bar
  4. For mixed solution: Total moles $= 6 \times 10^{-5}$ mol, Total volume $= 1.5$ L
  5. Calculate $z = \frac{6 \times 10^{-5}}{1.5} \times 0.083 \times 300 = 9.96 \times 10^{-4}$ bar
  6. Notice that all three solutions have the same concentration (moles/volume ratio), hence $x = y = z = 9.96 \times 10^{-4}$ bar

4 April 2026 Shift 1, Q75 · Q75 · Numerical · Solutions · Medium

A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is _______. (Nearest integer) [Given: boiling point of the pure solvent = 319.5 K, $K_b$ of the solvent = 0.3 K kg mol$^{-1}$]

Show answer and solution

Answer: 5

Key idea: Raoult's Law and Boiling Point Elevation

  1. Calculate boiling point elevation: $\Delta T_b = 320 - 319.5 = 0.5$ K
  2. Use $\Delta T_b = K_b \times m$ to find molality: $0.5 = 0.3 \times \frac{n_2}{0.04}$
  3. Solve for moles of solute: $n_2 = \frac{0.5 \times 0.04}{0.3} = \frac{1}{15}$ mol
  4. Apply Raoult's law: $\frac{P_0 - P}{P_0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solution
  5. Calculate: $\frac{760-750}{760} = \frac{1}{76} = \frac{n_2}{n_1}$
  6. Find moles of solvent: $n_1 = 76 \times n_2 = 76 \times \frac{1}{15} = \frac{76}{15} \approx 5.07$
  7. Round to nearest integer: $n_1 = 5$ mol

4 April 2026 Shift 2, Q55 · Q55 · MCQ · Solutions · Medium

At 27 °C, 0.1 M, 1 L K$_4$[Fe(CN)$_6$] aqueous solution and 0.1 M, 1 L FeCl$_3$ aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is *correct*? [Figure: A container divided by semi-permeable membrane AB. Left side 'x' contains 0.1 M, 1L K$_4$[Fe(CN)$_6$], right side 'y' contains 0.1 M, 1L FeCl$_3$]

  1. (1) Blue color is formed on both sides.
  2. (2) Ionic solutes in aqueous solution can pass through semi-permeable membrane.
  3. (3) Solution on side 'y' is hypotonic.
  4. (4) To cause the reverse flow of solvent during osmosis, external pressure (any value) should be applied to side 'x'.
Show answer and solution

Answer: Option (3)

Key idea: Osmotic Pressure and Colligative Properties - Van't Hoff Factor and Tonicity

  1. Calculate van't Hoff factor for K₄[Fe(CN)₆]: dissociates into 4K⁺ + [Fe(CN)₆]⁴⁻, so $i = 5$
  2. Calculate van't Hoff factor for FeCl₃: dissociates into Fe³⁺ + 3Cl⁻, so $i = 4$
  3. Find effective particle concentration: Side x: $i \times C = 5 \times 0.1 = 0.5$ M; Side y: $i \times C = 4 \times 0.1 = 0.4$ M
  4. Compare osmotic pressures: $\pi_x = 0.5RT > \pi_y = 0.4RT$, so side x has higher osmotic pressure
  5. Identify tonicity: Side y has lower osmotic pressure than side x, making it hypotonic
  6. Verify other options: Semi-permeable membrane blocks ionic solutes (no blue color formation), and reverse osmosis requires pressure > $\Delta\pi$ (not any value)

5 April 2026 Shift 1, Q54 · Q54 · MCQ · Solutions · Easy

What is the mole fraction of water in 10% by weight (w/w) of aqueous urea solution? [Given: Molar mass of H, O, C and N are 1, 16, 12 and 14 g mol$^{-1}$ respectively.]

  1. (1) 0.825
  2. (2) 0.032
  3. (3) 0.867
  4. (4) 0.967
Show answer and solution

Answer: Option (4)

Key idea: Mole fraction calculation in solutions

  1. For 10% w/w urea solution: 10 g urea + 90 g water = 100 g solution
  2. Calculate molar mass of urea (CH₄N₂O): $M_{urea} = 12 + 4(1) + 2(14) + 16 = 60$ g/mol
  3. Calculate moles: $n_{urea} = \frac{10}{60} = \frac{1}{6}$ mol and $n_{water} = \frac{90}{18} = 5$ mol
  4. Apply mole fraction formula: $X_{water} = \frac{n_{water}}{n_{water} + n_{urea}}$
  5. $X_{water} = \frac{5}{5 + \frac{1}{6}} = \frac{5}{5.167} = 0.967$

5 April 2026 Shift 1, Q60 · Q60 · MCQ · The Solid State · Medium

Which of the following is **NOT** a physical or chemical characteristics of interstitial compounds?

  1. (1) They have high melting points, higher than those of pure metals.
  2. (2) They are very soft and ionic in nature.
  3. (3) They retain metallic conductivity.
Show answer and solution

Answer: Option (2)

Key idea: Properties of Interstitial Compounds

  1. Recall that interstitial compounds form when small atoms (H, B, C, N) occupy voids in transition metal lattices
  2. Identify key properties: (i) Very **hard** (not soft), used as abrasives (ii) High melting points, often higher than pure metals (iii) Retain metallic conductivity
  3. Analyze Option (1): High melting points higher than pure metals - This is **TRUE** (e.g., TiC melts at 3140°C vs Ti at 1668°C)
  4. Analyze Option (2): Very soft and ionic - This is **FALSE**. Interstitial compounds are very **hard** with metallic/covalent character, not ionic
  5. Analyze Option (3): Retain metallic conductivity - This is **TRUE**
  6. Since the question asks what is **NOT** a characteristic, Option (2) is the answer

5 April 2026 Shift 2, Q73 · Q73 · Numerical · Solutions · Medium

20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water. The molar mass of hemoglobin is __________ kg mol$^{-1}$. (Nearest integer) (Given : $g = 10$ m s$^{-2}$, $R = 8.3$ kPa dm$^3$ K$^{-1}$mol$^{-1}$, density of solution = 1000 kg m$^{-3}$)

Show answer and solution

Answer: 62

Key idea: Osmotic Pressure and Colligative Properties

  1. Calculate osmotic pressure from height: $\pi = \rho gh = 1000 \times 10 \times 0.08 = 800$ Pa $= 0.8$ kPa
  2. Apply van't Hoff equation: $\pi = \frac{n}{V}RT$ where $n$ is moles of hemoglobin
  3. Calculate moles: $n = \frac{\pi V}{RT} = \frac{0.8 \times 1}{8.3 \times 300} = \frac{0.8}{2490} = 3.213 \times 10^{-4}$ mol
  4. Convert mass to kg: mass $= 20$ g $= 0.02$ kg
  5. Calculate molar mass: $M = \frac{\text{mass}}{n} = \frac{0.02}{3.213 \times 10^{-4}} = 62.24$ kg mol$^{-1}$
  6. Round to nearest integer: $M = 62$ kg mol$^{-1}$

6 April 2026 Shift 1, Q56 · Q56 · MCQ · Solutions · Easy

When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $x$ % of vapor pressure of pure solvent. What is $x$ %?

  1. (1) 50%
  2. (2) 60%
  3. (3) 70%
  4. (4) 80%
Show answer and solution

Answer: Option (4)

Key idea: Raoult's Law and Relative Lowering of Vapor Pressure

  1. Given: $n_{solute} = 0.25$ mol, $n_{solvent} = 1$ mol (non-volatile, non-ionizable solute)
  2. Calculate total moles: $n_{total} = 1 + 0.25 = 1.25$ mol
  3. Find mole fraction of solvent: $X_{solvent} = \frac{1}{1.25} = 0.8$
  4. Apply Raoult's Law: $P_{solution} = P°_{solvent} \times X_{solvent}$
  5. Therefore: $P_{solution} = P°_{solvent} \times 0.8 = 80\%$ of $P°_{solvent}$

6 April 2026 Shift 2, Q55 · Q55 · MCQ · Solutions · Medium

Given below are two statements: [Figure: Table showing Chamber 1 with 18 g glucose in 100 mL aqueous solution and Chamber 2 with 30 g glucose in 250 mL aqueous solution, separated by semi-permeable membrane] **Statement I:** $H_2O$ molecules move from the chamber 1 to chamber 2. **Statement II:** The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate (molar mass = 174 g/mol) in 2 L of water (at 27 °C) is 0.0107 bar. (Given: R = 0.083 dm³ bar K⁻¹ mol⁻¹ and assume complete dissociation of electrolyte) In the light of the above statements, choose the correct answer from the options given below:

  1. (1) Both **Statement I** and **Statement II** are true
  2. (2) Both **Statement I** and **Statement II** are false
  3. (3) **Statement I** is true but **Statement II** is false
  4. (4) **Statement I** is false but **Statement II** is true
Show answer and solution

Answer: Option (4)

Key idea: Osmotic Pressure and Colligative Properties

  1. Calculate molarity of Chamber 1: $M_1 = \frac{18/180}{0.1} = 1$ M
  2. Calculate molarity of Chamber 2: $M_2 = \frac{30/180}{0.25} = 0.668$ M
  3. Since $M_1 > M_2$, water moves from Chamber 2 to Chamber 1, so Statement I is **false**
  4. For Statement II, calculate moles of K₂SO₄: $n = \frac{0.050}{174} = 2.874 \times 10^{-4}$ mol
  5. K₂SO₄ dissociates into 3 ions, so van't Hoff factor $i = 3$
  6. Calculate osmotic pressure: $\pi = iCRT = 3 \times \frac{2.874 \times 10^{-4}}{2} \times 0.083 \times 300 = 0.0107$ bar
  7. Statement II is **true**, so answer is option (4)

8 April 2026 Shift 2, Q52 · Q52 · MCQ · Solutions · Medium

Given below are two statements: Given: Molar mass of C, H, O, Cl are 12, 1, 16 and 35.5 g mol$^{-1}$, respectively Statement I: In 30% (w/w) solution of methanol in $\text{CCl}_4$(at T K), the mole fraction of $\text{CCl}_4$ is equal to 0.33. Statement II: Mixture of methanol and $\text{CCl}_4$ shows positive deviation from Raoult's law. In the light of the above statements, choose the correct answer from the options given below:

  1. (1) Both Statement I and Statement II are true
  2. (2) Both Statement I and Statement II are false
  3. (3) Statement I is true but Statement II is false
  4. (4) Statement I is false but Statement II is true
Show answer and solution

Answer: Option (1)

Key idea: Mole fraction calculation and Raoult's law deviations

  1. Calculate molar masses: Methanol $(CH_3OH) = 32$ g/mol, $CCl_4 = 154$ g/mol
  2. In 30% (w/w) solution: 30 g methanol and 70 g $CCl_4$ per 100 g solution
  3. Calculate moles: $n_{CH_3OH} = 30/32 = 0.9375$ mol, $n_{CCl_4} = 70/154 = 0.4545$ mol
  4. Mole fraction of $CCl_4 = 0.4545/(0.9375 + 0.4545) = 0.33$ ✓ Statement I is TRUE
  5. Methanol is polar (H-bonding), $CCl_4$ is non-polar. Polar-nonpolar mixtures show weaker intermolecular forces than pure components
  6. This leads to higher vapor pressure, hence positive deviation from Raoult's law ✓ Statement II is TRUE

8 April 2026 Shift 2, Q54 · Q54 · MCQ · Solutions · Medium

Which of the following statements are **not correct**? A. For water, magnitude of $K_b$ is more than the magnitude of $K_f$. B. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point. C. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers. D. The dimerised form of benzoic acid in benzene is $C_6H_5-C-OH \cdots \cdots O=C-C_6H_5$ with O double bonded to C and OH group Choose the **correct** answer from the options given below:

  1. (1) A and B only
  2. (2) A and D only
  3. (3) A, B and D only
  4. (4) A, C and D only
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Colligative Properties - Ebullioscopic and Cryoscopic Constants

8 April 2026 Shift 2, Q62 · Q62 · MCQ · Solutions · Medium

Given below are two statements: **Statement I:** Vapours of the liquid with higher boiling point condense before vapours of the liquid with lower boiling points in fractional distillation. **Statement II:** The vapours rising up in the fractionating column become richer in high boiling component of the mixture. In the light of the above statements, choose the correct answer from the options given below:

  1. (1) Both Statement I and Statement II are true
Show answer and solution

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Fractional Distillation - Vapor Composition and Condensation

FAQs

Frequently asked questions

How many questions come from Solutions and Colligative Properties in JEE Main?

In the 9 shifts of JEE Main 2026 Session 2 we found 13 questions on it, about 1.4 per shift.

Where can I practise Solutions and Colligative Properties PYQs for free?

On this page, with solutions. For more, use NTA Abhyas and our free mock tests.

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