JEE Chemistry Solutions MCQs Set 35

Practice MCQs for JEE Chemistry Solutions

Access targeted multiple-choice questions for Solutions designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Solutions Questions and Solutions

Access the complete set of multiple-choice questions for Solutions below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: 18 g of glucose (C6H12O6) is added to 178.29 g of water. The vapour pressure of water for this aqueous solution at 100°C is:
(a) 759 torr
(b) 7.60 torr
(c) 76 torr
(d) 752.4 torr
Show Answer & Explanation

Answer: (d) 752.4 torr

Explanation:
1. Moles: glucose = 0.1; water = 178.29/18 = 9.905.
2. xglucose = 0.1/10.005 = 0.01.
3. Ps = 760 × (1 − 0.01) = 752.4 torr.
4. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 2: If the solubility of vinyl chloride (g) is 0.09 M at STP (1 bar), then the value of Henry's constant will be:
(a) 0.0016 bar
(b) 617.284 bar
(c) 6.17 × 10−2 bar
(d) 308.642 bar
Show Answer & Explanation

Answer: (b) 617.284 bar

Explanation:
1. Mole fraction x = 0.09/55.55 = 1.62 × 10−3 (1 L of water is 55.55 mol).
2. KH = p/x = 1/(1.62 × 10−3) = 617 bar.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 3: Which of the following statements is incorrect?
(a) in a solution of volatile liquids, the partial pressure of any component is directly proportional to its mole fraction
(b) components of an azeotropic mixture can be separated in pure form using fractional distillation
(c) total vapour pressure of a solution of binary volatile liquids decreases on increasing the mole fraction of the less volatile component
(d) at a given pressure, the higher the value of KH, the lower is the solubility of the gas in the liquid
Show Answer & Explanation

Answer: (b) components of an azeotropic mixture can be separated in pure form using fractional distillation

Explanation:
1. An azeotrope boils at constant composition, so fractional distillation cannot separate its components.
2. The other statements are correct.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 4: What will be the osmotic pressure of a 0.03 N solution of aluminium sulphate at 27°C, if the salt dissociates 90% in solution?
(a) 0.566 atm
(b) 0.677 atm
(c) 3.399 atm
(d) 4.065 atm
Show Answer & Explanation

Answer: (a) 0.566 atm

Explanation:
1. Molarity = 0.03/6 = 0.005 M.
2. i = 1 + 4(0.9) = 4.6.
3. π = iCRT = 4.6 × 0.005 × 0.0821 × 300 = 0.566 atm.
4. Answer: (a)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 5: What will be the amount of ice separated on cooling a solution of 40 g ethylene glycol in 400 g of water up to −9.3°C? (Kf for water = 1.86 K kg mol−1)
(a) 177.78 g
(b) 129.03 g
(c) 222.22 g
(d) 270.97 g
Show Answer & Explanation

Answer: (d) 270.97 g

Explanation:
1. At −9.3°C, molality = 9.3/1.86 = 5 m.
2. Glycol = 40/62 = 0.645 mol, so water remaining = 0.645/5 = 0.129 kg = 129.03 g.
3. Ice = 400 − 129.03 = 270.97 g.
4. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 6: Which one of the following solutions has the maximum vapour pressure at 27°C?
(a) 1 M Na2SO4
(b) 1 M AlCl3
(c) 1 M KBr
(d) 1 M MgCl2
Show Answer & Explanation

Answer: (c) 1 M KBr

Explanation:
1. Vapour pressure is highest where the particle concentration is lowest.
2. KBr gives 2 ions, against 3, 4 and 3 for the others.
3. Answer: (c)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 7: At 298 K, 1000 cm3 of a solution containing 4.34 g of solute shows an osmotic pressure of 2.55 atm. What is the molar mass of the solute? (R = 0.0821 L atm K−1 mol−1)
(a) 41.64 g mol−1
(b) 82.73 g mol−1
(c) 58.31 g mol−1
(d) 91.65 g mol−1
Show Answer & Explanation

Answer: (a) 41.64 g mol−1

Explanation:
1. M = wRT/(πV) = 4.34 × 0.0821 × 298/(2.55 × 1).
2. = 41.64 g mol−1.
3. Answer: (a)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 8: Which of the following will have the maximum depression in freezing point?
(a) 0.5 M Li2SO4
(b) 1 M KCl
(c) 0.5 M BaCl2
(d) 1 M Al2(SO4)3
Show Answer & Explanation

Answer: (d) 1 M Al2(SO4)3

Explanation:
1. Compare i × concentration: 1.5, 2, 1.5 and 5.
2. 1 M Al2(SO4)3 has the largest value.
3. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 9: At 40°C the vapour pressures of pure benzene and toluene are 160 mm Hg and 60 mm Hg respectively. If equimolar amounts of the liquids are mixed to form an ideal solution, the vapour pressure of the solution will be:
(a) 140 mm Hg
(b) 110 mm Hg
(c) 220 mm Hg
(d) 100 mm Hg
Show Answer & Explanation

Answer: (b) 110 mm Hg

Explanation:
1. P = xBP°B + xTP°T = 0.5 × 160 + 0.5 × 60.
2. = 110 mm Hg.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 10: A binary liquid solution is made by mixing n-heptane and ethanol. Which statement is correct regarding the behaviour of the solution?
(a) the solution formed is an ideal solution
(b) the solution is non-ideal, showing positive deviation from Raoult's law
(c) the solution is non-ideal, showing negative deviation from Raoult's law
(d) n-heptane shows positive deviation while ethanol shows negative deviation from Raoult's law
Show Answer & Explanation

Answer: (b) the solution is non-ideal, showing positive deviation from Raoult's law

Explanation:
1. Ethanol is hydrogen-bonded; heptane breaks these bonds and cannot replace them.
2. Interactions weaken, so the solution shows positive deviation.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 11: What is the molar mass of a non-ionising solid if 10 g of it dissolved in 100 g of water forms a solution which froze at −1.22°C? (Kf = 1.86 K kg mol−1)
(a) 265 g/mol
(b) 152 g/mol
(c) 130 g/mol
(d) 65 g/mol
Show Answer & Explanation

Answer: (b) 152 g/mol

Explanation:
1. M = Kf × w × 1000/(ΔTf × W) = 1.86 × 10 × 1000/(1.22 × 100).
2. = 152 g/mol.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 12: At 293 K, the vapour pressure of pure benzene is 75 mm Hg and that of pure toluene is 22 mm Hg. The vapour pressure of the solution containing 20 mol% benzene and 80 mol% toluene is:
(a) 32.6 mm Hg
(b) 64.4 mm Hg
(c) 97 mm Hg
(d) 3.26 mm Hg
Show Answer & Explanation

Answer: (a) 32.6 mm Hg

Explanation:
1. P = 0.2 × 75 + 0.8 × 22 = 15 + 17.6.
2. = 32.6 mm Hg.
3. Answer: (a)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 13: What will be the normality of a "20 V H2O2" solution?
(a) 1.78 N
(b) 3.56 N
(c) 10 N
(d) 0.28 N
Show Answer & Explanation

Answer: (b) 3.56 N

Explanation:
1. Normality = volume strength/5.6.
2. = 20/5.6 = 3.57 N ≈ 3.56 N.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 14: Which of the following solutions has the highest boiling point? (Assume all salts completely dissociate.)
(a) 0.1 M Al2(SO4)3
(b) 0.1 M BaCl2
(c) 0.1 M glucose
(d) 0.1 M AlCl3
Show Answer & Explanation

Answer: (a) 0.1 M Al2(SO4)3

Explanation:
1. Boiling point elevation depends on the particle concentration.
2. Al2(SO4)3 gives 5 ions per formula unit (0.5 M), the most.
3. Answer: (a)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 15: What will be the freezing point (in °C) of a solution obtained by dissolving 0.1 g of potassium ferricyanide (molar mass 329) in 100 g of water? Kf = 1.86 K kg mol−1 (assume complete dissociation)
(a) −2.3 × 10−2
(b) −5.7 × 10−2
(c) −5.7 × 10−3
(d) −1.2 × 10−2
Show Answer & Explanation

Answer: (a) −2.3 × 10−2

Explanation:
1. K3[Fe(CN)6] gives 4 ions, so i = 4. m = (0.1/329)/0.1 = 3.04 × 10−3.
2. ΔTf = 4 × 1.86 × 3.04 × 10−3 = 0.0226, so Tf = −2.3 × 10−2 °C.
3. Answer: (a)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 16: A solution of a substance containing 1.05 g per 100 mL was found to be isotonic with a 3% (w/v) glucose solution. The molecular mass of the substance is:
(a) 31.5
(b) 6.3
(c) 630
(d) 63
Show Answer & Explanation

Answer: (d) 63

Explanation:
1. Isotonic: 3/180 = 1.05/M.
2. M = 1.05 × 180/3 = 63.
3. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 17: The vapour pressure of benzene at 90°C is 1020 torr. A solution of 5 g of a solute in 58.5 g of benzene has a vapour pressure of 990 torr. The molar mass of the solute is:
(a) 78.2
(b) 178.2
(c) 200
(d) 220
Show Answer & Explanation

Answer: (d) 220

Explanation:
1. Relative lowering = 30/1020 = 0.0294 = n/(n + 0.75), where benzene = 58.5/78 = 0.75 mol.
2. n = 0.0227 mol, so M = 5/0.0227 = 220.
3. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 18: Which of the following has the minimum freezing point?
(a) 0.1 M K2Cr2O7
(b) 0.1 M NH4Cl
(c) 0.1 M BaSO4
(d) 0.1 M Al2(SO4)3
Show Answer & Explanation

Answer: (d) 0.1 M Al2(SO4)3

Explanation:
1. The minimum freezing point goes with the maximum number of particles.
2. Al2(SO4)3 gives 5 ions; K2Cr2O7 3, NH4Cl 2, and BaSO4 is practically insoluble.
3. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 19: What is the correct sequence of osmotic pressure of 0.01 M aqueous solutions of (a) Al2(SO4)3 (π1), (b) Na3PO4 (π2), (c) BaCl2 (π3), (d) glucose (π4)?
(a) π4 > π2 > π3 > π1
(b) π3 > π4 > π2 > π1
(c) π3 > π4 > π1 > π2
(d) π1 > π2 > π3 > π4
Show Answer & Explanation

Answer: (d) π1 > π2 > π3 > π4

Explanation:
1. Osmotic pressure follows the number of particles: Al2(SO4)3 = 5, Na3PO4 = 4, BaCl2 = 3, glucose = 1.
2. So π1 > π2 > π3 > π4.
3. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 20: A solution contains a non-volatile solute of molecular mass M2. Which of the following can be used to calculate the molecular mass of the solute in terms of osmotic pressure?
(a) M2 = (m2/π)VRT
(b) M2 = (m2/V)(RT/π)
(c) M2 = (m2/V)πRT
(d) M2 = (m2/V)(π/RT)
Show Answer & Explanation

Answer: (b) M2 = (m2/V)(RT/π)

Explanation:
1. πV = (m2/M2)RT.
2. M2 = m2RT/(πV) = (m2/V)(RT/π).
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 21: Normal saline solution is:
(a) 1 N salt solution
(b) 0.9% mass/volume NaCl solution
(c) 1 N NaCl solution
(d) normal salt solution
Show Answer & Explanation

Answer: (b) 0.9% mass/volume NaCl solution

Explanation:
1. Normal saline is isotonic with blood.
2. It is 0.9% (w/v) NaCl in water.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 22: A certain substance 'A' tetramerises in water to the extent of 80%. A solution of 2.5 g of A in 100 g of water lowers the freezing point by 0.3°C. The molar mass of A is (Kf = 1.86):
(a) 122
(b) 31
(c) 244
(d) 62
Show Answer & Explanation

Answer: (d) 62

Explanation:
1. 4A ⇌ A4: i = 1 − (3/4)α = 1 − 0.6 = 0.4.
2. ΔTf = i Kf (w/M)(1000/W): 0.3 = 0.4 × 1.86 × (2.5/M) × 10, so M = 62.
3. Answer: (d)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 23: The compound PdCl4·6H2O is a hydrated complex. A 1 molal aqueous solution of it has a freezing point of 269.28 K. Assuming 100% ionisation of the complex, find its molecular formula (Kf for water = 1.86 K kg mol−1, freezing point of water = 273 K).
(a) [Pd(H2O)6]Cl4
(b) [Pd(H2O)4Cl2]Cl2·2H2O
(c) [Pd(H2O)3Cl3]Cl·3H2O
(d) [Pd(H2O)2Cl4]·4H2O
Show Answer & Explanation

Answer: (c) [Pd(H2O)3Cl3]Cl·3H2O

Explanation:
1. ΔTf = 273 − 269.28 = 3.72 K, so i = 3.72/1.86 = 2.
2. Two ions form only when one Cl− is outside the sphere: [Pd(H2O)3Cl3]Cl·3H2O.
3. Answer: (c)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Question 24: A decimolar solution of potassium ferricyanide, K3[Fe(CN)6], has an osmotic pressure of 3.94 atm at 27°C. Hence the percent ionisation of the solute is:
(a) 10%
(b) 20%
(c) 30%
(d) 40%
Show Answer & Explanation

Answer: (b) 20%

Explanation:
1. i = π/(CRT) = 3.94/(0.1 × 0.0821 × 300) = 1.6.
2. i = 1 + 3α gives α = 0.2, i.e. 20%.
3. Answer: (b)

Teacher's Note:
1. Write the colligative-property relation first and decide the van't Hoff factor. Check the answer against the options.

Free study material for Chemistry

Solutions Objective Questions & Solutions for JEE Chemistry

Download Multiple Choice Questions: Solutions (JEE Chemistry)

Review structured objective questions for JEE Chemistry Solutions. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Concept Clarification for Solutions

Built using the official NCERT book for JEE, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Additional Study Resources for JEE Chemistry

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Chemistry Solutions MCQs Set 35?

You can get most exhaustive JEE Chemistry Solutions MCQs Set 35 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Solutions MCQs Set 35 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Solutions MCQs Set 35, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Solutions MCQs Set 35?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Solutions MCQs Set 35 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.