JEE Chemistry Solutions MCQs Set 29

Multiple Choice Questions (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.

Practice Solutions MCQs for JEE Chemistry

View or download the dedicated Solutions MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: The molarity of a solution of sodium chloride (molar mass 58.5 g mol−1) in water containing 5.85 g of sodium chloride in 500 mL of solution is:
(a) 0.25
(b) 2.0
(c) 1.0
(d) 0.2
Show Answer & Explanation

Answer: (d) 0.2

Explanation:
1. Moles = 5.85/58.5 = 0.1.
2. Molarity = 0.1/0.5 L = 0.2 M.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 2: 1000 g of an aqueous solution of CaCO3 contains 10 g of calcium carbonate. The concentration of the solution is:
(a) 10 ppm
(b) 100 ppm
(c) 1000 ppm
(d) 10,000 ppm
Show Answer & Explanation

Answer: (d) 10,000 ppm

Explanation:
1. ppm = (mass of solute/mass of solution) × 106 = (10/1000) × 106 = 10,000 ppm.
2. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 3: The normality of a 0.3 M phosphorous acid (H3PO3) solution is:
(a) 0.15
(b) 0.6
(c) 0.9
(d) 0.1
Show Answer & Explanation

Answer: (b) 0.6

Explanation:
1. H3PO3 is dibasic (only two replaceable H), so n-factor = 2.
2. N = 0.3 × 2 = 0.6.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 4: The molarity of 720 g of pure water is:
(a) 40 M
(b) 4 M
(c) 55.5 M
(d) Cannot be determined
Show Answer & Explanation

Answer: (c) 55.5 M

Explanation:
1. Pure water has 1000 g L−1 (density 1 g mL−1), i.e. 1000/18 = 55.5 mol L−1.
2. The amount taken does not change the molarity.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 5: Equal weights of NaCl and KCl are dissolved separately in equal volumes of solutions. The molarity of the two solutions will be:
(a) Equal
(b) That of NaCl will be less than that of KCl
(c) That of NaCl will be more than that of KCl
(d) That of NaCl will be half of that of KCl
Show Answer & Explanation

Answer: (c) That of NaCl will be more than that of KCl

Explanation:
1. Moles = mass/molar mass; NaCl (58.5) has the lower molar mass than KCl (74.5).
2. So for the same mass NaCl has more moles and a higher molarity.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 6: The mole fraction of oxygen in a mixture of 7 g of nitrogen and 8 g of oxygen is:
(a) 8/15
(b) 0.5
(c) 0.25
(d) 1.0
Show Answer & Explanation

Answer: (b) 0.5

Explanation:
1. n(N2) = 7/28 = 0.25; n(O2) = 8/32 = 0.25.
2. x(O2) = 0.25/0.5 = 0.5.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 7: The mole fraction of ethanol in an ethanol-water mixture is 0.25. The percentage concentration of ethanol by weight is:
(a) 25%
(b) 75%
(c) 46%
(d) 54%
Show Answer & Explanation

Answer: (c) 46%

Explanation:
1. Per mole of mixture: 0.25 × 46 = 11.5 g ethanol and 0.75 × 18 = 13.5 g water.
2. Weight % = 11.5/25 × 100 = 46%.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 8: The normality of 10 mL of a '20 V' H2O2 solution is:
(a) 1.79
(b) 3.58
(c) 60.86
(d) 6.086
Show Answer & Explanation

Answer: (b) 3.58

Explanation:
1. 20 V means 1 mL releases 20 mL O2 at STP.
2. N = volume strength/5.6 = 20/5.6 = 3.57 ≈ 3.58.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 9: H2O2 solution used for hair bleaching is sold as a solution of approximately 5.0 g H2O2 per 100 mL. The molecular mass of H2O2 is 34. The molarity of this solution is approximately:
(a) 0.15 M
(b) 1.5 M
(c) 3.0 M
(d) 3.4 M
Show Answer & Explanation

Answer: (b) 1.5 M

Explanation:
1. Per litre: 50 g.
2. M = 50/34 = 1.47 ≈ 1.5 M.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 10: Two bottles A and B contain 1 M and 1 m aqueous solutions of sulphuric acid respectively (density ≈ 1 g mL−1). Then:
(a) A is more concentrated than B
(b) B is more concentrated than A
(c) Concentration of A = concentration of B
(d) It is not possible to compare the concentration
Show Answer & Explanation

Answer: (a) A is more concentrated than B

Explanation:
1. 1 M has 98 g of H2SO4 in 1 L of solution, which has less than 1000 g of water, so its molality is above 1.
2. 1 m has 98 g in 1000 g water, whose volume is above 1 L, so its molarity is below 1.
3. Hence A is more concentrated.
4. Answer: (a)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 11: For a dilute aqueous solution (density about 1 g mL−1), the molar concentration of the solution is:
(a) Always equal to normality
(b) More than the molality of the solution
(c) Equal to the molality of the solution
(d) Less than the molality of the solution
Show Answer & Explanation

Answer: (d) Less than the molality of the solution

Explanation:
1. M = 1000 m d/(1000 + m M2), where M2 is the molar mass of the solute.
2. With d near 1, the denominator exceeds 1000, so M is slightly smaller than m.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 12: The Henry's law constant for dissolution of CH4 in benzene at 298 K is 2 × 105 mm Hg. The solubility (mole fraction) of CH4 in benzene at 298 K under 760 mm Hg is:
(a) 1.2 × 10−5
(b) 3.8 × 10−3
(c) 4 × 10−7
(d) 1 × 10−2
Show Answer & Explanation

Answer: (b) 3.8 × 10−3

Explanation:
1. p = KH x, so x = 760/(2 × 105) = 3.8 × 10−3.
2. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 13: 1 mol of heptane (V.P. = 92 mm Hg) was mixed with 4 mol of octane (V.P. = 31 mm Hg). The vapour pressure of the resulting ideal solution is:
(a) 46.2 mm Hg
(b) 40.0 mm Hg
(c) 43.2 mm Hg
(d) 38.4 mm Hg
Show Answer & Explanation

Answer: (c) 43.2 mm Hg

Explanation:
1. x(heptane) = 0.2, x(octane) = 0.8.
2. P = 0.2 × 92 + 0.8 × 31 = 18.4 + 24.8 = 43.2 mm Hg.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 14: If P0A and P0B are 108 and 36 torr respectively, the mole fraction of A in the vapour phase, if B has mole fraction 0.5 in solution, is:
(a) 0.25
(b) 0.75
(c) 0.60
(d) 0.35
Show Answer & Explanation

Answer: (b) 0.75

Explanation:
1. PA = 0.5 × 108 = 54; PB = 0.5 × 36 = 18.
2. yA = 54/72 = 0.75.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 15: Which condition is NOT satisfied by an ideal solution?
(a) ΔHmix = 0
(b) ΔVmix = 0
(c) ΔSmix = 0
(d) Obeyance of Raoult's law
Show Answer & Explanation

Answer: (c) ΔSmix = 0

Explanation:
1. An ideal solution has ΔH = 0 and ΔV = 0 and obeys Raoult's law.
2. Mixing always increases randomness, so ΔSmix is positive, not zero.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 16: Which of the following pairs does NOT form an ideal solution?
(a) C6H6 and C6H5CH3
(b) C2H5Cl and C2H5OH
(c) C6H5Cl and C6H5Br
(d) C2H5Br and C2H5I
Show Answer & Explanation

Answer: (b) C2H5Cl and C2H5OH

Explanation:
1. Ideal pairs have similar molecular structure and polarity.
2. Ethanol is hydrogen bonded while chloroethane is not, so they deviate from ideality.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 17: An azeotropic mixture of two liquids has a boiling point lower than either of them when it:
(a) shows a positive deviation from Raoult's law
(b) shows no deviation from Raoult's law
(c) shows a positive deviation from Henry's law
(d) shows a negative deviation from Henry's law
Show Answer & Explanation

Answer: (a) shows a positive deviation from Raoult's law

Explanation:
1. Positive deviation raises the vapour pressure, so the boiling point falls below that of both components (minimum boiling azeotrope).
2. Answer: (a)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 18: Which is NOT a colligative property?
(a) Osmotic pressure
(b) Relative lowering of vapour pressure
(c) Depression in freezing point
(d) Refractive index
Show Answer & Explanation

Answer: (d) Refractive index

Explanation:
1. Colligative properties depend only on the number of solute particles.
2. Refractive index depends on the nature of the substance.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 19: The vapour pressure of a solution having a solid solute and a liquid solvent is:
(a) Directly proportional to the mole fraction of the solvent
(b) Inversely proportional to the mole fraction of the solvent
(c) Directly proportional to the mole fraction of the solute
(d) Inversely proportional to the mole fraction of the solute
Show Answer & Explanation

Answer: (a) Directly proportional to the mole fraction of the solvent

Explanation:
1. Raoult's law: Ps = xsolvent P0.
2. Answer: (a)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 20: The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is:
(a) 23.9 mm Hg
(b) 24.2 mm Hg
(c) 21.42 mm Hg
(d) 31.44 mm Hg
Show Answer & Explanation

Answer: (c) 21.42 mm Hg

Explanation:
1. Mole fraction of water = 0.9.
2. P = 0.9 × 23.8 = 21.42 mm Hg.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 21: The vapour pressure of pure A is 10 torr and at the same temperature when 1 g of B is dissolved in 20 g of A, its vapour pressure is reduced to 9.0 torr. If the molecular mass of A is 200 amu, the molecular mass of B is:
(a) 100 amu
(b) 90 amu
(c) 75 amu
(d) 120 amu
Show Answer & Explanation

Answer: (b) 90 amu

Explanation:
1. Relative lowering = 1/10 = xB, and nA = 20/200 = 0.1 mol.
2. nB/(nB + 0.1) = 0.1 gives nB = 0.0111 mol.
3. MB = 1/0.0111 = 90.
4. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 22: The boiling points of C6H6, CH3OH, C6H5NH2 and C6H5NO2 are 80°C, 65°C, 184°C and 212°C respectively. Which will show the highest vapour pressure at room temperature?
(a) C6H6
(b) CH3OH
(c) C6H5NH2
(d) C6H5NO2
Show Answer & Explanation

Answer: (b) CH3OH

Explanation:
1. The liquid with the lowest boiling point is the most volatile.
2. CH3OH (65°C) has the highest vapour pressure.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 23: The vapour pressure of a pure liquid A is 70 torr at 27°C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and the total vapour pressure of the solution is 84 torr at 27°C. The vapour pressure of pure liquid B at 27°C is:
(a) 14
(b) 56
(c) 140
(d) 70
Show Answer & Explanation

Answer: (c) 140

Explanation:
1. 84 = 0.8 × 70 + 0.2 PB0 = 56 + 0.2 PB0.
2. PB0 = 28/0.2 = 140 torr.
3. Answer: (c)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 24: The molal elevation constant is the ratio of the elevation in boiling point to:
(a) Molarity
(b) Molality
(c) Mole fraction of solute
(d) Mole fraction of solvent
Show Answer & Explanation

Answer: (b) Molality

Explanation:
1. ΔTb = Kb m, so Kb = ΔTb/m (molality).
2. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 25: An aqueous solution containing 1 g of urea boils at 100.25°C. An aqueous solution containing 3 g of glucose in the same volume will boil at:
(a) 100.75°C
(b) 100.5°C
(c) 100°C
(d) 100.25°C
Show Answer & Explanation

Answer: (d) 100.25°C

Explanation:
1. Moles of urea = 1/60 = 0.0167; moles of glucose = 3/180 = 0.0167.
2. Equal moles in the same volume of water give the same elevation, so 100.25°C.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 26: An aqueous solution freezes at −0.186°C (Kf = 1.86 K kg mol−1, Kb = 0.512 K kg mol−1). The elevation in boiling point is:
(a) 0.186
(b) 0.512
(c) 0.512/1.86
(d) 0.0512
Show Answer & Explanation

Answer: (d) 0.0512

Explanation:
1. m = 0.186/1.86 = 0.1.
2. ΔTb = 0.512 × 0.1 = 0.0512.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 27: Camphor is used as the solvent to determine the molecular weight of a non-volatile solute by the Rast method because camphor:
(a) Has a high molal depression constant
(b) Has a high melting point
(c) Is cheap
(d) All of these
Show Answer & Explanation

Answer: (a) Has a high molal depression constant

Explanation:
1. Camphor has an unusually large Kf (about 40 K kg mol−1), which gives a measurable depression even for small amounts of solute.
2. Its high melting point and cost are not the reason.
3. Answer: (a)

Teacher's Note:
1. Rast method uses molten camphor.
2. Large Kf amplifies the measured depression.

Question 28: In a depression of freezing point experiment, it is found that:
(a) The vapour pressure of the solution is more than that of pure H2O
(b) The vapour pressure of the solution is more than that of the pure solute
(c) Only solute molecules solidify at the freezing point
(d) Only solvent molecules solidify at the freezing point
Show Answer & Explanation

Answer: (d) Only solvent molecules solidify at the freezing point

Explanation:
1. At the freezing point pure solvent separates as solid ice.
2. The solution has a lower vapour pressure than the pure solvent.
3. Answer: (d)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

Question 29: In osmosis, the net flow of solvent molecules takes place from:
(a) A solution of higher solute concentration to one of lower solute concentration
(b) A solution of lower solute concentration to one of higher solute concentration
(c) Neither; only solute molecules move across the membrane
(d) The solution to the pure solvent
Show Answer & Explanation

Answer: (b) A solution of lower solute concentration to one of higher solute concentration

Explanation:
1. A semipermeable membrane allows only solvent molecules through.
2. Net flow is towards the side with higher solute concentration, which has the lower solvent concentration.
3. Answer: (b)

Teacher's Note:
1. Write the formula first and substitute carefully.
2. Check units before comparing options.

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