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Class XII · Chemistry

Solutions: NEET MCQ practice

407 practice questions on Solutions, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

407

Practice questions

174

Easy

222

Medium

11

Hard

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What Solutions covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Methods of expressing concentration- molality, molarity, mole fraction, percentage
  • Vapour pressure of solutions, Raoult's law- ideal and non-ideal solutions
  • Colligative properties- lowering of vapour pressure, depression of freezing point, elevation of boiling point, osmotic pressure
  • Determination of molecular mass using colligative properties
  • Abnormal molar mass, van't Hoff factor and its significance

Sample questions on Solutions

Try each one, then open the answer and explanation.

Question 1 · Practice question

In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 10% solution(all percentages are mass/mass percentages):

  1. A.7 : 3
  2. B.5 : 7
  3. C.7 : 5
  4. D.7 : 10
Show answer

Answer: C. 7 : 5

Alligation gives the mixing ratio directly as (desired % – weaker %) : (stronger % – desired %) = (10 – 3) : (15 – 10) = 7 : 5. Option (a) 7:3 would give a higher concentration because it uses relatively more of the stronger solution; option (b) 5:7 would give a lower concentration because it uses relatively more of the weaker solution; option (d) 7:10 would be even more diluted.

Question 2 · Practice question

105 ml of pure water at 4°C saturated with NH₃ gas yielded a solution of density 0.9 g/ml and containing 30% NH by mass. Find the volume of resulting NH₃ solution.

  1. A.66.67 ml
  2. B.166.67 ml
  3. C.133.33 ml
  4. D.266.67 ml
Show answer

Answer: B. 166.67 ml

The key is using mass conservation: mass of solution = mass of water + mass NH₃, and given 30% NH₃ by mass, solve for mass NH₃ = 45 g. Then total solution mass = 150 g, divide by density 0.9 g/ml to get volume = 166.67 ml. The tempting wrong answers arise if one mistakenly uses volume percentages or ignores density.

Question 3 · Practice question

An excess of NaOH was added to 100 mL of a ferric chloride solution. This caused the precipitation of 1,425 g of Fe(OH) 3. Calculate the normality of the ferric chloride solution

  1. A.0.20 N
  2. B.0.50 N
  3. C.0.25 N
  4. D.0.40 N
Show answer

Answer: D. 0.40 N

This is a stoichiometry problem involving normality calculation. First, find moles of Fe(OH)₃ precipitated (mass/molar mass = 1.425/107 ≈ 0.0133 mol). Since 1 mole of FeCl₃ gives 1 mole Fe(OH)₃, moles of FeCl₃ = 0.0133 mol. For FeCl₃, n-factor is 3, so equivalents = 0.0133 × 3 = 0.04 eq in 100 mL (0.1 L), giving normality 0.04/0.1 = 0.4 N. Wrong options arise if n-factor is incorrectly taken as 1 (giving 0.133 N, not listed) or if volume is not converted to liters (giving 3.995 N, also not listed).

Question 4 · Practice question

Though covalent in nature, methanol is soluble in water, why?

  1. A.Methanol is transparent like water
  2. B.Due to hydrogen bonding between methanol and water molecules
  3. C.Due to van der Waals' forces between methanoland water
  4. D.Due to covalent attraction forces
Show answer

Answer: B. Due to hydrogen bonding between methanol and water molecules

Methanol (CH₃OH) has a polar -OH group that forms strong hydrogen bonds with water molecules, making it highly soluble. While van der Waals forces exist, they are not the main driving force; the key is hydrogen bonding. Covalent attraction forces are intramolecular and do not explain solubility between different molecules.

Question 5 · Practice question

What is molarity in terms of volume strength of hydrogen peroxide?

  1. A.Volume strength = 1.2 Molarity
  2. B.Volume strength = 11.2 Molarity
  3. C.Volume strength = 11 Molarity
  4. D.Volume strength = 112 Molarity
Show answer

Answer: B. Volume strength = 11.2 Molarity

The correct relation between volume strength and molarity comes from the decomposition of hydrogen peroxide: 2H2O₂ → 2H₂O + O₂. One mole of H₂O₂ in one liter (1 M) produces half a mole of O₂, which at STP occupies 11.2 liters. Hence, volume strength = 11.2 × molarity. Option (a) is too small, (c) and (d) are incorrect common mis-memories (11 instead of 11.2 or 112 instead of 11.2).

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