Definition
Concentration units
Mass percent (w/w) = (mass of solute/mass of solution)×100. Molarity (M) = moles of solute / L of solution. Molality (m) = moles of solute / kg of solvent. Mole fraction χ.
-- NCERT Class 12 Chemistry, Ch. 1, p. 4Concentration Units
8 MCQs6 revision cards9-step worked example
Source: NCERT SolutionsPYQ coverage: NEET 2022, 2025Official key: NTA-verifiedLast updated: 26 Sep 2026
Concentration Units, explained for NEET
The topic "Methods of expressing concentration" is deceptively simple — it is the gateway skill for every colligative-property calculation in NEET Chapter 5. The common failure mode is not that aspirants forget the formulas, but that they confuse the denominator: solvent vs. solution, mass vs. volume, and when each unit is appropriate.
NCERT Class 12 Chemistry Chapter 1, page 1, defines a solution as a homogeneous mixture of two or more substances. Concentration describes how much solute is present relative to the solution or solvent. NEET expects fluency in these seven concentration expressions:
Mass percentage (w/w%): (mass of solute / mass of solution) × 100. Independent of temperature.
Volume percentage (v/v%): (volume of solute / volume of solution) × 100. Temperature-dependent.
Mass by volume percentage (w/v%): (mass of solute in g / volume of solution in mL) × 100. Common in pharmacy.
Parts per million (ppm): (mass of solute / mass of solution) × 10⁶. Used for trace quantities — pollutant concentrations, dissolved oxygen.
Mole fraction (x): x₁ = n₁ / (n₁ + n₂). Dimensionless. x₁ + x₂ = 1 for a binary solution. Used in Raoult's law and vapour-pressure calculations (topics covered in separate lessons).
Molarity (M): moles of solute per litre of solution. Temperature-dependent because volume changes with temperature.
Molality (m): moles of solute per kilogram of solvent. Temperature-independent because mass does not change with temperature.
The critical distinction: Molarity uses solution volume; molality uses solvent mass. NEET questions routinely test whether you substitute the correct denominator. A second high-frequency confusion: mole fraction vs. mass fraction — these are not interchangeable and require molar-mass conversion between them.
Watch out: when a problem gives you mass of solution (not solvent), you must subtract the solute mass before computing molality.
Can you answer these Concentration Units MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following concentration terms is independent of temperature?
Show answer and why every option is right or wrong
Answer: D. Molality is defined as moles of solute per kg of solvent. Since mass does not change with temperature, molality is temperature-independent (NCERT Class 12 Chemistry Chapter 1, page 5).
Why A is wrong: A is wrong because molarity depends on solution volume, which changes with temperature.
Why B is wrong: B is wrong because normality, like molarity, is defined per litre of solution, making it temperature-dependent.
Why C is wrong: C is wrong because volume percentage depends on volume of solute and solution, both of which change with temperature.
MCQ 2Easy RecallPractice
The molality of a solution is defined as the number of moles of solute dissolved in:
Show answer and why every option is right or wrong
Answer: C. Molality = moles of solute per kilogram of solvent (not solution). This is the defining feature that distinguishes it from molarity (NCERT Class 12 Chemistry Chapter 1, page 5).
Why A is wrong: A is wrong because moles per litre of solution defines molarity, not molality.
Why B is wrong: B is wrong because the denominator for molality is mass of solvent, not mass of solution. Using solution mass would include solute mass incorrectly.
Why D is wrong: D is wrong because litres of solvent is not a standard concentration unit. Molality uses mass (kg), not volume.
MCQ 3Easy RecallPractice
For a binary solution, if the mole fraction of solute is 0.2, the mole fraction of solvent is:
Show answer and why every option is right or wrong
Answer: B. In a binary solution, x₁ + x₂ = 1. If x_solute = 0.2, then x_solvent = 1 − 0.2 = 0.8 (NCERT Class 12 Chemistry Chapter 1, page 3).
Why A is wrong: A is wrong because this assumes solvent and solute have equal mole fractions, which contradicts the given data.
Why C is wrong: C is wrong because 0.4 = 2 × 0.2, which has no basis — mole fractions of a binary solution sum to 1, not to any multiple.
Why D is wrong: D is wrong because a mole fraction cannot exceed 1 by definition (it is a ratio of parts to whole).
MCQ 4Direct ApplicationPractice
5.85 g of NaCl (molar mass = 58.5 g/mol) is dissolved in 500 g of water. The molality of the solution is:
Show answer and why every option is right or wrong
Answer: B. Moles of NaCl = 5.85 / 58.5 = 0.1 mol. Mass of solvent = 500 g = 0.5 kg. Molality = 0.1 / 0.5 = 0.2 mol/kg.
Why A is wrong: A is wrong because this uses 1 kg as the solvent mass instead of the given 0.5 kg (500 g). m = 0.1/1.0 = 0.1 is the error.
Why C is wrong: C is wrong because this likely results from dividing mass of solute (5.85) by molar mass incorrectly or using grams instead of kg in the wrong step.
Why D is wrong: D is wrong because 0.5 would result from using 0.1 mol / 0.2 kg, a denominator error — the solvent mass is 0.5 kg, not 0.2 kg.
MCQ 5Direct ApplicationPractice
A solution is prepared by dissolving 18 g of glucose (molar mass = 180 g/mol) in 178.2 g of water to make 200 mL of solution. Which pair of values is correct?
(i) Molality
(ii) Molarity
Show answer and why every option is right or wrong
Answer: D. Moles of glucose = 18/180 = 0.1 mol. Molality = 0.1 mol / 0.1782 kg solvent = 0.561 mol/kg. Molarity = 0.1 mol / 0.200 L solution = 0.50 mol/L. The denominators differ — solvent mass vs. solution volume.
Why A is wrong: A is wrong because it uses solution volume (or approximates mass as 0.2 kg) for both. Molality requires solvent mass, which is 0.1782 kg, not 0.200 kg.
Why B is wrong: B is wrong because it swaps the two values. Molality uses solvent mass (0.1782 kg → 0.561), while molarity uses solution volume (0.200 L → 0.50). Confusing the denominators reverses the answers.
Why C is wrong: C is wrong because it uses the same denominator for both, likely using solvent mass for both calculations. Molarity requires solution volume, not solvent mass.
MCQ 6Direct ApplicationPractice
A solution contains 10 g of solute in 90 g of water. The mass percentage of the solute is:
Show answer and why every option is right or wrong
Answer: A. Mass percentage = (mass of solute / mass of solution) × 100 = (10 / (10 + 90)) × 100 = 10%. The denominator is mass of solution (solute + solvent), not solvent alone.
Why B is wrong: B is wrong because 11.1% = (10/90) × 100, which uses mass of solvent as denominator instead of mass of solution. Mass percentage always uses total solution mass.
Why C is wrong: C is wrong because 9% = (90/1000) × 100 or a miscalculation. There is no standard path to this value from the given data.
Why D is wrong: D is wrong because 90% would be the mass percentage of the solvent (water), not the solute.
MCQ 7CalculationPractice
A 1.0 molal aqueous solution of urea (molar mass = 60 g/mol) is prepared. The mass of solution that contains 1 mole of urea is:
Show answer and why every option is right or wrong
Answer: A. 1.0 molal = 1 mol urea per 1 kg solvent. Mass of urea = 1 × 60 = 60 g. Total mass of solution = 1000 g (solvent) + 60 g (solute) = 1060 g. The key step: molality defines solute per kg solvent, so solution mass = solvent mass + solute mass.
Why B is wrong: B is wrong because 1000 g is the solvent mass only. The solution includes both solvent and solute: 1000 + 60 = 1060 g.
Why C is wrong: C is wrong because 940 = 1000 − 60, which subtracts solute mass from solvent mass — a meaningless operation. Solution mass is always solvent + solute.
Why D is wrong: D is wrong because 60 g is the mass of the solute alone, not the solution.
MCQ 8CalculationPractice
0.5 moles of ethanol (C₂H₅OH, molar mass = 46 g/mol) is mixed with 2.5 moles of water (molar mass = 18 g/mol). The mole fraction of ethanol and the molality of ethanol in this solution are, respectively:
Show answer and why every option is right or wrong
Answer: C. Mole fraction of ethanol = 0.5 / (0.5 + 2.5) = 0.5/3.0 = 0.167. Mass of solvent (water) = 2.5 × 18 = 45 g = 0.045 kg. Molality = 0.5 / 0.045 = 11.1 mol/kg. Two distinct calculations: mole fraction uses total moles; molality uses solvent mass in kg.
Why A is wrong: A is wrong because m = 0.20 likely results from dividing 0.5 mol by 2.5 (treating moles of water as kg). Solvent mass must be converted: 2.5 mol × 18 g/mol = 45 g = 0.045 kg.
Why B is wrong: B is wrong because x = 0.20 = 0.5/2.5, which divides by moles of solvent only instead of total moles (0.5 + 2.5 = 3.0). Mole fraction denominator is TOTAL moles.
Why D is wrong: D is wrong because both values are swapped or miscalculated. x = 0.20 uses wrong denominator; m = 0.167 has no valid calculation path from the given data.
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Concentration Units: quick recall before you leave
How do you solve a Concentration Units question? A worked example
- 1
Given
• Mass of glucose = 36 g• Molar mass of glucose = 180 g/mol (exact, defined by molecular formula)• Mass of solvent (water) = 500 g• Density of solution = 1.072 g/mL
- 2
Required
(a) Molality (m) in mol/kg
(b) Molarity (M) in mol/L - 3
Concept
Molality uses solvent mass in the denominator; molarity uses solution volume. For molarity, we need the solution volume, which requires knowing the total solution mass and its density.
- 4
Formula
• m = n / m_solvent (kg)• M = n / V_solution (L)
- 5
Substitution
• Moles of glucose: n = 36 / 180 = 0.20 mol• Mass of solvent = 500 g = 0.500 kg• Mass of solution = 500 + 36 = 536 g• Volume of solution = 536 / 1.072 = 500 mL = 0.500 L
- 6
Calculation
(a) m = 0.20 / 0.500 = 0.40 mol/kg
(b) M = 0.20 / 0.500 = 0.40 mol/L
Note: The molar mass (180 g/mol) is exact by definition of the molecular formula and does not limit significant figures. The given masses (36 g, 500 g) and density (1.072 g/mL) are the measured values that govern precision. - 7
Final answer
(a) Molality = 0.40 mol/kg
(b) Molarity = 0.40 mol/L
The numerical coincidence (both = 0.40) is specific to this problem's data. In general, molality ≠ molarity because their denominators differ. - 8
Common trap
Using solution mass (536 g) instead of solvent mass (500 g) for molality. This gives m = 0.20/0.536 = 0.373, a wrong answer that appears plausible.
- 9
Similar NEET-style question
"4.5 g of urea (molar mass = 60 g/mol) is dissolved in 250 g of water. The solution density is 1.004 g/mL. Find the molality and molarity of the solution." (Answer: m = 0.30 mol/kg; M ≈ 0.295 mol/L)
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What to remember before solving Concentration Units questions
Which Concentration Units formulas do you need for NEET?
Molality
Molal concentration: moles of solute per kg of solvent. Temperature-independent.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m | molality | mol/kg |
| n | moles solute | mol |
Valid when
- Mass of SOLVENT (not solution)
Molarity
Molar concentration: moles of solute per litre of solution.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | molarity | mol/L |
| n | moles solute | mol |
| V | solution volume | L |
Valid when
- Volume of SOLUTION not solvent
- Temperature dependent (volume changes with T)
More in Solutions: 4 exam traps and mistakes · 6 formulas · 3 question patterns from its other lessons.
Concentration Units questions from past NEET papers
2 questions from NEET 2022, 2025. Answers verified against NTA official keys.
NEET 2025
1A-III, B-II, C-I, D-IV
2A-II, B-IV, C-I,D-III
3A-II, B-I, C-IV, D-III
4A-III, B-I, C-IV, D-II
NTA Answer: Option 3(final)
NEET 2022
In one molal solution that contains 0.5 mole of a solute, there is
11000 g of solvent
2500 mL of solvent
3500 g of solvent
4100 mL of solvent
NTA Answer: Option 3(final)
Sources
NCERT refs: Class 12 Chemistry Chapter 1, p.1
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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