The Molar Concentration Formula gives the number of moles of solute dissolved per litre of solution, expressed as \( C = n/V \), and is a foundational concept in NCERT Class 11 Chemistry (Chapter 1 — Some Basic Concepts of Chemistry). This formula is equally critical for NEET and JEE Main aspirants, where solution chemistry problems appear every year. In this article, you will find the complete derivation, a comprehensive formula sheet, three progressively difficult solved examples, CBSE exam tips for 2025-26, common mistakes to avoid, and JEE/NEET application patterns.
Key Molar Concentration Formulas at a Glance
Quick reference for the most important molarity and concentration formulas.
- Molarity: \( C = \dfrac{n}{V} \)
- Moles from mass: \( n = \dfrac{w}{M} \)
- Combined formula: \( C = \dfrac{w}{M \times V} \)
- Dilution law: \( C_1 V_1 = C_2 V_2 \)
- Molality: \( m = \dfrac{n_{\text{solute}}}{W_{\text{solvent}}(\text{kg})} \)
- Mole fraction: \( \chi_A = \dfrac{n_A}{n_A + n_B} \)
- Parts per million: \( \text{ppm} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^6 \)
What is Molar Concentration Formula?
The Molar Concentration Formula defines how concentrated a solution is. Specifically, it measures the number of moles of a solute present in exactly one litre of solution. Chemists also call this quantity molarity, and it is denoted by the symbol C or M.
According to NCERT Class 11 Chemistry, Chapter 1 (Some Basic Concepts of Chemistry), molarity is the most widely used unit of concentration in laboratory work. It allows chemists to compare solution strengths and perform stoichiometric calculations with precision.
The SI unit of molar concentration is mol L−1 (moles per litre), also written as M (molar). A 1 M solution contains exactly 1 mole of solute dissolved in 1 litre of solution.
Molar concentration is temperature-dependent. This is because volume changes with temperature, while mass does not. Students must keep this distinction in mind when comparing molarity with molality.
CBSE Class 11 and Class 12 both test this concept in board examinations. NEET and JEE Main include at least one direct molarity calculation or dilution problem in every session. Understanding the Molar Concentration Formula thoroughly gives students a significant scoring advantage.
Molar Concentration Formula — Expression and Variables
The standard expression for molar concentration (molarity) is:
\[ C = \frac{n}{V} \]
When mass of solute and molar mass are given instead of moles, the formula expands to:
\[ C = \frac{w}{M \times V} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| C | Molar Concentration (Molarity) | mol L−1 (M) |
| n | Number of moles of solute | mol |
| V | Volume of solution | Litre (L) |
| w | Mass of solute | gram (g) |
| M | Molar mass of solute | g mol−1 |
Derivation of the Molar Concentration Formula
The derivation follows directly from the definition of a mole and the concept of concentration.
Step 1: By definition, concentration = amount of solute ÷ volume of solution.
Step 2: The “amount” of a substance is measured in moles. So, \( C = n / V \).
Step 3: The number of moles is related to mass by \( n = w / M \), where w is the mass in grams and M is the molar mass in g mol−1.
Step 4: Substituting Step 3 into Step 2 gives the expanded formula:
\[ C = \frac{w}{M \times V} \]
This derivation is directly aligned with NCERT Class 11 Chemistry, Chapter 1, and is frequently asked as a short-answer question in CBSE board exams.
Complete Chemistry Concentration Formula Sheet
The table below covers all major concentration-related formulas from NCERT Class 11 and Class 12 Chemistry. Use this as your revision sheet before board exams and competitive tests.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Molarity (Molar Concentration) | \( C = n/V \) | n = moles, V = volume (L) | mol L−1 | Class 11, Ch 1 |
| Molarity from mass | \( C = w/(M \times V) \) | w = mass (g), M = molar mass (g/mol), V = volume (L) | mol L−1 | Class 11, Ch 1 |
| Dilution Law | \( C_1 V_1 = C_2 V_2 \) | C = molarity, V = volume before/after dilution | mol L−1, L | Class 11, Ch 1 |
| Molality | \( m = n_{\text{solute}} / W_{\text{solvent}} \) | n = moles of solute, W = mass of solvent (kg) | mol kg−1 | Class 12, Ch 2 |
| Mole Fraction (solute) | \( \chi_A = n_A / (n_A + n_B) \) | n_A = moles of solute, n_B = moles of solvent | Dimensionless | Class 12, Ch 2 |
| Mass Percentage (w/w) | \( \% = (w_{\text{solute}}/w_{\text{solution}}) \times 100 \) | w = mass in grams | % | Class 11, Ch 1 |
| Volume Percentage (v/v) | \( \% = (V_{\text{solute}}/V_{\text{solution}}) \times 100 \) | V = volume in mL or L | % | Class 11, Ch 1 |
| Parts Per Million (ppm) | \( \text{ppm} = (m_{\text{solute}}/m_{\text{solution}}) \times 10^6 \) | m = mass in same units | ppm | Class 11, Ch 1 |
| Normality | \( N = \text{equivalents} / V \) | equivalents = n × n-factor, V = volume (L) | eq L−1 | Class 11, Ch 1 |
| Molarity–Molality Relation | \( m = \frac{1000 \times C}{1000 \times d – C \times M} \) | d = density (g/mL), M = molar mass, C = molarity | mol kg−1 | Class 12, Ch 2 |
Molar Concentration Formula — Solved Examples
The following three examples progress from Class 9–10 level to JEE/NEET difficulty. Work through each one carefully to build confidence.
Example 1 (Class 9–10 Level — Direct Application)
Problem: Calculate the molarity of a solution prepared by dissolving 4 g of sodium hydroxide (NaOH) in enough water to make 500 mL of solution. (Molar mass of NaOH = 40 g mol−1)
Given:
- Mass of NaOH, w = 4 g
- Molar mass of NaOH, M = 40 g mol−1
- Volume of solution, V = 500 mL = 0.5 L
Step 1: Calculate moles of NaOH using \( n = w/M \).
\[ n = \frac{4}{40} = 0.1 \text{ mol} \]
Step 2: Apply the Molar Concentration Formula \( C = n/V \).
\[ C = \frac{0.1}{0.5} = 0.2 \text{ mol L}^{-1} \]
Answer
The molarity of the NaOH solution = 0.2 M
Example 2 (Class 11–12 Level — Dilution Problem)
Problem: A student has 200 mL of a 5 M hydrochloric acid (HCl) solution. She dilutes it with distilled water to make a final volume of 1000 mL. What is the molarity of the diluted solution?
Given:
- Initial molarity, C₁ = 5 M
- Initial volume, V₁ = 200 mL
- Final volume, V₂ = 1000 mL
- Final molarity, C₂ = ?
Step 1: Apply the dilution law \( C_1 V_1 = C_2 V_2 \).
\[ 5 \times 200 = C_2 \times 1000 \]
Step 2: Solve for C₂.
\[ C_2 = \frac{5 \times 200}{1000} = \frac{1000}{1000} = 1 \text{ M} \]
Step 3: Verify by checking moles. Initial moles = 5 × 0.2 = 1 mol. Final moles = 1 × 1 = 1 mol. Moles are conserved. The answer is consistent.
Answer
The molarity of the diluted HCl solution = 1 M
Example 3 (JEE/NEET Level — Density-Based Concentration)
Problem: A sulphuric acid (H₂SO₄) solution has a mass percentage of 49% and a density of 1.38 g mL−1. Calculate the molarity of this solution. (Molar mass of H₂SO₄ = 98 g mol−1)
Given:
- Mass percentage = 49%
- Density, d = 1.38 g mL−1
- Molar mass, M = 98 g mol−1
Step 1: Assume 1 litre (1000 mL) of solution as the basis.
Step 2: Calculate the mass of 1 litre of solution using density.
\[ \text{Mass of solution} = 1000 \times 1.38 = 1380 \text{ g} \]
Step 3: Calculate the mass of H₂SO₄ in this solution.
\[ \text{Mass of H}_2\text{SO}_4 = \frac{49}{100} \times 1380 = 676.2 \text{ g} \]
Step 4: Calculate moles of H₂SO₄.
\[ n = \frac{676.2}{98} = 6.9 \text{ mol} \]
Step 5: Apply the Molar Concentration Formula. Volume = 1 L (our assumed basis).
\[ C = \frac{n}{V} = \frac{6.9}{1} = 6.9 \text{ mol L}^{-1} \]
Answer
The molarity of the H₂SO₄ solution ≈ 6.9 M
CBSE Exam Tips 2025-26
- Always convert volume to litres before substituting into the Molar Concentration Formula. The most common error is using mL instead of L, which gives answers 1000 times too large.
- Memorise the dilution law \( C_1 V_1 = C_2 V_2 \) separately. CBSE Class 11 and Class 12 board papers frequently include a 2-mark dilution question. It is easy marks if you remember this relation.
- For density-based problems, always assume 1 litre of solution as your basis. This simplifies the calculation and avoids confusion. We recommend practising at least five such problems before the exam.
- Write units at every step in your answer. CBSE examiners award marks for correct units. Missing the unit “mol L−1” in your final answer can cost you half a mark.
- Distinguish molarity from molality. Molarity uses volume of solution (in L); molality uses mass of solvent (in kg). In CBSE 2025-26 papers, a definition question on this distinction appears almost every year.
- Practise NCERT Exercise questions from Chapter 1 (Class 11) and Chapter 2 (Class 12) thoroughly. Our experts suggest solving all in-text and back-exercise problems at least twice.
Common Mistakes to Avoid
Students lose marks on molar concentration problems due to a small set of recurring errors. Recognising these mistakes in advance saves you valuable exam time.
- Mistake 1 — Using mL instead of L: The Molar Concentration Formula requires volume in litres. Always divide mL values by 1000 before substituting. For example, 250 mL = 0.25 L, not 250 L.
- Mistake 2 — Confusing moles of solute with moles of solution: Molarity counts moles of the dissolved substance (solute) only. Do not include moles of solvent in your numerator.
- Mistake 3 — Using mass of solvent instead of volume of solution: The denominator in the molarity formula is the total volume of the solution, not the volume or mass of the solvent alone. This is a very common error in density-based problems.
- Mistake 4 — Incorrect molar mass calculation: Always double-check the molar mass by adding atomic masses carefully. For example, the molar mass of Na₂CO₃ is (2 × 23) + 12 + (3 × 16) = 106 g mol−1, not 84 g mol−1.
- Mistake 5 — Applying dilution law when solute is added: The formula \( C_1 V_1 = C_2 V_2 \) applies only when a solution is diluted with pure solvent. If more solute is added, you must recalculate moles from scratch.
JEE/NEET Application of Molar Concentration Formula
In our experience, JEE aspirants encounter the Molar Concentration Formula in at least two to three questions per session. NEET also tests this concept regularly in the Physical Chemistry section. Understanding the patterns below will help you score confidently.
Pattern 1 — Density and Mass Percentage to Molarity
JEE Main frequently provides the density (g/mL) and mass percentage of a solution and asks for its molarity. The approach is always the same: assume 1 L of solution, find total mass using density, extract solute mass using percentage, convert to moles, and apply \( C = n/V \). This is the most tested pattern.
Pattern 2 — Mixing Two Solutions of Different Concentrations
A classic JEE problem mixes V₁ litres of a C₁ molar solution with V₂ litres of a C₂ molar solution of the same solute. The final molarity is:
\[ C_{\text{mix}} = \frac{C_1 V_1 + C_2 V_2}{V_1 + V_2} \]
This formula comes directly from the conservation of moles. NEET also uses this pattern in titration-based questions.
Pattern 3 — Molarity–Molality Interconversion
JEE Advanced sometimes asks students to convert molarity to molality or vice versa when density is given. The relation is:
\[ m = \frac{1000 \times C}{1000 \times d – C \times M} \]
Here, d is the density of the solution in g/mL, C is molarity, M is molar mass of solute, and m is molality. Deriving this relation from first principles is a common 4-mark JEE Advanced question. We recommend practising this derivation until it becomes automatic.
FAQs on Molar Concentration Formula
Explore More Chemistry and Physics Formulas
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- Visit our Complete Physics Formulas hub for a full collection of NCERT-aligned formula sheets.
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- Revise the Flow Rate Formula for fluid mechanics problems that appear in JEE Main and Class 11 Physics.
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For official NCERT textbook content, refer to the NCERT official website to download the latest Class 11 and Class 12 Chemistry PDFs.