Molarity Formula Explained: Definition, Units and Solved Examples
Molarity is the single most used concentration unit in chemistry. It appears in Class 11 solutions, Class 12 electrochemistry and kinetics, every titration you perform in the lab, and in JEE, NEET, IIT-JAM, GATE and CSIR-NET question papers. The formula itself is one line. The marks are lost in the details — litres versus millilitres, and solution volume versus solvent volume. This guide fixes both.
Definition and formula
Molarity is the number of moles of solute dissolved in one litre of solution.
Combined: M = mass(g) ÷ [molar mass(g mol⁻¹) × V(L)]
What each symbol means
| Symbol | Meaning | Unit |
|---|---|---|
| M | Molarity (molar concentration) | mol L⁻¹, written M or mol/dm³ |
| n | Moles of solute | mol |
| V | Volume of the final solution, not of the solvent | L (litre) |
| mass | Mass of solute weighed out | g |
| molar mass | Sum of atomic masses in the formula | g mol⁻¹ |
The word "solution" in the denominator is doing real work. When you dissolve 5.85 g of NaCl and top up to the 500 mL mark of a volumetric flask, the 500 mL is solute + solvent together. You never measure 500 mL of water and add salt to it — that gives a slightly different, unknown volume.
Worked example 1 — Molarity from a weighed mass
5.85 g of NaCl is dissolved and made up to 500 mL of solution. Find the molarity.
Molar mass of NaCl = 22.990 + 35.45 = 58.44 g mol⁻¹
n = 5.85 ÷ 58.44 = 0.1001 mol
V = 500 mL = 0.500 L
M = 0.1001 ÷ 0.500 = 0.200 mol L⁻¹ (0.200 M)
Worked example 2 — How much to weigh out
This is the practical version, and the one your lab record needs.
How many grams of NaOH are needed to prepare 250 mL of 0.100 M solution?
Molar mass of NaOH = 22.990 + 15.999 + 1.008 = 39.997 ≈ 40.00 g mol⁻¹
V = 250 mL = 0.250 L
n = M × V = 0.100 × 0.250 = 0.0250 mol
mass = n × molar mass = 0.0250 × 40.00 = 1.00 g
So: weigh 1.00 g NaOH, dissolve in a little water, transfer to a 250 mL volumetric flask and make up to the mark.
Worked example 3 — Molarity of concentrated sulphuric acid
Reagent bottles give percentage by mass and density, not molarity. Converting is a standard entrance-exam question.
Concentrated H₂SO₄ is 98% by mass and has density 1.84 g mL⁻¹. Find its molarity.
Take exactly 1 L = 1000 mL of the acid.
Mass of solution = 1000 × 1.84 = 1840 g
Mass of pure H₂SO₄ = 98% of 1840 = 0.98 × 1840 = 1803.2 g
Molar mass of H₂SO₄ = 2(1.008) + 32.06 + 4(15.999) = 2.016 + 32.06 + 63.996 =
98.07 g mol⁻¹
n = 1803.2 ÷ 98.07 = 18.39 mol
This is in exactly 1 L, so M = 18.4 mol L⁻¹
That 18.4 M figure is worth memorising — it is the starting point for almost every acid dilution problem set in Indian syllabi.
Worked example 4 — Reading the question carefully
0.20 mol of glucose is dissolved to give 800 mL of solution. Find M.
V = 800 mL = 0.800 L
M = 0.20 ÷ 0.800 = 0.25 mol L⁻¹
No molar mass is needed here because moles were given directly. Students often compute the molar mass of glucose out of habit and waste ninety seconds in an exam.
Molarity vs molality — one word decides it
Molality (m) is moles of solute per kilogram of solvent, not per litre of solution.
Because volume expands with temperature but mass does not, molarity changes when the solution is heated and molality does not. That is exactly why colligative-property formulas — elevation of boiling point, depression of freezing point — are written with molality, while titration and kinetics use molarity.
Common mistakes that cost marks
- Leaving the volume in millilitres. 500 mL is 0.500 L. Forgetting to divide by 1000 makes the answer 1000 times too small. This is the single most common error.
- Using the volume of solvent instead of the volume of solution. "Dissolved in 500 mL of water" and "made up to 500 mL of solution" are different statements.
- Using mass number instead of atomic mass. Cl is 35.45, not 35; S is 32.06, not 32. The difference shows up in three-significant-figure answers.
- Confusing molarity with normality. For H₂SO₄, normality = 2 × molarity, because one mole supplies two replaceable H⁺ ions.
- Assuming molarity stays fixed with temperature. It does not; molality does.
- Rounding the molar mass too early. Carry three decimals through the working, round only at the end.
Where molarity appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Solutions chapter, preparing standard solutions in the lab record |
| CBSE/ICSE Class 12 | Electrochemistry, chemical kinetics rate expressions, colligative properties |
| JEE / NEET | Percent-to-molarity conversion, dilution, titration numericals |
| IIT-JAM / CUET-PG | Volumetric analysis, buffer preparation |
| GATE / CSIR-NET | Reaction kinetics, spectroscopy sample preparation, analytical chemistry |
Stop second-guessing your unit conversions. The Concentration & Molarity calculator works in either direction — mass to molarity, or molarity to the mass you must weigh out — and handles mL/L conversion for you.
Open the Molarity / Concentration Calculator →If solutions and stoichiometry are where your marks leak, ABC Chemistry teaches Class 11–12 chemistry at the Gurugram coaching centre and through online classes across India, with home tuition available in Delhi-NCR — see abcchemistry.in.