Molarity
M = mol ÷ L
Uses M = n/V with volume in litres. Methodology & sources →
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Molarity is the most common way to express concentration: moles of solute per litre of solution.
Volume must be in litres, so divide millilitres by 1000 first. If you only know the mass of solute, convert to moles with n = m ÷ M, where M is the molar mass — use the grams option above and the calculator does it for you. For step-by-step examples, see How to Approach Stoichiometry or the Molar Mass Calculator.
Worked Example — Molarity from Mass, and Moles in a Sample
Question: 18.3 g of KNO₃ is dissolved in enough water to make 250.0 mL of solution. Find the molarity, then find how many moles are in a 35.0 mL portion of it.
Step 1 — molar mass of KNO₃: M = 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
Step 2 — moles of KNO₃: n = 18.3 ÷ 101.11 = 0.1810 mol
Step 3 — molarity: M = 0.1810 mol ÷ 0.2500 L = 0.724 M
Step 4 — moles in 35.0 mL: n = M × V = 0.724 mol/L × 0.0350 L = 0.0253 mol
Answer: 0.724 M, containing 0.0253 mol in a 35.0 mL portion.
Common Mistakes
- Volume in mL instead of L. Molarity is defined per litre. 250.0 mL is 0.2500 L — divide by 1000 before dividing.
- Mass confused with moles. Grams must be converted to moles with the molar mass first; molarity is never grams per litre.
- Molarity vs. molality. Molarity (M, mol/L) is based on solution volume; molality (m, mol/kg) is based on solvent mass. They are not interchangeable, and only molality is temperature-independent.
Frequently Asked Questions
Molarity equals moles of solute divided by litres of solution: M = n / V. Rearranged: n = M × V and V = n / M.
Divide the mass in grams by the molar mass in g/mol: n = m / M. Switch the amount toggle to grams and enter the solute formula — the calculator looks up the molar mass and converts automatically.
Molarity (M) is moles of solute per litre of solution. Molality (m) is moles of solute per kilogram of solvent. Molarity changes slightly with temperature because volume expands or contracts; molality does not, because mass is unaffected by temperature.
Yes, slightly. As temperature rises, the solution volume expands a little, so the same amount of solute is now in a larger volume, giving a marginally lower molarity. The effect is small for dilute aqueous solutions but matters for precise volumetric work.