Molarity Calculator

Molarity (M) is the most widely used unit of concentration in chemistry, defined as the number of moles of solute dissolved per litre of solution. This calculator works with the core relationship M = n / V — moles of solute divided by volume of solution in litres — and lets you solve for whichever quantity you need. Pick a mode (find molarity, moles, or volume), enter the two values you know, and the result appears instantly with the formula shown step by step.

Concentration calculations are a daily task in any chemistry setting, so this tool is built for school and college students working through stoichiometry and solution problems, for lab technicians and researchers preparing reagents and buffers, and for teachers checking answers. Knowing the molarity of a solution tells you exactly how much reactive material is present in a given volume, which is what makes reactions reproducible and titrations accurate.

The formula and its rearrangements

The defining equation is M = n / V, where M is molarity in mol/L (also written "M" for molar), n is the amount of solute in moles, and V is the volume of the solutionin litres. Rearranging the same equation gives the other two forms this calculator uses:

• n = M × V — moles of solute = molarity × volume
• V = n / M — volume of solution = moles ÷ molarity

When you start from a mass rather than moles, first convert: moles = mass (g) ÷ molar mass (g/mol). Substituting that into the definition gives the practical chained form M = (mass ÷ molar mass) ÷ volume, which is how most bench calculations actually begin.

Worked example

Suppose you dissolve 36 g of glucose (C₆H₁₂O₆, molar mass 180.16 g/mol) and make the solution up to 2 litres. First find the moles: 36 ÷ 180.16 = 0.1998 mol, about 0.2 mol. Then apply the formula: M = n / V = 0.2 ÷ 2 = 0.1 mol/L. The glucose solution is therefore 0.1 M. To go the other way, if you needed 0.5 mol of solute at 0.1 M, the volume would be V = n / M = 0.5 ÷ 0.1 = 5 litres.

Common reference points

A few everyday concentrations help build intuition: seawater is roughly 0.6 M in NaCl, blood plasma sits near 0.15 M NaCl (the basis of "normal saline"), stomach acid is about 0.1 M HCl, and a 1 M NaOH solution is a strong base routinely used in laboratories. Very dilute solutions in biology and medicine are quoted in millimolar (mM = 0.001 M) or micromolar (μM = 0.000001 M).

Why molarity matters

Chemistry happens between particles, and the mole is how chemists count particles in bulk — one mole is 6.022 × 10²³ entities, Avogadro's number. Because reactions consume reactants in fixed mole ratios, expressing concentration in moles per litre lets you scale a recipe up or down and still combine reactants in exactly the right proportions. This is what makes molarity central to stoichiometry: from a balanced equation and the molarities and volumes of your solutions, you can predict how much product forms or which reactant runs out first.

Molarity is equally important in titration and analytical work. In an acid–base titration you measure the volume of a solution of known molarity needed to neutralise a sample, then back-calculate the unknown concentration from the mole ratio at the equivalence point. The same thinking applies when diluting a stock solution, mixing a buffer to a target pH, or preparing a standard for instrument calibration — in every case, knowing moles per litre is what turns a vague "some solute in some water" into a precise, repeatable quantity.

Tips and common mistakes

• Molarity uses the volume of the final solution, not the volume of solvent you started with. To make 1 L of solution, add solute to a flask and top up to the 1 L mark — do not add 1 L of water to the solute.
• Keep volume in litres. If you measured in millilitres, divide by 1000 first (500 mL = 0.5 L).
• Molarity is temperature-dependent because the solution's volume expands or contracts with temperature; molality (moles per kilogram of solvent) does not, which is why colligative-property problems prefer molality.
• Molarity (M) is not the same as normality (N). Normality counts reactive equivalents, so for a diprotic acid like H₂SO₄, 1 M equals 2 N.
• To dilute a stock solution, use M₁V₁ = M₂V₂ rather than recomputing from scratch.