Molar Concentration Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/5/2026
The molar concentration of a solution is calculated with C = n ÷ V, where n is the amount of solute in moles and V is the volume of the solution in liters. For 0.5 mol dissolved in 2 L, the concentration is 0.25 mol/L.
Explanation
Molar concentration (or molarity), denoted C, measures the amount of solute dissolved per liter of solution. It is expressed in moles per liter (mol/L), a unit also written "M" in some textbooks (a "1 M" solution means 1 mol/L). Be careful not to confuse the volume of the final solution (the denominator of the formula) with the volume of solvent added alone: if you dissolve a solute and then top up to a given volume with water, it is that final, topped-up volume that counts, not just the water added. The amount of substance n, if not known directly, can be calculated from the mass of solute divided by its molar mass (n = m ÷ M), a calculation not covered by this tool.
Example: 0.5 mol dissolved in 2 L of solution
Inputs
Amount of substance: 0.5 mol. Volume: 2 L.
Calculation
Concentration = amount of substance ÷ volume = 0.5 ÷ 2 = 0.25.
Result
The molar concentration of this solution is 0.25 mol/L.
Frequently asked questions
What's the difference between molarity and mass concentration?
Molarity (mol/L) counts the number of particles (moles) of solute per liter, while mass concentration (g/L) counts their mass. Both measure a concentration, but molarity is directly tied to the number of molecules or ions in solution, which is more relevant for predicting a chemical reaction.
How do I calculate the amount of substance n if I only know the mass of the solute?
Divide the mass of the solute (in grams) by its molar mass (in g/mol), a value specific to each substance found on its label or in a reference table. For example, the molar mass of sodium chloride (NaCl) is about 58.44 g/mol.
What happens if I dilute a solution?
Diluting a solution (adding solvent without adding solute) increases the volume V without changing the amount of substance n, which mechanically lowers the concentration C = n ÷ V. This is the principle used to prepare a less concentrated solution from a stock solution.