How it’s calculated
Adding solvent does not change the amount of solute, so concentration × volume is the same before and after.
Example (OpenStax Chemistry 2e): 0.850 L of 5.00 M copper nitrate diluted to 1.80 L gives C2 = 0.850 × 5.00 ÷ 1.80 = 2.36 M. You add 1.80 − 0.850 = 0.95 L of water.
Concentrations can be in any unit as long as C1 and C2 match. The equation assumes volumes are additive, which is a good approximation for dilute aqueous solutions.
Frequently asked questions
How much stock do I need?
Choose "Stock volume needed (V1)": V1 = C2 × V2 ÷ C1. To make 500 mL of 0.1 M from 2 M stock you need 0.1 × 500 ÷ 2 = 25 mL, topped up to 500 mL.
What is a dilution factor?
C1 ÷ C2, or equivalently V2 ÷ V1. A 1:10 dilution reduces concentration tenfold.
Is it the same as M1V1 = M2V2?
Yes, M1V1 = M2V2 is the same equation written with molarity.
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Sources
Formulas are taken from the free public references above. Results are provided “as is” for informational and educational purposes only. See our disclaimer.
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