Molarity Calculator
With mass of solute 58.44 g, molar mass 58.44 g/mol, final volume 1 liters, or a concentration you want 0.1 mol/L and 1 more field, molarity comes to 1.0000 mol/L — molarity. It is reached in 8 steps, the last of which is 1 / 1, and each one is printed on the page with its numbers filled in. The formula is the one published by NIST Special Publication 811, not an approximation fitted to it.
Molarity from mass and volume, the mass needed for a target concentration, and the dilution that gets you there.
Formula and sources checked · How we check
Mass of solute 58.44 g, Molar mass 58.44 g/mol, Final volume 1 liters, Or a concentration you want 0.1 mol/L
1.0000 mol/L
Molarity for the example below. Editing a field recomputes the calculator below; this figure holds the answer the page was loaded with.
It is written into the HTML rather than drawn by a script, so a search engine reading this page without running JavaScript still finds an answer.
- Moles of solute
58.44 / 58.441 mol- Molarity
1 / 11 mol/L- In millimolar
1 * 10001,000 mM- Grams per liter
58.44 / 158.44 g/L- Moles needed for your target
0.1 * 0.50.05 mol- Mass to weigh out
0.05 * 58.442.922 g- Volume of this stock to dilute instead
0.05 / 1 * 100050 mL- Solvent to add to it
max(0, 0.5 * 1000 - 50)450 mL
Worked example
58.44 g of sodium chloride is exactly one mole, so in a liter it makes a 1 molar solution. To get 500 mL of 0.1 M from it, take 50 mL of the stock and add 450 mL of water — that is C₁V₁ = C₂V₂ with the numbers filled in.
How to work it out yourself
- 1.Molarity is moles per liter of final solution, not per liter of solvent. Dissolve the solid first and then make up to the mark — adding a liter of water to the solid gives a slightly different volume and therefore a slightly wrong concentration.
- 2.For a dilution, work in the two fields at the bottom: the stock volume and the solvent to add come from C₁V₁ = C₂V₂.
- 3.Always add acid to water rather than water to acid. The arithmetic is the same and the consequences are not.
The formula
- Moles of solute
58.44 / 58.44 - Molarity
1 / 1 - In millimolar
1 * 1000 - Grams per liter
58.44 / 1 - Moles needed for your target
0.1 * 0.5 - Mass to weigh out
0.05 * 58.44 - Volume of this stock to dilute instead
0.05 / 1 * 1000 - Solvent to add to it
max(0, 0.5 * 1000 - 50)
Source: NIST Special Publication 811 — Guide for the Use of the International System of Units
Questions people actually ask
- What is a mole?
- Exactly 6.02214076 × 10²³ entities, fixed by definition since the 2019 redefinition of the SI. The useful consequence is that one mole of a substance weighs its molar mass in grams — 58.44 g of sodium chloride, 18.02 g of water — which is what makes the arithmetic on this page a division rather than a table lookup.
- What is the difference between molarity and molality?
- Molarity is moles per liter of solution; molality is moles per kilogram of solvent. Molarity changes with temperature because the volume does, molality does not — which is why molality is used for freezing-point and boiling-point work and molarity for everything else.
- How do I dilute a stock solution?
- C₁V₁ = C₂V₂: the moles you take out of the stock are the moles that end up in the new solution. For 500 mL of 0.1 M from a 1 M stock, that is 50 mL of stock made up to 500 mL total — not 50 mL plus 500 mL, which is the mistake that turns a tenfold dilution into an elevenfold one.
- Why does the final volume matter rather than the water added?
- Because dissolving something changes the volume. Adding a mole of salt to a liter of water gives more than a liter of solution, so a solution made that way is under 1 M. Volumetric flasks exist precisely to make up to a known final volume rather than to measure what went in.
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