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Diluting 1 M diluted 10x

The exact resulting concentration when diluting 1 M diluted 10x, using C1V1=C2V2. Adjust any field below to try your own numbers.

Calculation Type

Mass & Volume

The mass of solute (the substance being dissolved), in grams.
g
The mass of one mole of the substance, in grams per mole — found on the periodic table (for elements) or by summing atomic masses (for compounds).
g/mol

Moles & Volume

The amount of solute in moles — use this directly if you already know the mole count instead of the mass.
mol
The total volume of the final solution (solute plus solvent), in liters — used by both the mass-based and moles-based calculation.
L

Dilution (C1V1 = C2V2)

Fill in any 3 of the 4 fields below and leave the one you want solved for blank.

M
M

V1 and V2 can be in any consistent unit (mL, L) — just use the same one for both.

Solved For

C1

1 M

V1

10

C2

0.1 M

V2

100

Result

C1V1 = C2V2: 1 × 10 = 0.1 × 100 — solving for Final Concentration (C2) gives the value shown above.

Resulting concentration (M)

What is Molarity?

Molarity is a measure of solution concentration — the amount of a substance (the solute) dissolved per unit volume of the total solution, expressed in moles per liter (mol/L), written with the symbol M. It's the standard concentration unit throughout chemistry, from classroom titrations to pharmaceutical dosing to industrial process chemistry, because it directly ties a measurable volume to a mole-based reaction quantity.

This calculator handles the two most common molarity tasks: finding a solution's molarity from either its moles (or mass and molar mass) and volume, and solving a dilution using the C1V1 = C2V2 relationship — the standard way to figure out how to dilute a concentrated stock solution down to a working concentration.

Resulting Concentration at Different Dilution Factors

Holding the initial concentration at 1 M fixed, here's the resulting concentration across a range of common dilution factors.

Dilution Factor Resulting Concentration
2x 0.5 M
5x 0.2 M
10x 0.1 M
20x 0.05 M
50x 0.02 M
100x 0.01 M

Molarity Formulas

Molarity (M) = Moles of Solute (mol) ÷ Volume of Solution (L)
Moles = Mass (g) ÷ Molar Mass (g/mol)
Dilution: C1 × V1 = C2 × V2

Molarity vs. Molality vs. Normality

These three concentration units are easy to confuse. Molarity (M) is moles of solute per liter of solution — the total volume after mixing, which this calculator uses. Molality (m) is moles of solute per kilogram of solvent only, which doesn't change with temperature the way molarity does (since volume expands slightly with heat but mass doesn't) — useful in precise physical chemistry work. Normality (N) accounts for a substance's reactive capacity (like how many H+ ions an acid can donate), so it can differ from molarity by a whole-number multiple depending on the reaction.

Why C1V1 = C2V2 Works

Diluting a solution adds solvent but doesn't add or remove any solute — the total moles of solute stay exactly the same before and after. Since moles = concentration × volume, that means concentration-times-volume before dilution must equal concentration-times-volume after: C1V1 = C2V2. This is why the volume units on both sides just need to match each other (both mL, or both L) — the equation still balances regardless of which absolute unit you pick.

Molar Mass and the Periodic Table

Molar mass — grams per mole — comes directly from the periodic table for a single element (its atomic weight, in g/mol), or by summing the atomic weights of every atom in a compound's formula for a molecule. Table salt (NaCl) has a molar mass of about 58.44 g/mol (22.99 for sodium plus 35.45 for chlorine), which is why the default example above dissolves 58.44 g of salt to make exactly 1 mole of solute.

Example — Your Current Inputs

C1V1 = C2V2: 1 × 10 = 0.1 × 100 — solving for Final Concentration (C2) gives the value shown above.

Additional Example — Making a Lab Buffer

A lab has a 5 M stock solution of a buffer salt and needs 200 mL of a 0.5 M working solution. Using C1V1 = C2V2: 5 M × V1 = 0.5 M × 200 mL, so V1 = 20 mL of stock solution diluted up to a final volume of 200 mL with solvent gives exactly the target 0.5 M concentration.

About These Parameters

Mass / Molar Mass / Moles
Use mass and molar mass together if you're weighing out a solid to dissolve; use moles directly if you already know the mole count (for example, from a reaction stoichiometry calculation).
Volume of Solution
The total final volume of the solution — solute plus solvent together — not just the solvent added. This is what molarity is measured against.
C1, V1, C2, V2 (Dilution)
The initial concentration and volume (before dilution) and the final concentration and volume (after adding solvent). Leave exactly one blank to solve for it; V1 and V2 just need to share the same unit as each other (both mL or both L).

Frequently Asked Questions

Do V1 and V2 have to be in liters?

No — any volume unit works as long as V1 and V2 use the same unit as each other (both mL, or both L), since the ratio between them is what matters in the C1V1=C2V2 equation, not the absolute unit.

What's the difference between molarity and concentration?

Molarity is one specific type of concentration measurement (moles per liter of solution) — "concentration" more broadly can also refer to molality, mass percent, parts per million, or normality, each useful in different contexts.

How do I find a compound's molar mass?

Add up the atomic weights (from the periodic table) of every atom in the compound's chemical formula. Water (H₂O) is 2 × 1.008 (hydrogen) + 16.00 (oxygen) ≈ 18.02 g/mol.

Why does diluting a solution lower its molarity but not its total moles?

Adding solvent increases the volume the same fixed amount of solute is spread across, so the concentration (moles per liter) drops — but the actual number of moles of solute present never changes just from adding more solvent, which is exactly what C1V1 = C2V2 captures.

See also