# Molarity and dilution

Canonical: https://duckyhelper.com/learn/chemistry/molarity-and-dilution/
Updated: 2026-10-01

Molarity (M) is the concentration of a solution: moles of solute per liter of solution. A 1.0 M solution has 1.0 mol of solute in every liter. Molarity = moles ÷ liters. When you dilute a solution by adding water, the moles of solute stay the same, so \(M_1V_1 = M_2V_2\). Convert milliliters to liters before you use the molarity formula.

## Key ideas

$$
M = \frac{n}{V}
$$

- \(M\) is molarity, in moles per liter (mol/L), written M and read "molar".
- \(n\) is the amount of solute, in moles (mol). The solute is what gets dissolved.
- \(V\) is the volume of the whole solution, in liters (L). Divide milliliters by 1000 to get liters.

Dilution adds solvent (usually water) but no solute. The moles of solute before, \(M_1V_1\), equal the moles after, \(M_2V_2\):

$$
M_1 V_1 = M_2 V_2
$$

- \(M_1\) and \(V_1\) are the concentration (M) and volume of the concentrated stock solution you start with.
- \(M_2\) and \(V_2\) are the concentration (M) and the **total** final volume after diluting.
- \(V_1\) and \(V_2\) can be in mL or L, as long as both use the same unit, because the unit cancels.

**Which formula to use**

| The problem gives you | It asks for | Use |
| --- | --- | --- |
| grams of solute and volume | molarity | grams to moles, then \(M = n/V\) |
| molarity and volume | grams of solute | \(n = MV\), then moles to grams |
| a stock solution and a target | volume or concentration | \(M_1V_1 = M_2V_2\) |

> **Note: Atomic masses used on this page**
>
> H 1.008, C 12.01, O 16.00, Na 22.99, Cl 35.45 (g/mol).

## Worked examples

**Example 1: find the molarity**

Problem: A student dissolves 5.85 g of NaCl in water and adds water until the total volume is 250.0 mL. What is the molarity?

1. Grams to moles. \(M_{\mathrm{NaCl}} = 22.99 + 35.45 = 58.44\) g/mol.

   $$
   n = \frac{5.85\ \text{g}}{58.44\ \text{g/mol}} = 0.10010\ \text{mol}
   $$
2. Milliliters to liters.

   $$
   V = \frac{250.0\ \text{mL}}{1000\ \text{mL/L}} = 0.2500\ \text{L}
   $$
3. Divide moles by liters.

   $$
   M = \frac{0.10010\ \text{mol}}{0.2500\ \text{L}} = 0.4004\ \text{M}
   $$
4. 5.85 g has 3 significant figures, so round to 3.

Answer: 0.400 M NaCl

**Example 2: how much solute to weigh out**

Problem: How many grams of glucose, \(\mathrm{C_6H_{12}O_6}\), do you need to make 500. mL of a 0.150 M solution?

1. Rearrange \(M = n/V\) to \(n = MV\). Use liters.

   $$
   n = 0.150\ \text{mol/L} \times 0.500\ \text{L} = 0.0750\ \text{mol}
   $$
2. Moles to grams. \(M_{\mathrm{C_6H_{12}O_6}} = 180.16\) g/mol.

   $$
   0.0750\ \text{mol} \times 180.16\ \text{g/mol} = 13.51\ \text{g}
   $$

Answer: 13.5 g of glucose

**Example 3: diluting a stock solution**

Problem: You dilute 25.0 mL of 6.00 M HCl with water to a total volume of 150.0 mL. What is the new concentration?

1. Write \(M_1V_1 = M_2V_2\) and solve for \(M_2\).

   $$
   M_2 = \frac{M_1 V_1}{V_2}
   $$
2. Plug in. Both volumes are in mL, so the unit cancels.

   $$
   M_2 = \frac{6.00\ \text{M} \times 25.0\ \text{mL}}{150.0\ \text{mL}} = 1.00\ \text{M}
   $$

Answer: 1.00 M HCl

**Example 4: how much stock to measure**

Problem: How many milliliters of 12.0 M HCl do you need to make 2.00 L of 0.500 M HCl?

1. Solve \(M_1V_1 = M_2V_2\) for \(V_1\).

   $$
   V_1 = \frac{M_2 V_2}{M_1} = \frac{0.500\ \text{M} \times 2.00\ \text{L}}{12.0\ \text{M}} = 0.08333\ \text{L}
   $$
2. Liters to milliliters.

   $$
   0.08333\ \text{L} \times 1000\ \text{mL/L} = 83.33\ \text{mL}
   $$
3. Safety: add the acid to most of the water, then fill to 2.00 L. Never add water to concentrated acid.

Answer: 83.3 mL of 12.0 M HCl

## Common mistakes and how to fix them

- **Using milliliters in \(M = n/V\).** 0.100 mol in 250 mL is 0.400 M, not 0.000400 M. Fix: convert mL to L first.
- **Using grams instead of moles.** Molarity counts moles. Fix: divide grams by molar mass before dividing by volume.
- **Using the water added as \(V_2\).** \(V_2\) is the total final volume. Fix: water added = \(V_2 - V_1\).
- **Using the volume of solvent instead of solution.** Molarity uses the volume of the finished solution. That is why labs dissolve the solid, then fill to the line of a volumetric flask.

**Practice problems**

1. What is the molarity of a solution with 0.300 mol of solute in 1.50 L of solution?
   A. 0.200 M
   B. 0.450 M
   C. 5.00 M
   D. 0.300 M

   Answer: 0.200 M. \(M = n/V = 0.300 \div 1.50 = 0.200\) M. 0.450 comes from multiplying, and 5.00 from dividing the wrong way.

2. 10.0 g of NaOH is dissolved to make 0.500 L of solution. What is the molarity?

   Answer: 0.500 M. \(M_{\mathrm{NaOH}} = 22.99 + 16.00 + 1.008 = 40.00\) g/mol. \(10.0 \div 40.00 = 0.250\) mol, and \(0.250 \div 0.500 = 0.500\) M.

3. How many moles of solute are in 75.0 mL of 0.200 M KCl?

   Answer: 0.0150 mol. \(n = MV = 0.200\ \text{mol/L} \times 0.0750\ \text{L} = 0.0150\) mol.

4. You dilute 50.0 mL of 2.00 M NaCl to 0.250 M. What is the final volume?
   A. 0.00625 L
   B. 0.400 L
   C. 0.350 L
   D. 4.00 L

   Answer: 0.400 L. \(V_2 = M_1V_1 / M_2 = 2.00 \times 50.0 \div 0.250 = 400.\) mL, which is 0.400 L.

5. In the dilution above, how much water do you add?

   Answer: 0.350 L (350. mL). Water added = final volume minus starting volume: \(0.400 - 0.0500 = 0.350\) L.

## Frequently asked questions

### What is the difference between molarity and molality?

Molarity (M) is moles of solute per liter of solution. Molality (m) is moles of solute per kilogram of solvent. Molarity changes a little with temperature because liquids expand. Molality does not, so it is used for freezing point and boiling point problems.

### Why is it okay to use mL in \(M_1V_1 = M_2V_2\)?

The volume appears on both sides, so whatever unit you use cancels out. You only need both volumes in the same unit. In \(M = n/V\) there is nothing to cancel, so the volume must be in liters to give mol/L.

### What does a square bracket like [NaCl] mean?

Square brackets mean the molar concentration of whatever is inside. [NaCl] = 0.400 M means 0.400 mol of NaCl per liter. You will see this a lot in [acids, bases and pH](https://duckyhelper.com/learn/chemistry/acids-bases-and-ph/) and equilibrium.

## Sources

- [OpenStax Chemistry 2e, 3.3 Molarity](https://openstax.org/books/chemistry-2e/pages/3-3-molarity), accessed 2026-10-01

## Related

- [The mole and molar mass](https://duckyhelper.com/learn/chemistry/mole-and-molar-mass/)
- [Acids, bases and pH](https://duckyhelper.com/learn/chemistry/acids-bases-and-ph/)
- [Stoichiometry: how much reacts and how much forms](https://duckyhelper.com/learn/chemistry/stoichiometry/)
- [Chemistry study guides](https://duckyhelper.com/learn/chemistry/)

## Try asking Ducky

- "Check my dilution problem. Did I use the total volume or the water I added?"
- "I keep forgetting to convert mL to L. Give me three quick molarity problems to practice."
- "Look at my lab plan for making 0.100 M copper sulfate. Are my masses right?"

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