Molarity and dilution (M1V1 = M2V2) with examples

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	$M_1V_1 = M_2V_2$

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?

Grams to moles. $M_{\mathrm{NaCl}} = 22.99 + 35.45 = 58.44$ g/mol.
$n = \frac{5.85\ \text{g}}{58.44\ \text{g/mol}} = 0.10010\ \text{mol}$
Milliliters to liters.
$V = \frac{250.0\ \text{mL}}{1000\ \text{mL/L}} = 0.2500\ \text{L}$
Divide moles by liters.
$M = \frac{0.10010\ \text{mol}}{0.2500\ \text{L}} = 0.4004\ \text{M}$
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?

Rearrange $M = n/V$ to $n = MV$ . Use liters.
$n = 0.150\ \text{mol/L} \times 0.500\ \text{L} = 0.0750\ \text{mol}$
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?

Write $M_1V_1 = M_2V_2$ and solve for $M_2$ .
$M_2 = \frac{M_1 V_1}{V_2}$
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?

Solve $M_1V_1 = M_2V_2$ for $V_1$ .
$V_1 = \frac{M_2 V_2}{M_1} = \frac{0.500\ \text{M} \times 2.00\ \text{L}}{12.0\ \text{M}} = 0.08333\ \text{L}$
Liters to milliliters.
$0.08333\ \text{L} \times 1000\ \text{mL/L} = 83.33\ \text{mL}$
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

What is the molarity of a solution with 0.300 mol of solute in 1.50 L of solution?
1. 0.200 M
2. 0.450 M
3. 5.00 M
4. 0.300 M
Show answer
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.
10.0 g of NaOH is dissolved to make 0.500 L of solution. What is the molarity?
Show answer
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.
How many moles of solute are in 75.0 mL of 0.200 M KCl?
Show answer
Answer: 0.0150 mol
$n = MV = 0.200\ \text{mol/L} \times 0.0750\ \text{L} = 0.0150$ mol.
You dilute 50.0 mL of 2.00 M NaCl to 0.250 M. What is the final volume?
1. 0.00625 L
2. 0.400 L
3. 0.350 L
4. 4.00 L
Show answer
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.
In the dilution above, how much water do you add?
Show answer
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 and equilibrium.

Sources

OpenStax Chemistry 2e, 3.3 Molarity, accessed October 1, 2026

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

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