Molarity is a measure of concentration equal to the number of moles of solute divided by the volume of solution in litres (M = mol/L). A 1 M solution contains 1 mole of solute per litre.

Why does M₁V₁ = M₂V₂ work for dilution?

Diluting a solution adds solvent but no extra solute, so the moles of solute stay the same. Since moles = M × V, the product M × V is unchanged: M₁V₁ (before) equals M₂V₂ (after).

Solution Stoichiometry & Molarity

Most reactions happen in solution — so we measure amount by concentration and volume, not by weighing.

High schoolIntro Gen ChemUni Year 1

⏱️ About 20 min

In the lab you rarely weigh out reactants — you pour them. Chemistry happens in solution, so the natural way to measure 'how much' is concentration times volume. One number, molarity, ties a measured volume straight back to moles, and from there every stoichiometry tool you already have still works.

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The big idea: Molarity is moles of solute per litre of solution (M = mol/L). It converts a volume you can measure with a pipette into moles you can do stoichiometry with. Diluting adds solvent but not solute, so the moles stay fixed — which is exactly what M₁V₁ = M₂V₂ captures.

🎯 By the end, you'll be able to

Define molarity and calculate it from moles and volume
Convert between molarity, volume, and moles of solute
Use M₁V₁ = M₂V₂ for dilution problems
Apply mole ratios to a titration-style neutralisation

📎 Helpful to know first

Molarity: moles per litre

Molarity (M) is the concentration of a solution: how many moles of dissolved substance (the solute) sit in each litre of solution. A 2 M solution has 2 moles of solute per litre.

Its whole purpose is convenience. You can't easily weigh a dissolved solid, but you can measure a volume precisely. Molarity turns that measured volume straight into moles — the currency of every reaction calculation.

\[ M = \frac{n\ (\text{mol solute})}{V\ (\text{L solution})} \]

Molarity = moles of solute ÷ litres of solution. Rearranged: moles = M × V.

🔑 Rearrange to get moles

The form you'll use most is n = M × V: multiply concentration (mol/L) by volume (L) to get moles of solute. Keep volume in litres — if you're given millilitres, divide by 1000 first (250 mL = 0.250 L).

📝 Worked example: Dissolve 58.44 g of NaCl in enough water to make 0.500 L of solution. What is the molarity?

First convert grams to moles. Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol.
n = 58.44 g ÷ 58.44 g/mol = 1.00 mol NaCl.
Now apply M = n ÷ V = 1.00 mol ÷ 0.500 L.
= 2.00 M.

✓ 2.00 M (2.00 mol/L) sodium chloride.

✨ Concentration is not the same as amount

A drop and a bucket of the same 1 M solution have the same concentration but very different amounts of solute — the bucket holds far more moles. Concentration (M) is an intensive property (it doesn't depend on how much you have); the number of moles (M × V) is what scales with volume. Keep the two ideas separate.

Dilution: same solute, more solvent

To dilute a solution you add solvent (usually water). Crucially, you're adding no more solute — so the number of moles of solute doesn't change; they're just spread through a larger volume, lowering the concentration.

Because moles = M × V stays constant, the moles before equal the moles after. That's the whole content of the dilution formula:

\[ M_1 V_1 = M_2 V_2 \]

Moles of solute before (M₁V₁) = moles after (M₂V₂). Only concentration and volume change; the solute amount is fixed.

📝 Worked example: A titration: 25.0 mL of hydrochloric acid (HCl) is exactly neutralised by 30.0 mL of 0.100 M sodium hydroxide (NaOH). Find the concentration of the HCl. The reaction is HCl + NaOH → NaCl + H₂O.

Find moles of the known solution (NaOH): n = M × V = 0.100 mol/L × 0.0300 L = 0.00300 mol NaOH.
Use the mole ratio. HCl + NaOH → NaCl + H₂O is 1 : 1, so moles of HCl = moles of NaOH = 0.00300 mol.
Now find the HCl concentration: M = n ÷ V = 0.00300 mol ÷ 0.0250 L.
= 0.120 M HCl.

✓ 0.120 M hydrochloric acid.

✏️ Practice: How many moles of solute are in 0.250 L of a 0.40 M solution?

mol

Solution

Rearrange molarity to moles: n = M × V.
n = 0.40 mol/L × 0.250 L.
= 0.10 mol of solute.

✏️ Practice: You dilute 50.0 mL of 6.0 M HCl by adding water until the total volume is 300.0 mL. What is the new concentration?

Solution

Use M₁V₁ = M₂V₂, solving for M₂ = M₁V₁ ÷ V₂.
M₂ = (6.0 M × 50.0 mL) ÷ 300.0 mL. (The mL cancel, so no need to convert to litres here.)
= 300 ÷ 300 = 1.0 M. The solute moles never changed — the volume grew 6×, so the concentration fell 6×.

Check your understanding

1. What does a 3.0 M solution mean?

Molarity is moles of solute per litre of solution. A 3.0 M solution has 3.0 mol of dissolved solute in every litre.

2. When you dilute a solution with water, what stays the same?

Adding water changes the volume and lowers the concentration, but you add no solute — so the moles of solute are unchanged. That's why M₁V₁ = M₂V₂ works.

3. Two beakers hold the same 1 M salt solution: one has 10 mL, the other 500 mL. Which is true?

Concentration (1 M) is identical, but moles = M × V, so the larger volume contains far more moles of solute. Concentration and amount are different things.

✅ Key takeaways

Molarity (M) = moles of solute ÷ litres of solution — it turns a measured volume into moles.
Rearrange to n = M × V; always keep the volume in litres.
Dilution adds solvent, not solute, so moles are conserved: M₁V₁ = M₂V₂.
Concentration is intensive (doesn't depend on amount); moles (M × V) scale with volume.
In a titration, find moles of the known solution, apply the mole ratio, then divide by the unknown's volume.

➡️ You now have the full stoichiometric toolkit — counting particles, balancing reactions, and measuring them out in solution. These same moles and ratios reappear everywhere ahead, from gas laws to acid–base titration curves.

Want to test yourself on this? Try the Chemistry practice test →