Molarity is moles over liters
Molarity (M) is the standard lab unit: moles of solute per liter of solution. If you start from a weighed mass, first convert to moles by dividing by the molar mass, then divide by the volume in liters. The classic slip is leaving volume in milliliters — 500 mL is 0.5 L, not 500.
The units, and where each is used
| Unit | Definition | Used for |
|---|---|---|
| Molarity (M) | mol ÷ L of solution | Most lab work, stoichiometry |
| Molality (m) | mol ÷ kg of solvent | Freezing / boiling point |
| Mass % (w/w) | mass solute ÷ mass solution × 100 | Product labels |
| Volume % (v/v) | vol solute ÷ vol solution × 100 | Alcohols, solvents |
| ppm | mg ÷ L (in water) | Trace levels, water quality |
| Normality (N) | M × equivalents per mole | Acid–base titration |
Quick chain for water: 1% by mass ≈ 10,000 ppm ≈ 10,000,000 ppb. And 1000 ppb = 1 ppm.
Diluting a stock solution
To thin a concentrated solution, the moles of solute do not change — only the volume grows. That is the whole content of C1V1 = C2V2.
- Dilute 500 mL of 2 M to 0.5 M → V₂ = (2 × 500) ÷ 0.5 = 2,000 mL, so add 1,500 mL of water.
- Make 100 mL of 2 M from 12 M stock → V₁ = (2 × 100) ÷ 12 = 16.7 mL of stock, then top up to 100 mL.
- For acids, add the acid to the water, never the reverse — mixing is strongly exothermic and can spatter.
Common questions
What is the difference between molarity and molality?
Molarity is moles of solute per liter of solution. Molality is moles per kilogram of solvent. Molarity shifts with temperature because volume expands and contracts; molality does not, since mass stays fixed. Labs use molarity; freezing- and boiling-point work uses molality.
How do I convert mass percent to ppm?
For dilute water-based solutions, multiply the mass percent by 10,000. So 0.005% by mass is about 50 ppm. This holds because ppm equals milligrams per liter when the density is close to 1 g/mL.
How does the dilution formula C1V1 = C2V2 work?
It keeps the amount of solute constant. Multiply the starting concentration by its volume, set it equal to the target concentration times the final volume, and solve for the unknown. To make 100 mL of 2 M from 12 M stock you need 16.7 mL of stock, topped up with water.
