Molarity is moles of solute per liter of solution, M=Vn. The volume is the whole solution, not just the water you started with.
Worked example. You dissolve 5.85 g of NaCl and add water until the solution reaches 250 mL. Moles of NaCl =5.85 g÷58.44 g/mol=0.100 mol. Volume =0.250 L. Molarity =0.100÷0.250=0.400 M.
Molality is moles of solute per kilogram of solvent. Mass does not change with temperature the way volume does, so molality is used for boiling point elevation and freezing point depression calculations.
To dilute, use M1V1=M2V2. Adding water doesn't change the moles of solute, only the volume they're spread through. To make 500 mL of 0.50 M HCl from 2.0 M stock, solve V1=(0.50)(500)÷2.0=125 mL. Put some water in a 500 mL volumetric flask, carefully add the 125 mL of HCl stock to that water, and swirl to mix. Let the solution cool to room temperature before adding water to the 500 mL mark. Acid goes into water, not the other way around, because mixing releases heat.
Percent shows up two ways. Mass percent of a solution is grams of solute over grams of solution, times 100, so 60 g glycerin in a 150 g lotion is 40%. Percent composition of a compound is the mass of one element over the molar mass, times 100. Hydrogen in water is 2.016÷18.02×100=11.2%.
- Molarity (M): Moles of solute divided by liters of solution.
- Molality (m): Moles of solute divided by kilograms of solvent.
- Dilution: Adding solvent to lower concentration while the moles of solute stay the same.
- Parts per million (ppm): Mass of solute divided by mass of solution, times 10⁶, used for very dilute solutions like trace lead in water.
Check yourself. How many grams of glucose (C₆H₁₂O₆, 180.2 g/mol) do you need to make 0.500 L of a 0.200 M solution? You need 18.0 g. Moles = 0.200 mol/L × 0.500 L = 0.100 mol, and 0.100 mol × 180.2 g/mol = 18.0 g.
| Unit | Numerator | Denominator | Typical use |
|---|
| Molarity (M) | Moles of solute | Liters of solution | Lab prep, titrations, stoichiometry |
| Molality (m) | Moles of solute | Kilograms of solvent | Boiling point and freezing point problems |
| Mass percent | Grams of solute | Grams of solution | Product labels and everyday mixtures (multiply the ratio by 100) |
| ppm | Grams of solute | Grams of solution | Pollutants and trace amounts (multiply the ratio by 10⁶) |