Bottles of concentrated acid are labeled with mass percent and density; a recipe asks for molarity; a freezing-point problem needs molality. Molality, Mole Fraction and Parts per Million defined each one, and Density is the bridge between mass and volume that lets you switch.
Key idea
Pick a convenient amount and split it into solute and solvent. For molarity, take 1 L of solution: its mass is $1000 \times d$ grams. For mass percent or molality, take 100 g of solution or 1 kg of solvent. Then find moles of solute, mass of solvent and volume of solution, and divide as each unit asks.
Why it works
Concentration is a ratio, so it doesn't depend on how much solution you imagine. Choosing 1 L or 100 g makes the arithmetic easy. The only extra fact needed is how mass and volume are related for this solution, its density. Molarity has volume of solution on the bottom; molality has mass of solvent; so going between them you must account for the mass of the solute itself, which is large in concentrated solutions.
Worked example: Mass percent to molarity
Concentrated hydrochloric acid is 37.0% HCl by mass and has a density of $1.18\ \text{g/mL}$. Find its molarity (HCl $36.46\ \text{g/mol}$).
- Take 1.000 L: mass $= 1000\ \text{mL} \times 1.18\ \text{g/mL} = 1180\ \text{g}$ of solution.
- HCl: $0.370 \times 1180 = 436.6\ \text{g}$, which is $436.6/36.46 = 11.97\ \text{mol}$.
- Molarity: $11.97\ \text{mol}/1.000\ \text{L} = 12.0\ \text{M}$.
Answer: $12.0\ \text{M}$
Worked example: Mass percent to molality
Find the molality of the same 37.0% HCl.
- Take 100.0 g of solution: 37.0 g HCl and $100.0 - 37.0 = 63.0\ \text{g}$ of water.
- HCl: $37.0/36.46 = 1.015\ \text{mol}$.
- Molality: $1.015/0.0630 = 16.1\ \text{mol/kg}$. Much larger than 12.0 M: in a concentrated solution, a liter of solution holds much less than a kilogram of water.
Answer: $16.1\ \text{mol/kg}$
Worked example: Molarity to molality
A $1.00\ \text{M}$ NaCl solution has a density of $1.040\ \text{g/mL}$. Find its molality.
- 1.000 L of solution: $1040\ \text{g}$, containing $1.00\ \text{mol} = 58.44\ \text{g}$ of NaCl.
- Water: $1040 - 58.44 = 981.6\ \text{g} = 0.9816\ \text{kg}$.
- $m = 1.00/0.9816 = 1.02\ \text{mol/kg}$.
Answer: $1.02\ \text{mol/kg}$
Common mistake
Treating the volume of solution in milliliters as the mass of solvent in grams, so that 1 L of 12 M HCl seems to hold 1000 g of water. It holds only about 743 g (1180 g of solution minus 436.6 g of HCl). The shortcut "molarity ≈ molality" works only for dilute water solutions. Check: always subtract the solute's mass to get the solvent's.