Molarity from Molality: Step-by-Step Guide
This guide shows how to convert a solution’s molality (moles of solute per kilogram of solvent) into molarity (moles of solute per liter of solution). You need the molality value, the molar mass of the solute, and the density of the final solution. The process involves calculating the mass of solute and solvent, finding the total solution volume from density, then dividing moles by that volume.
What You’ll Need
- Molality of the solution (mol/kg)
- Molar mass of the solute (g/mol)
- Density of the solution (g/mL or g/cm³)
- Calculator for arithmetic
Step-by-Step
- Write down the given values. For example, suppose you have a 2.0 mol/kg solution of NaCl (molar mass = 58.44 g/mol) with a density of 1.10 g/mL.
- Assume a convenient mass of solvent. Since molality is per kilogram of solvent, take exactly 1.000 kg (1000 g) of solvent. This makes calculations straightforward.
- Calculate moles of solute. Multiply molality by the mass of solvent in kg: moles = molality × kg solvent. In the example: 2.0 mol/kg × 1.000 kg = 2.0 mol.
- Find the mass of solute. Multiply moles by molar mass: mass solute = moles × molar mass. For NaCl: 2.0 mol × 58.44 g/mol = 116.88 g.
- Determine the total mass of the solution. Add the mass of solvent (1000 g) and mass of solute: 1000 g + 116.88 g = 1116.88 g.
- Convert solution mass to volume using density. Volume (mL) = mass (g) ÷ density (g/mL). Here: 1116.88 g ÷ 1.10 g/mL = 1015.35 mL.
- Convert volume to liters. Divide mL by 1000: 1015.35 mL ÷ 1000 = 1.01535 L.
- Divide moles of solute by volume in liters. Molarity = moles ÷ liters. So 2.0 mol ÷ 1.01535 L = 1.97 M (rounded to two significant figures).
Common Mistakes
- Using the solvent volume instead of solution volume. Molarity uses the total solution volume, not just the solvent. Always include the solute’s contribution to the volume.
- Forgetting to convert density to consistent units. If density is in g/mL, keep mass in grams and volume in mL, then convert to liters. Mixing kg with g/mL leads to errors.
- Assuming density of water (1 g/mL) for the solution. The density of a solution often differs from pure water, especially at higher concentrations. Always use the given solution density.
- Rounding too early. Keep extra digits during intermediate steps and round only at the final answer to avoid rounding errors.
Tips
- Always start with 1.000 kg of solvent to simplify the math—molality is defined per kg, so this makes moles equal to the molality number.
- Check units at each step: molality (mol/kg) × kg = mol; molar mass (g/mol) × mol = g; g ÷ (g/mL) = mL; mL ÷ 1000 = L; mol ÷ L = mol/L (M).
- If the density is given in g/cm³, note that 1 g/cm³ = 1 g/mL, so no conversion is needed.
- For dilute solutions (molality < 0.1), molarity and molality are numerically close, but this shortcut fails for concentrated solutions.
Troubleshooting
Your answer seems too high or too low. Recheck the density units. If density is in kg/L, you must convert to g/mL (divide by 1000) or adjust mass units accordingly.
You don’t have the solution density. Without density, you cannot accurately convert molality to molarity. You can approximate by using the density of pure solvent if the solution is very dilute, but this introduces error.
The solute is a hydrate. If the solute contains water of crystallization (e.g., CuSO₄·5H₂O), use the molar mass of the hydrated form, and remember that the water in the hydrate becomes part of the solvent—adjust the solvent mass accordingly.
Your calculator gives a different result. Double-check that you divided moles by liters, not by milliliters. A common slip is dividing by volume in mL, which gives a value 1000 times too large.
Frequently Asked Questions
Q: Can I convert molality to molarity without density?
A: No. Density is essential because it links mass of solution to volume. Without it, you can only estimate for very dilute solutions where density ≈ 1 g/mL.
Q: Why is molality used instead of molarity in some contexts?
A: Molality depends only on masses, which don’t change with temperature. Molarity depends on volume, which expands or contracts with temperature, so molality is preferred in colligative property studies.
Q: Is molarity always larger than molality?
A: Not necessarily. If the solution density is greater than 1 g/mL, molarity can be larger. For dilute aqueous solutions, molarity is slightly higher because 1 L of solution weighs more than 1 kg.
Q: What if I have more than one solute?
A: Calculate the total moles of all solutes and the total mass of all solutes to find the total solution mass. Then use the combined mass to find volume and divide total moles by that volume.
Q: How do I convert molarity back to molality?
A: Use the same relationship in reverse: find mass of solute from molarity and volume, subtract from total solution mass to get solvent mass in kg, then divide moles by that mass.
Q: Does the formula change for non-aqueous solvents?
A: No, the logic is the same. Just use the density of the actual solution and the molar mass of the solute, regardless of solvent.
Related
Method: the standard formula for this calculation, rounded as each result states. See how the calculators work or report an error.