Molarity to Molality Converter
Convert molarity (M) to molality (m) using solution density and solute molar mass. Get instant conversions.
Molarity to Molality Converter
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Formula: m = M / (d - M * MM / 1000)
Worked example โ 1 M NaCl = 1.0188 molal
Formula
m = M / (d - M * MM / 1000)
Molality equals molarity divided by the quantity (solution density minus molarity times molar mass divided by 1000). The denominator represents the mass of solvent in kilograms per liter of solution: total solution mass (density times 1000 g) minus solute mass (molarity times molar mass in grams), all divided by 1000 to convert to kg.
Worked Examples
Example 1: NaCl Solution Conversion
Problem:Convert 1 M NaCl solution (molar mass 58.44 g/mol, density 1.04 g/mL) to molality.
Solution:Solution mass per L = 1.04 x 1000 = 1040 g Solute mass = 1 x 58.44 = 58.44 g Solvent mass = 1040 - 58.44 = 981.56 g = 0.98156 kg Molality = 1 / 0.98156 = 1.0188 mol/kg
Result:1 M NaCl = 1.0188 molal
Example 2: Concentrated H2SO4 Conversion
Problem:Convert 5 M H2SO4 (molar mass 98.079, density 1.29 g/mL) to molality.
Solution:Solution mass = 1.29 x 1000 = 1290 g Solute mass = 5 x 98.079 = 490.395 g Solvent mass = 1290 - 490.395 = 799.605 g = 0.7996 kg Molality = 5 / 0.7996 = 6.253 mol/kg
Result:5 M H2SO4 = 6.253 molal
Frequently Asked Questions
What is the difference between molarity and molality?
Molarity (M) is defined as moles of solute per liter of solution, while molality (m) is defined as moles of solute per kilogram of solvent. The key distinction is that molarity uses total solution volume while molality uses only solvent mass. Molality is independent of temperature because mass does not change with temperature, whereas molarity varies because solution volume expands or contracts. This makes molality preferred for precise thermodynamic calculations and colligative property measurements.
Why do you need solution density to convert molarity to molality?
Solution density is required because molarity is volume-based (moles per liter of solution) while molality is mass-based (moles per kilogram of solvent). To bridge this gap, you need to know the mass of one liter of solution, which is the density times 1000 (converting kg/L to g/L). From the total mass, subtract the solute mass (molarity times molar mass) to find the solvent mass. Without density, there is no way to determine the solvent mass from a volume measurement alone.
When is molality preferred over molarity in practical applications?
Molality is preferred in situations involving temperature changes, precise thermodynamic measurements, and colligative property calculations. Since molality depends on mass (which is invariant with temperature), it remains constant whether the experiment is conducted at 20 degrees or 80 degrees Celsius. It is used for boiling point elevation, freezing point depression, and osmotic pressure calculations. In contrast, molarity is more practical for volumetric lab work since measuring volumes with graduated cylinders or pipettes is straightforward.
For dilute aqueous solutions, are molarity and molality approximately equal?
Yes, for very dilute aqueous solutions at room temperature, molarity and molality are nearly identical. This is because the density of a dilute aqueous solution is close to 1.0 g/mL (similar to pure water), and the solute mass is negligible compared to the solvent mass. In such cases, one liter of solution weighs approximately 1000 grams and contains nearly 1 kilogram of water. However, for concentrated solutions, the difference becomes significant because the solute contributes substantial mass to the solution.
How do density and molar mass build the solvent mass in the denominator?
Take one litre of solution. Its total mass in grams is the density in g/mL multiplied by 1,000. The dissolved solute accounts for molarity times molar mass grams of that. Subtracting leaves the solvent mass in grams, and dividing by 1,000 converts it to kilograms. Molality is then the same number of moles divided by that solvent mass, which is exactly the denominator d minus M times MM over 1,000.
What if my density is given in kg/m3 instead of g/mL?
Divide by 1,000 first, because 1 g/mL equals 1,000 kg/m3. Water near room temperature is about 998 kg/m3, which is 0.998 g/mL. Entering the kg/m3 figure directly makes the denominator enormous and drives the molality to near zero, a mistake that is easy to spot from the implausible result. Density in g/cm3 is numerically identical to g/mL and can be entered as it stands.
Why does molarity change with temperature while molality does not?
Molarity is moles per litre of solution, and volume expands as the solution warms, so the same amount of solute occupies more litres and the molarity falls. Molality is moles per kilogram of solvent, and mass is unaffected by temperature. That is why molality is the correct concentration measure for experiments run over a temperature range, and why the density used in this conversion must be the density at the working temperature.
How do I convert molality back into molarity?
Rearrange the relationship. Starting from one kilogram of solvent, you have m moles of solute weighing m times MM grams, so the solution masses 1,000 plus m times MM grams and occupies that mass divided by the density. Expressed compactly, M equals m times d divided by the quantity one plus m times MM over 1,000, with MM in g/mol and d in g/mL. The same three inputs serve both directions.
How does molality relate to mole fraction?
Mole fraction is moles of solute divided by total moles of solute plus solvent. Since molality already fixes the solvent at one kilogram, the moles of solvent are 1,000 divided by its molar mass, which for water is about 55.5 moles per kilogram. The mole fraction is then m divided by m plus 55.5. Mole fraction is dimensionless and is the natural variable for vapour-pressure and Raoult's law work.
Why is molality the right basis for colligative property calculations?
Freezing-point depression and boiling-point elevation are proportional to the molal concentration of dissolved particles, with a constant specific to the solvent and a factor for how many particles each formula unit produces on dissolution. Because those experiments deliberately change the temperature, a volume-based concentration would drift as the measurement proceeds. Molality stays fixed, so the proportionality holds across the whole temperature range being observed.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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