Molarity and Molality

Introduction

Molarity and molality both use moles of solute, but they answer different concentration questions. Molarity divides by the volume of the final solution ; molality divides by the mass of solvent . Confusing those denominators can produce a plausible-looking answer with the wrong chemical meaning. Their distinction becomes particularly important when temperature changes or when colligative-property formulas call for molality.

Core explanation

Molarity is c = nsolute/Vsolution, commonly in mol L⁻¹ or M. If 0.50 mol of a solute is present in exactly 2.00 L of final solution, c = 0.250 mol L⁻¹. The denominator is not 2.00 L of water added before dissolving, unless that is also the measured final volume. Mixing and dissolution can change volume, and liquids may contract or expand on mixing. A volumetric flask sets final solution volume at a stated temperature, which is why it is suited to preparing a specified molarity.

Molality is m = nsolute/msolvent, where solvent mass is expressed in kilograms, so its unit is mol kg⁻¹. If 0.50 mol solute is dissolved in 2.00 kg solvent, m = 0.250 mol kg⁻¹. This numerical equality with the previous molarity is coincidental; litres of final solution and kilograms of solvent are different physical quantities. The total solution mass includes solute and solvent, but the molality denominator is solvent alone.

For 5.85 g NaCl dissolved in 0.500 kg water, use a molar mass near 58.5 g mol⁻¹ to find n = 5.85/58.5 = 0.100 mol. Molality is 0.100/0.500 = 0.200 mol kg⁻¹. Molarity cannot be calculated from these data without final solution volume or enough additional information, such as solution density and mass. Assuming the final volume is exactly 0.500 L solely because the solvent mass is 0.500 kg is an unjustified shortcut.

Temperature can change the volume of a liquid solution, so molarity of a fixed amount of solute may change slightly with temperature. The mass of solvent does not change merely by warming a closed sample, so its molality remains fixed as long as no solvent evaporates or chemical reaction changes the count of solute formula units. This is why molality is convenient in boiling-point elevation and freezing-point depression equations: temperature variation does not alter its mass denominator.

Neither unit automatically counts all independently moving dissolved particles. A solution prepared with 0.100 mol NaCl may have a formula-unit molarity or molality based on NaCl added; an idealised particle model has about 0.200 mol ions after complete dissociation. For colligative properties, one must account for particle multiplication separately, often through a van 't Hoff factor. State whether a concentration describes analytical formula units or actual species when that distinction affects a conclusion.

Dilution calculations most directly use molarity: if no solute is lost, c₁V₁ = c₂V₂, because both sides represent solute moles. This equation requires consistent volume units and a final solution volume. It is not a universal equation for changing molality by adding solvent mass. For molality, track nsolute and kilograms of solvent explicitly; adding 0.500 kg to 0.500 kg solvent halves m if solute amount remains fixed.

Step-by-step reasoning

1. Convert solute mass to moles if moles are not given. 2. Read the required unit: mol L⁻¹ needs final solution volume; mol kg⁻¹ needs solvent mass. 3. Convert millilitres to litres or grams of solvent to kilograms before division. 4. Record whether the solution is heated, diluted or evaporated; identify which denominator changes. 5. If the property depends on particles, distinguish formula-unit amount from dissociated species.

Visual explanation

Draw a central box labelled “nsolute.” Send one arrow to a blue flask labelled “divide by Vsolution in L → c” and another to a balance labelled “divide by msolvent in kg → m.” On the flask, mark thermal expansion changing its scale. On the balance, mark closed-system mass as unchanged with temperature. Underline the word “solution” on one denominator and “solvent” on the other.

Real-world analogy

Counting students per classroom volume differs from counting students per mass of furniture. Both ratios use the same number of students, but the denominator tells a different story. Likewise two concentrations can share a numerical value by chance without representing the same measured basis.

Real-world example

A laboratory prepares a standard aqueous solution in a volumetric flask to a marked final volume for a titration, so molarity directly supplies moles from a measured aliquot. A freezing-point experiment instead tracks a weighed mass of solvent and a weighed solute; molality is convenient because changing temperature during the experiment does not change the mass denominator.

Why?

Why is a 0.200 m NaCl solution not automatically 0.200 M? Molality's denominator is 1 kg of solvent; molarity's denominator is 1 L of final solution. They are related only if solution density, composition and volume information are supplied.

Common misconception

“One kilogram of water always produces one litre of solution.” Solute adds mass and may alter the final volume; water density itself varies with temperature. Measuring solvent mass is not equivalent to measuring final solution volume.

Worked example

Mix 0.150 mol glucose with 0.750 kg water. The molality is m = 0.150 mol / 0.750 kg = 0.200 mol kg⁻¹. After mixing, the final solution is measured as 0.800 L at the stated temperature. Its molarity is c = 0.150 mol / 0.800 L = 0.1875 mol L⁻¹, or 0.188 M to three significant figures. The two values differ because 0.750 kg solvent and 0.800 L solution are not the same denominator. If the final volume had not been measured, only molality would be supported.

Quick check

1. Which mass goes into molality: mass of entire solution or mass of solvent alone? Answer: Mass of solvent alone, expressed in kilograms; solute mass is excluded from that denominator.

Exam focus

Write definitions with denominators before inserting numbers. A statement of “0.1 concentration” is incomplete without a unit. Use final solution volume for molarity, solvent kilograms for molality, and keep particle corrections separate from analytical concentration.

Advanced insight

At very dilute aqueous concentrations, numerical molarity and molality can be close because water density is near 1 kg L⁻¹ and solute contributes little volume. That approximation fails for concentrated solutions or different solvents. The exact conversion requires density and molar mass, which the next pages develop.

Summary

Molarity c = n/Vsolution uses litres of final solution and can vary with temperature as volume changes. Molality m = n/msolvent uses kilograms of solvent and remains constant for a closed sample without composition change. Select the definition required by the question rather than assuming their values match.

Practice questions

1. What is the molality of 0.30 mol solute in 1.50 kg solvent? Answer: 0.30/1.50 = 0.20 mol kg⁻¹. 2. What is the molarity of 0.30 mol solute in 1.50 L final solution? Answer: 0.30/1.50 = 0.20 mol L⁻¹; the same number here has a different unit. 3. Why can warming a closed solution alter molarity but leave molality unchanged? Answer: The solution volume can expand, while solvent mass remains constant if nothing evaporates or reacts.