Molarity, Molality and Mass Fraction Together

Introduction

The word “concentration” hides several different denominators. Molarity divides solute moles by total solution volume. Molality divides those same solute moles by solvent mass. Mass fraction divides solute mass by total solution mass. A numerical problem can report all three for one sample, but only if it keeps each denominator distinct and has enough information, often including solution density.

Core explanation

Let n s be moles of dissolved solute, m s its mass, m w the solvent mass, V sol the final solution volume, and m sol = m s + m w the total solution mass for a two-component solution. Molarity c = n s/V sol with volume in litres. Molality b = n s/(m w in kilograms). Mass fraction w s = m s/m sol, usually multiplied by 100 for mass percent. These definitions use different physical measures. The volume of a solution is not generally the sum of separate component volumes, so do not substitute solvent volume for V sol unless the problem justifies it.

Density links total mass and solution volume: ρ = m sol/V sol. For a sample described by mass fraction and density, choose a convenient solution volume such as 1.000 L. Density then gives solution mass, mass fraction gives solute mass, and subtraction gives solvent mass. Convert solute mass to moles with its molar mass and compute both c and b. The choice of 1 L is a calculation basis, not a claim that the sample was prepared in a one-litre flask.

For example, a 10.0% by-mass NaCl solution with density 1.07 g mL⁻¹ has 1070 g solution per 1.000 L. It contains 107 g NaCl and 963 g water. Using M(NaCl) = 58.44 g mol⁻¹, n s = 107/58.44 = 1.83 mol. Molarity is about 1.83 mol L⁻¹. Molality is 1.83/0.963 = 1.90 mol kg⁻¹. The numerical values differ because a litre of this solution is not a kilogram of solvent. Mass fraction remains 0.100 by construction.

If only molarity is supplied, molality generally cannot be found without information about solution density or solvent mass. If only molality is supplied, molarity needs the solution volume or density. A mass fraction alone gives the ratio of component masses but no volume, so it cannot uniquely determine molarity. This is an information issue, not an algebra trick.

Temperature can change solution volume and density, so molarity can vary with temperature even if no solute is gained or lost. Molality and mass fraction are based on masses, which normally remain unchanged with simple heating in a closed sample. They can still change if solvent evaporates or material reacts. For mixtures with multiple solutes, solvent mass excludes all solute masses, and total solution mass includes every component.

Step-by-step reasoning

1. Write the definitions with their denominators: litre of solution, kilogram of solvent, and total solution mass. 2. Choose a convenient basis such as 1.000 L solution or 100.0 g solution according to the given data. 3. Use density only to convert between total solution mass and final solution volume. 4. Split total mass into solute and solvent using mass fraction; convert solute mass to moles. 5. Calculate each requested concentration using its own denominator, then check units and plausibility.

Visual explanation

Draw one beaker labelled “final solution.” Under it place three arrows: one points to solution volume V sol for molarity, one to only the water mass m w for molality, and one to total solution mass m sol for mass fraction. The numerator on the first two arrows is n s, while the numerator on the third is m s. This shows why the three measures cannot be interchanged by renaming units.

Real-world analogy

A school might report students per classroom, students per teacher, or girls as a fraction of all students. These statistics share a population but use different denominators and answer different questions. Concentration measures likewise describe the same sample from different bases; the denominator must travel with the number.

Real-world example

A laboratory recipe may specify molarity because measured solution volumes are convenient for dispensing. A freezing-point experiment uses molality because its particle-count relation is based on solvent mass. A product label may give mass percent because producers can weigh ingredients without relying on volume at a particular temperature.

Why?

Why does density permit a mass-fraction-to-molarity conversion? Mass fraction gives solute grams per gram of solution, while density gives solution grams per unit volume. Multiplying them gives solute grams per volume; dividing by solute molar mass gives moles per volume.

Common misconception

“One litre of solution contains one kilogram of solvent.” Neither is generally true. Density concerns the whole solution , and the solvent mass is smaller than total solution mass when solute is present. Use subtraction rather than replacing volume with a guessed mass.

Worked example

A 500.0 mL portion of a 12.0% by-mass glucose solution has density 1.04 g mL⁻¹. Its total mass is (500.0 mL)(1.04 g mL⁻¹) = 520 g. Glucose mass is 0.120(520) = 62.4 g; water mass is 457.6 g = 0.4576 kg. With M(glucose) = 180.16 g mol⁻¹, glucose amount is 62.4/180.16 = 0.346 mol. Molarity is 0.346/0.5000 = 0.693 mol L⁻¹; molality is 0.346/0.4576 = 0.758 mol kg⁻¹. The mass fraction is 0.120. Rounding at the end preserves the calculation relationship.

Quick check

1. Which denominator is used for molality: total solution mass or solvent mass? Answer: Molality uses kilograms of solvent, excluding the mass of every dissolved solute.

Exam focus

Label every mass as solute, solvent or solution and every volume as final solution volume. Draw a mass inventory before converting. If density is missing, explain why a requested mass-to-volume concentration conversion may be underdetermined.

Advanced insight

For one solute of molar mass M s and solution density ρ in g L⁻¹, c = w sρ/M s. The formula is only a compressed version of the mass-basis method. For multiple solutes, b for one solute uses the same solvent mass but a separate mole numerator for that solute.

Summary

Molarity uses solute moles per solution litre, molality uses solute moles per solvent kilogram, and mass fraction uses solute mass per total mass. Density converts solution volume to total mass. Choose a basis, construct the mass inventory, and keep denominators explicit.

Practice questions

1. A 200 g solution is 5.00% solute by mass. Find solute and solvent masses. Answer: Solute is 0.0500 × 200 = 10.0 g; solvent is 190 g. 2. If that solute has M = 50.0 g mol⁻¹, find its molality. Answer: n s = 10.0/50.0 = 0.200 mol; b = 0.200/0.190 = 1.05 mol kg⁻¹. 3. What extra property is needed to obtain its molarity from those masses? Answer: Final solution volume or solution density, which converts total solution mass to volume.