Solutions · Lesson 1 of 11

Solutions and Ways to Express Composition

“Identify homogeneous mixtures and express their composition using percentages, parts per million and mole fractions.”

Learning Objectives

• Classify solutions by the states of their components. • Distinguish solvent, solute and a binary solution. • Calculate mass, volume and mass-by-volume percentages and ppm. • Calculate mole fractions and explain why they sum to one.

What makes a mixture a solution?

A spoonful of sugar disappears into water, yet the water tastes sweet throughout. The sugar has not vanished or become water: its molecules are dispersed among water molecules. A solution is a homogeneous mixture of two or more components, with uniform composition and properties on the molecular scale. A clear appearance alone is insufficient evidence: some colloids can look clear while remaining heterogeneous at a smaller scale.

Definition

Solvent and solute

The solvent is the component used as the dissolving medium and determines the solution’s physical state. It is generally present in the larger amount. The other component or components are solutes. A binary solution contains exactly two components.

For ethanol in a large amount of water, water is the solvent and ethanol the solute. For a gas mixture, the more abundant component is usually designated solvent. The terms describe roles in a mixture; a substance is not permanently a solute or a solvent. The final state, rather than the original state of a solute, is used to classify the solution.

Solution state	Solute before mixing	Solvent	Example
Gas	Gas	Gas	Oxygen mixed with nitrogen
Gas	Liquid	Gas	Chloroform vapour dispersed in nitrogen
Gas	Solid	Gas	Camphor vapour dispersed in nitrogen
Liquid	Gas	Liquid	Oxygen dissolved in water
Liquid	Liquid	Liquid	Ethanol in water
Liquid	Solid	Liquid	Glucose in water
Solid	Gas	Solid	Hydrogen in palladium
Solid	Liquid	Solid	Mercury incorporated into a solid sodium amalgam
Solid	Solid	Solid	Copper in gold within a homogeneous alloy phase

The original liquid or solid has to enter the gaseous state to make a gaseous solution; droplets in a mist or dust in air do not count as molecular gas solutions. Similarly, an alloy is a solid solution only over a composition range in which it forms a homogeneous phase. A large piece of undissolved metal in a liquid is not a solid solution.

Percentages: always specify the denominator

“Dilute” and “concentrated” are useful comparisons, but they do not tell us exactly how much solute is present. Concentration needs both an amount and a reference amount. Mass percent compares solute mass with total solution mass, not solvent mass. Total solution mass is the sum of the component masses when no material is lost.

Mass percentageLaTeX

mass percent = \frac{w _{2}}{w _{1} + w _{2}} \times 100

w₁ is solvent mass and w₂ solute mass, expressed in the same unit. The percentage is a dimensionless ratio multiplied by 100.

A 10% glucose solution

Problem
Interpret 10% glucose by mass and calculate the mass of water in 250 g of solution.

1.A 10% mass fraction means 10 g glucose in each 100 g of final solution, leaving 90 g water.
2.In 250 g solution, glucose mass = 0.10 × 250 = 25 g. Water mass = 250 − 25 = 225 g.
3.Adding 10 g glucose to 100 g water would instead give 10/110 × 100 = 9.09%, because the denominator would be 110 g.

Volume and mass-by-volume percentagesLaTeX

% (v / v) = \frac{V _{2}}{V _{solution}} \times 100, % (w / v) = \frac{w _{2} ( g )}{V _{solution} ( mL )} \times 100

For v/v, both volumes use the same unit and specified temperature. A 5% w/v solution contains 5 g solute per 100 mL of final solution.

A 10% v/v ethanol solution is prepared from 10 mL ethanol and enough water to reach a final volume of 100 mL at the stated temperature. It does not mean that mixing 10 mL ethanol with exactly 90 mL water necessarily yields 100 mL. Molecules can pack differently on mixing, so liquid volumes need not add. Mass-by-volume percentage combines two kinds of units and must never be silently interpreted as mass percent. A 0.9% w/v sodium chloride solution contains 0.9 g per 100 mL solution.

Trace composition and parts per million

Mass-based parts per millionLaTeX

ppm (w / w) = \frac{w _{2}}{w _{solution}} \times 1 0^{6}

1 ppm by mass = 1 mg solute per kg solution. State whether a ppm value is based on mass or volume.

Small amounts matter in water quality, atmospheric composition and materials processing. A mass-based ppm value is dimensionless; mg L⁻¹ is a mass concentration. They are numerically close for a very dilute aqueous solution of density approximately 1 kg L⁻¹, but are not generally interchangeable. For seawater, density must be included rather than assuming one litre has a mass of one kilogram.

Dissolved oxygen in seawater

Problem
One litre of seawater has mass 1030 g and contains 0.0060 g O₂. Express this as mass-based ppm.

1.Use total seawater mass because ppm here is a mass ratio.
2.ppm = (0.0060 g / 1030 g) × 10⁶ = 5.825, or about 5.8 ppm.
3.The mass concentration is 6.0 mg L⁻¹. The difference from 5.8 ppm comes from the seawater density.

Mole fraction measures relative particle amounts

Mass fractions do not directly give the fraction of molecules. One gram of a light molecule contains more molecules than one gram of a heavy molecule. Convert each mass to moles using its own molar mass before comparing particle amounts. Here a mole fraction counts the specified chemical entities in a mixture; later we will distinguish analytical solute amounts from particles formed by dissociation.

Mole fractionLaTeX

x_{i} = \frac{n _{i}}{\sum _{j} n _{j}}, x_{1} + x_{2} = 1

nᵢ is the amount of component i in moles. Mole fractions are dimensionless and lie between zero and one.

The sum is one because adding every numerator gives the same total number of moles as the common denominator. In a binary mixture, calculating one fraction immediately gives the other by subtraction from one. A mole fraction is independent of the overall sample size: doubling every component doubles numerator and denominator together.

Ethylene glycol mole fraction

Problem
Find the mole fractions in a 20% by mass ethylene glycol solution in water. Use molar masses 62 and 18 g mol⁻¹.

1.Choose a 100 g solution basis: 20 g glycol and 80 g water. This convenient basis does not restrict the result to a 100 g sample.
2.Glycol amount = 20/62 = 0.322581 mol; water amount = 80/18 = 4.444444 mol.
3.Total amount = 4.767025 mol. x(glycol) = 0.322581/4.767025 = 0.0676692 ≈ 0.0677; x(water) = 0.9323.
4.The glycol mass fraction is 0.20, but its mole fraction is much smaller because glycol molecules have greater mass than water molecules.

Temperature and composition

For a closed mixture without reaction or phase loss, mass fractions, mass-based ppm and mole fractions remain unchanged on heating. Concentrations involving measured liquid volumes may change. Do not extend a temperature-independence statement to every possible meaning of ppm or percentage.

Quick Quiz

Quick check

Which sample is a liquid solution of a gas?

Explanation: Water is the liquid solvent and oxygen is molecularly dissolved. The other homogeneous examples have a solid solvent; droplets in air are heterogeneous.

Quick check

A 10% w/w solution of glucose contains 20 g glucose. What is the total solution mass?

Explanation: 20 g is 10% of the total, so total mass = 20/0.10 = 200 g, including 180 g water.

Quick check

What is the correct meaning of 2% w/v?

Explanation: w/v specifies grams of solute in 100 mL of final solution; it is neither mass percent nor molarity.

Quick check

A mixture contains 1 mol A and 3 mol B. What is x(A)?

Explanation: Divide the amount of A by total amount: 1/(1+3)=0.25.

Quick check

Which conversion is always valid for mass-based ppm?

Explanation: Mass-based ppm is a mass ratio of 10⁻⁶, exactly 1 mg per kg. A litre conversion requires density.

Practice Problems

A mixture contains 22 g benzene and 122 g carbon tetrachloride. Find both mass percentages.
Find the benzene mole fraction in a 30% by mass benzene–carbon tetrachloride mixture. Use molar masses 78 and 154 g mol⁻¹.
Water contains 15 ppm chloroform by mass. Express this as mass percent and identify the solute and solvent masses in 1 kg solution.
Find the mass percentage when 6.5 g aspirin is dissolved in 450 g acetonitrile.
Explain the nine solute–solvent state combinations and identify a gas-in-solid example.
How would you prepare 250 mL of a 4% v/v ethanol solution, and why is 240 mL water not a guaranteed amount?

Add the masses: total = 144 g. Benzene percent = 22/144 × 100 = 15.2778%, or 15.28%. Carbon tetrachloride percent = 122/144 × 100 = 84.72%. The percentages add to 100, providing a check.

Key Takeaways

• A solution is a homogeneous mixture; its solvent determines its physical state. • There are nine component-state combinations, grouped into gas, liquid and solid solutions. • Mass percent uses final solution mass; volume percent uses final solution volume. • Mass-by-volume percent means grams per 100 mL solution. • Mass-based ppm is milligrams per kilogram solution. • Mole fraction uses amounts in moles, and all component mole fractions sum to one. • Volume-based concentrations require attention to temperature and non-additive volumes.

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Next · Lesson 2

Molarity, Molality and Concentration Calculations