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Unit I · Solutions
12 min readConcept sheet1-mark MCQ on which term changes with temperature; 2 or 3-mark numericals on mole fraction, molarity and molality; also a NEET favourite

Types of solutions and ways to express concentration

A solution is a homogeneous mixture. Its concentration can be given as percentage, ppm, mole fraction, molarity or molality; molality does not change with temperature, molarity does.

Molarity (M)Moles of solute per litre of solution: M = n₂ ÷ V (in L); unit mol L⁻¹
Molality (m)Moles of solute per kilogram of solvent: m = n₂ ÷ w₁ (in kg); unit mol kg⁻¹
Mole fraction (x)x₁ = n₁ ÷ (n₁ + n₂); no unit; the mole fractions in a solution add up to 1
Parts per million (ppm)(Mass of component ÷ total mass of solution) × 10⁶; used for very small amounts

Solute and solvent

A solution is a homogeneous mixture: its composition and properties are the same all through. A binary solution has two components. The solvent is the component present in the larger amount, and it decides the physical state of the solution (solid, liquid or gas). The other component is the solute. Most of this chapter is about liquid solutions.

Types of solutions

Type of solutionSoluteSolventExample
GaseousGasGasOxygen and nitrogen mixed together (air)
GaseousLiquidGasChloroform vapour mixed with nitrogen
GaseousSolidGasCamphor vapour in nitrogen
LiquidGasLiquidOxygen dissolved in water
LiquidLiquidLiquidEthanol dissolved in water
LiquidSolidLiquidGlucose or sugar dissolved in water
SolidGasSolidHydrogen dissolved in palladium
SolidLiquidSolidAmalgam of mercury with sodium
SolidSolidSolidCopper dissolved in gold (gold jewellery)

Ways to express concentration

TermHow it is worked outUnitChanges with temperature?
Mass percentage (w/w)(Mass of component ÷ total mass of solution) × 100%No
Volume percentage (V/V)(Volume of component ÷ total volume of solution) × 100%Yes
Mass by volume percentage (w/V)Mass of solute in g per 100 mL of solutiong per 100 mLYes
Parts per million (ppm)(Parts of component ÷ total parts of solution) × 10⁶ppmNo, when by mass
Mole fraction (x)Moles of the component ÷ total molesNoneNo
Molarity (M)Moles of solute ÷ volume of solution in Lmol L⁻¹Yes
Molality (m)Moles of solute ÷ mass of solvent in kgmol kg⁻¹No

Molarity vs molality

Molarity (M)Moles of solute per litre of solution. It depends on volume. A liquid expands when warmed, so the same solution has a lower molarity at a higher temperature.
Molality (m)Moles of solute per kilogram of solvent. It depends only on masses, and mass does not change with temperature, so molality is independent of temperature.

That is why the boiling point and freezing point formulas use molality.

Worked example: one solution, three ways

Problem

23 g of ethanol (C₂H₅OH, M = 46 g mol⁻¹) is dissolved in 90 g of water (M = 18 g mol⁻¹). Find (a) the mass percentage of ethanol, (b) the mole fraction of each component and (c) the molality of the solution.

Working
  1. Moles of ethanol, n₂ = 23 g ÷ 46 g mol⁻¹ = 0.5 mol. Moles of water, n₁ = 90 g ÷ 18 g mol⁻¹ = 5 mol.
  2. (a) Mass percentage = 23 ÷ (23 + 90) × 100 = 23 ÷ 113 × 100 = 20.4%.
  3. (b) x(ethanol) = 0.5 ÷ (0.5 + 5) = 0.5 ÷ 5.5 = 0.091.
  4. x(water) = 5 ÷ 5.5 = 0.909. Check: 0.091 + 0.909 = 1.
  5. (c) Molality = 0.5 mol ÷ 0.090 kg = 5.56 mol kg⁻¹.
Answer

(a) 20.4% ethanol by mass; (b) x(ethanol) = 0.091, x(water) = 0.909; (c) 5.56 mol kg⁻¹

Worked example: molarity and ppm

Problem

(a) 2.0 g of NaOH (M = 40 g mol⁻¹) is dissolved in water and the volume is made up to 250 mL. Find the molarity. (b) A 500 g sample of well water contains 0.75 mg of fluoride ions. Express this in ppm.

Working
  1. (a) n = 2.0 g ÷ 40 g mol⁻¹ = 0.050 mol. V = 250 mL = 0.250 L.
  2. M = 0.050 mol ÷ 0.250 L = 0.20 mol L⁻¹.
  3. (b) Mass of F⁻ = 0.75 mg = 0.75 × 10⁻³ g. Mass of sample = 500 g.
  4. ppm = (0.75 × 10⁻³ g ÷ 500 g) × 10⁶ = 1.5 × 10⁻⁶ × 10⁶ = 1.5 ppm.
Answer

(a) 0.20 M; (b) 1.5 ppm of fluoride

In real life

A hospital drip of normal saline is a 0.9% (mass by volume) solution of NaCl, and the fluoride in drinking water is reported in ppm. Every label on a medicine bottle is a concentration statement.

Exam angle

2 marks: “Define molarity and molality. Which of the two is preferred when the temperature changes, and why?” Molarity = moles of solute per litre of solution; molality = moles of solute per kilogram of solvent. Molality is preferred because it depends on mass, which does not change with temperature; volume does.

3 marks: “18 g of glucose (C₆H₁₂O₆, M = 180 g mol⁻¹) is dissolved in 90 g of water. Find the mole fraction of glucose and the molality of the solution.” n(glucose) = 18 ÷ 180 = 0.1 mol; n(water) = 90 ÷ 18 = 5 mol. x(glucose) = 0.1 ÷ 5.1 = 0.0196. Molality = 0.1 mol ÷ 0.090 kg = 1.11 mol kg⁻¹.

1 marks: “Name two ways of expressing concentration that do not depend on temperature.” Any two of: mass percentage, mole fraction, molality, ppm by mass.

Remember it

Capital M for the Measuring flask (litres of solution); small m for mass (kilograms of solvent). Masses never change with temperature, so mass %, mole fraction and molality stay fixed.