Molarity and Dilution

Introduction

Molarity expresses solute amount per final volume of solution, making it especially useful for reactions carried out by measured liquid volumes. Preparing or diluting a solution requires attention to the word final : the relevant volume is the completed mixture, not the volume of solvent poured in.

Core explanation

Molarity c = n(solute)/V(solution), with common unit mol L⁻¹ or M. A solution containing 0.500 mol NaCl in a final volume of 2.00 L has c = 0.250 mol L⁻¹. The volume must be in litres for this unit: 250 mL is 0.250 L. If mass rather than solute amount is given, divide mass by molar mass before using the molarity equation.

To prepare a standard solution, weigh a calculated amount of solute, dissolve it in less than the target volume of solvent, then add solvent until the final solution reaches the calibration mark. Adding “one litre of water” to solute does not necessarily create exactly one litre of solution. Dissolution and mixing can change volume, and the solute occupies or influences volume.

For a dilution with no chemical loss of solute, n before = n after. Since n = cV, c₁V₁ = c₂V₂ when volumes and concentrations use compatible units and refer to the same solute. For example, to make 250 mL of 0.100 M solution from 1.00 M stock, stock volume V₁ = (0.100 × 250)/1.00 = 25.0 mL. Place that stock portion in a vessel and bring the final mixture to 250 mL. Do not simply add 250 mL water to the stock aliquot.

Molarity can vary with temperature because solution volume expands or contracts while solute amount remains the same. The effect may be small for routine work but matters in precise preparation. Molality uses solvent mass instead and has a different temperature dependence. Likewise the concentration of ions after dissolution may differ from the formula-unit molarity because each formula unit can supply multiple ions.

Mixing solutions with different concentrations is best handled through mole balances: add cᵢVᵢ for each portion, then divide total solute moles by final measured volume. Volumes of mixtures are not always exactly additive, so high-precision work uses final volume rather than assuming it equals the sum of starting volumes.

Step-by-step reasoning

1. Identify solute amount and final solution volume; convert mL to L. 2. Use c = n/V or n = cV for a single solution. 3. In dilution, equate initial and final solute moles. 4. Solve c₁V₁ = c₂V₂ and interpret V₂ as final volume.

Visual explanation

Draw a concentrated small cylinder transferring an aliquot into a larger volumetric flask. The flask mark labels final volume; added solvent fills the remainder below that mark.

Real-world analogy

Adding water to a concentrated drink spreads the same amount of flavoring through more liquid. The flavoring count is conserved, while concentration falls as final volume rises.

Real-world example

Laboratories prepare calibration solutions by pipetting a measured stock aliquot into a volumetric flask and filling to the mark. The known stock concentration and two volumes determine the final molarity.

Why?

Why does c₁V₁ = c₂V₂ work for a simple dilution? Each product cV is the same solute mole amount before and after solvent addition, provided none reacts or escapes.

Common misconception

“A 250 mL final solution needs 250 mL of added solvent.” The solute and any stock solution contribute to final volume, so fill to 250 mL total.

Worked example

How many grams NaOH make 500.0 mL of 0.200 M solution? Required n = 0.200 mol L⁻¹ × 0.5000 L = 0.100 mol. With M(NaOH) ≈ 40.00 g mol⁻¹, mass = 4.00 g. Dissolve safely, allow temperature to settle as needed, and bring final volume to 500.0 mL.

Quick check

1. What stock volume of 2.00 M solution is needed for 100 mL of 0.500 M solution? Answer: V₁ = (0.500 × 100)/2.00 = 25.0 mL stock, diluted to 100 mL final volume.

Exam focus

Write “L of solution” in the denominator. Use c₁V₁ = c₂V₂ only when solute amount is conserved and both sides describe the same solute.

Advanced insight

At high concentration, solution volume may differ appreciably from the sum of the component volumes. The operational definition of molarity avoids that ambiguity by referring to the measured final solution volume.

Summary

Molarity is moles of solute per litre of final solution. In a simple dilution, conserved solute amount gives c₁V₁ = c₂V₂, and the target volume is reached by filling the mixture to its final mark.

Practice questions

1. Find molarity for 0.0500 mol solute in 250 mL solution. Answer: 0.0500/0.250 = 0.200 M. 2. Does adding water change solute moles in an ordinary dilution? Answer: No, provided no solute reacts or is lost. 3. Why can 100 mL stock plus 100 mL solvent be an imprecise way to specify final volume? Answer: Mixed volumes need not add exactly; measure the final solution volume directly.