Molarity, Solution Preparation and Dilution

Molarity, symbol M, equals moles of solute divided by litres of solution. The denominator is the total volume of the final solution, which is why solutions are made up to a mark rather than by adding a measured volume of solvent.

The solute is the dissolved substance, the solvent is the dissolving medium, and the solution is the homogeneous mixture of both. In a dilute aqueous NaCl solution, water is the solvent and NaCl supplies the solute formula units. Volumes need not add exactly because mixing changes intermolecular arrangements. A 250.0 mL solution therefore is not prepared by adding 250.0 mL water to a weighed solid.

For complete dissociation of 0.100 M CaCl₂, the formula concentration is 0.100 mol CaCl₂ per liter, while [Ca²⁺] = 0.100 M and [Cl⁻] = 0.200 M. These ion concentrations follow subscripts, assuming complete dissolution and dissociation. Molarity of a weak acid does not by itself equal its H⁺ concentration because dissociation is incomplete.

To make 250.0 mL of 0.100 M NaCl, first find the moles needed, which is 0.2500 L x 0.100 mol/L = 0.0250 mol. Multiply by the molar mass to get 1.46 g. Weigh that mass, dissolve it in less than 250 mL of water, transfer to a 250.0 mL volumetric flask, then add water to the calibration mark and invert to mix.

Transfer the weighed solid quantitatively: dissolve it in a suitable portion of solvent, rinse the weighing or transfer vessel so all solute reaches the volumetric flask, and approach the final mark carefully. Mix thoroughly after making up to volume. Read the appropriate meniscus at eye level and use the flask at its calibration temperature; do not make up a hot solution and assume its cooled volume is unchanged. Overshooting the mark adds solvent but not solute, giving a concentration below the target.

Adding solvent spreads the same moles of solute through a larger volume. Because moles do not change, M₁V₁ = M₂V₂. The volumes may be in any unit as long as both sides match, which makes millilitres perfectly acceptable here.

Derive the dilution equation from n_before = n_after: M₁V₁ = M₂V₂. The aliquot V₁ is the stock solution volume transferred, and V₂ is the final total solution volume, not the water added. A 10.0 mL aliquot diluted to 100.0 mL has a dilution factor of ten and one-tenth the concentration. Taking a second 10.0 mL aliquot and diluting it to 100.0 mL gives a second factor of ten, for an overall factor of one hundred. The stock bottle itself is not diluted merely by removing an aliquot.

Molarity uses liters of solution, molality uses kilograms of solvent, and mass percent uses mass of solute divided by mass of solution times 100. Molarity can change with temperature because volume changes; molality does not change merely because the solution thermally expands.

Parts per million by mass means mg solute per kg solution. For a very dilute aqueous solution with density close to 1.00 kg/L, this is approximately mg/L; the equivalence is not universal for other solvents or dense solutions. Read the denominator before converting concentration units. In a molarity calculation n = MV requires liters when M is mol/L, while a dilution equation may use mL on both sides because the same conversion factor cancels.

When two solutions of the same solute are mixed, add their solute moles before dividing by final solution volume. M₁V₁ = M₂V₂ describes dilution of one sample with solvent; it does not account for solute already in a second solution. For a simple classroom calculation, assume volumes are additive when the problem says so.

If two solutions react, first calculate each solute's moles, apply stoichiometry and find what remains; M₁V₁ = M₂V₂ is not a reaction equation. Even for the same solute with no reaction, use (M₁V₁ + M₂V₂)/V_final, with consistent units. Equal-volume mixing gives the arithmetic mean of the two concentrations; unequal volumes require a weighted result. A final concentration between the starting values is a useful check when neither reaction nor solvent loss occurs.