Solutions · Lesson 2 of 11
Molarity, Molality and Concentration Calculations
“Convert between solution composition measures using conservation of mass, moles and carefully defined volumes.”
• Distinguish molarity from molality and explain their temperature dependence. • Convert mass percentages into molarity, molality and mole fraction. • Use solute conservation for dilution and mass balance for mixing. • Combine concentration calculations with balanced reactions.
Two concentrations with different reference amounts
Two solutions may contain the same mass of solute but behave differently because that solute is dispersed through different amounts of medium. Molarity compares solute amount with final solution volume. Molality compares it with solvent mass. Their symbols look similar, but their denominators answer different questions.
Problem
Find the molarity of 5.0 g NaOH in 450 mL final solution. Molar mass = 40 g mol⁻¹.
- 1.Convert mass to moles because molarity counts chemical amounts: n = 5.0/40 = 0.125 mol.
- 2.Convert volume to litres: 450 mL = 0.450 L.
- 3.C = 0.125/0.450 = 0.277778 mol L⁻¹ ≈ 0.278 M. This does not describe 450 mL of water before dissolution.
Problem
Find the molality of 2.5 g ethanoic acid, molar mass 60 g mol⁻¹, in 75 g benzene.
- 1.Acid amount = 2.5/60 = 0.0416667 mol.
- 2.Benzene mass = 75/1000 = 0.075 kg.
- 3.m = 0.0416667/0.075 = 0.555556 mol kg⁻¹ ≈ 0.556 mol kg⁻¹. This analytical molality is calculated before considering association of acid molecules.
Heating a closed solution usually expands it without changing its solute amount. Molarity therefore usually decreases on heating. Molality, mass fraction and the composition mole fractions remain constant provided there is no loss, reaction or phase separation. Molarity and molality can be numerically similar for very dilute water solutions near density 1 g mL⁻¹, but that approximation must be justified rather than assumed.
A reliable conversion method
Choose a convenient basis first. For a mass percentage, 100 g solution is usually simplest. Calculate the separate solvent and solute masses, convert each to moles when needed, and calculate volume only if density is given. Density is mass of solution divided by solution volume, so it converts total solution mass into final volume. It does not change solvent mass into solution mass.
Problem
A 10% w/w glucose solution has density 1.20 g mL⁻¹. Find molality, molarity and component mole fractions. Use glucose 180 and water 18 g mol⁻¹.
- 1.Choose 100 g solution: glucose = 10 g, water = 90 g. Glucose amount = 10/180 = 0.0555556 mol; water amount = 90/18 = 5.000 mol.
- 2.Molality = 0.0555556/0.090 = 0.617284 mol kg⁻¹. The denominator is solvent mass.
- 3.Volume of solution = 100/1.20 = 83.3333 mL = 0.0833333 L. Molarity = 0.0555556/0.0833333 = 0.666667 mol L⁻¹.
- 4.x(glucose) = 0.0555556/5.0555556 = 0.0109890; x(water) = 0.989011. Different denominators explain the different numerical values.
To derive the first expression, choose 1 kg solvent. Solute amount is m moles and its mass is mM₂ grams. Total solution mass is 1000 + mM₂ grams. Divide this mass by ρ to obtain mL, then by 1000 to obtain litres. Dividing m moles by that volume gives the displayed expression. For the reverse expression, multiply both sides by 1000 + mM₂, collect the terms containing m, and divide by 1000ρ − CM₂. A non-positive denominator signals inconsistent data or units.
Problem
How much urea, molar mass 60 g mol⁻¹, is required for 2.5 kg of a 0.25 mol kg⁻¹ aqueous solution?
- 1.Let urea mass be w grams. Water mass is (2500 − w) g because 2.5 kg refers to the final solution.
- 2.Use the definition: 0.25 = (w/60)/[(2500 − w)/1000].
- 3.Multiply to obtain 15(2500 − w) = 1000w, so 37500 = 1015w and w = 36.9458 g.
- 4.The required mass is about 36.95 g. Treating 2.5 kg as water would give a different amount and the wrong final solution mass.
Dilution conserves solute; mixing conserves component amounts
Adding solvent increases the denominator of molarity while leaving the solute amount unchanged. Volumes are final measured solution volumes, so the dilution equation does not require assuming that component volumes add. When two solutions are mixed, add their actual solute amounts; if mass percentages are given, add solute masses and total masses. Use a weighted average rather than a simple average unless the reference amounts are equal.
Problem
Mix 300 g of a 25% w/w solution with 400 g of a 40% w/w solution of the same solute. Find the new mass percent.
- 1.First portion contains 0.25 × 300 = 75 g solute; second contains 0.40 × 400 = 160 g.
- 2.Total solute mass = 235 g; total solution mass = 700 g.
- 3.New percent = 235/700 × 100 = 33.5714% ≈ 33.57%. The answer is closer to 40% because more of the stronger solution was used.
Use the balanced reaction before using concentration
Problem
A 1.00 g mixture contains equimolar Na₂CO₃ and NaHCO₃. Find the volume of 0.100 M HCl needed for complete reaction. Use molar masses 106 and 84 g mol⁻¹.
- 1.Balanced reactions: Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O; NaHCO₃ + HCl → NaCl + CO₂ + H₂O. They require two and one moles of acid per mole, respectively.
- 2.Let each salt amount be n. Then 106n + 84n = 1.00, giving n = 1/190 mol. Equimolar does not mean equal masses.
- 3.Required HCl amount = 2n + n = 3/190 = 0.0157895 mol.
- 4.V = n(HCl)/C = 0.0157895/0.100 = 0.157895 L ≈ 158 mL. Each equation conserves all atoms and net charge.
“92 g sodium ions per kg of water” specifies solvent mass, whereas “per kg of lake water” may mean total solution mass. Those are different denominators. Counterions also contribute mass. Do not force a unique molarity from ambiguous wording when density and composition do not specify the full solution mass.
Quick Quiz
Quiz
Which denominator defines molality?
Explanation: Molality is amount of solute divided by kilograms of solvent, whereas molarity uses litres of final solution.
A closed solution expands on warming without reaction. What usually happens?
Explanation: Expansion increases solution volume. Amount and solvent mass stay unchanged, so only the volume-based concentration changes.
30.0 mL of 0.500 M acid is diluted to 500 mL. What is its final molarity?
Explanation: C₂ = C₁V₁/V₂ = 0.500 × 30.0/500 = 0.0300 mol L⁻¹.
A 100 g solution has density 1.25 g mL⁻¹. What volume should be used for molarity?
Explanation: V = mass/density = 100/1.25 = 80 mL = 0.080 L.
An equimolar mixture of Na₂CO₃ and NaHCO₃ has n moles of each. What HCl amount is required for complete reaction?
Explanation: Carbonate consumes 2n moles HCl and hydrogen carbonate consumes n, giving 3n altogether.
Practice Problems
- Find the molarity of 68% by mass nitric acid of density 1.504 g mL⁻¹. Use 63 g mol⁻¹.
- Find molality, molarity and KI mole fraction for 20% w/w aqueous KI of density 1.202 g mL⁻¹. Use KI 166 and water 18 g mol⁻¹.
- Calculate the molarity of 30 g Co(NO₃)₂·6H₂O in 4.3 L solution and the final concentration when 30 mL of 0.5 M H₂SO₄ is diluted to 500 mL.
- An antifreeze mixture contains 222.6 g ethylene glycol and 200 g water; its density is 1.072 g mL⁻¹. Find molality and molarity, using 62 g mol⁻¹.
- A hypothetical aqueous solute of molar mass 311 g mol⁻¹ has molality 1.5 × 10⁻³ mol kg⁻¹. What mass of solution contains 1.5 mg solute?
- What mass of benzoic acid, 122 g mol⁻¹, is needed to prepare 250 mL of 0.15 M solution in methanol?
- A water sample has 15 ppm chloroform by mass. Calculate its molality using chloroform 119.5 g mol⁻¹.
- A lake-water sample contains 92 g Na⁺ per kg of final solution and has density 1.25 g mL⁻¹. Calculate Na⁺ molarity. What if the kilogram meant pure solvent instead?
Choose 100 g solution. HNO₃ amount = 68/63 = 1.079365 mol. Solution volume = 100/1.504 = 66.48936 mL = 0.06648936 L. Divide to obtain C = 16.23365 mol L⁻¹ ≈ 16.2 M. Density refers to the complete solution.
Key Takeaways
• Molarity counts moles per litre of final solution; molality counts moles per kilogram of solvent. • Use the full molar mass of a weighed hydrate. • Density connects final solution mass with final solution volume. • A chosen 100 g or 1 kg basis simplifies concentration conversions. • Dilution conserves solute amount when there is no reaction or loss. • Mixing requires adding component masses or amounts before dividing. • Balanced equations determine the stoichiometric amount required in a reaction. • Ambiguous solvent-versus-solution wording can prevent a unique numerical answer.