Part 2: Solutions, dilutions and gases

Keep units attached to every substitution when converting concentrations or applying the ideal gas equation.

Amount equals concentration times solution volume

For c in mol dm⁻³, volume must be in dm³. Divide cm³ by 1000. Mass concentration in g dm⁻³ = molar concentration × molar mass. Use final solution volume, not the amount of solvent poured in.

Worked example: 2.65 g Na₂CO₃ (M = 106) made up to 250.0 cm³ gives n = 0.0250 mol and c = 0.0250/0.2500 = 0.100 mol dm⁻³, or 10.6 g dm⁻³. Dissociation gives [Na⁺] = 0.200 mol dm⁻³.

Dilution conserves solute moles: c₁V₁ = c₂V₂. Diluting 20.0 cm³ of 0.500 mol dm⁻³ solution to 250.0 cm³ gives 0.0400 mol dm⁻³. The final volume is 250.0 cm³, not 270.0 cm³. Standard solution preparation is developed in the Acids topic.

Dilution and mixing: keep a mole inventory

When diluting 10.0 cm³ of 0.600 mol dm⁻³ NaCl to 150.0 cm³, initial amount is 0.600 × 0.0100 = 0.00600 mol. Final concentration is 0.00600/0.1500 = 0.0400 mol dm⁻³. The flask must be made up to the final mark; adding 150 cm³ water would make too much solution.

To mix two solutions of the same solute without reaction, add their amounts then divide by the final volume. Combining 20.0 cm³ of 0.200 mol dm⁻³ NaCl with 30.0 cm³ of 0.100 mol dm⁻³ NaCl gives 0.00400 + 0.00300 = 0.00700 mol. Assuming volumes add, c = 0.00700/0.0500 = 0.140 mol dm⁻³. The simple mean 0.150 is wrong because the volumes differ.

If the solutions react, perform reaction stoichiometry before calculating leftover concentrations. c₁V₁ = c₂V₂ describes conservation of the same solute in dilution; it does not replace the balanced equation for an acid–base reaction.

RTP volumes and reacting ratios

OCR’s data sheet uses a molar gas volume of 24.0 dm³ mol⁻¹ at room temperature and pressure. Use it when RTP is specified; it is an approximation, not a universal volume at every temperature. At the same temperature and pressure, gas volumes follow their mole ratios.

For N₂ + 3H₂ → 2NH₃, 30 cm³ N₂ requires 90 cm³ H₂ and could produce 60 cm³ gaseous NH₃ at the same conditions if reaction is complete. Real equilibrium yield may be lower.

Worked example: 72.0 cm³ H₂ at RTP represents 72.0/24000 = 0.00300 mol. Mg + 2HCl → MgCl₂ + H₂ means it came from 0.00600 mol HCl. The coefficient ratio comes before any concentration calculation.

Count all gaseous products, then any excess gas

Given 2KNO₃(s) → 2KNO₂(s) + O₂(g), 0.0600 mol nitrate forms 0.0300 mol gas, not 0.0600 mol gas. Only gases contribute to the gas volume. In an equation producing two different gases, add their amounts before using V = nVm or pV = nRT.

Worked mixture: 20.0 cm³ CH₄ reacts with 50.0 cm³ O₂ at the same conditions. CH₄ + 2O₂ → CO₂ + 2H₂O uses 40.0 cm³ oxygen, leaves 10.0 cm³ oxygen and forms 20.0 cm³ CO₂. After cooling so water condenses, the dry gas volume is 30.0 cm³. If water remains gaseous, its contribution must instead be included at the stated temperature and pressure.

For an ideal gas sample with unchanged amount, pV/T remains constant. Compressing 120 cm³ at 100 kPa to 80 cm³ at unchanged temperature gives p₂ = 100 × 120/80 = 150 kPa. This relationship cannot be used unchanged if reaction or leakage changes the gas amount.

Use SI units in pV = nRT

With R = 8.314 J K⁻¹ mol⁻¹, use pressure in Pa, volume in m³ and temperature in K. Convert kPa × 1000, cm³ ÷ 1 000 000 and °C + 273 (or 273.15 where justified). This model approximates real gases best at lower pressures and higher temperatures.

Worked example: 100 cm³ gas at 100 kPa and 300 K contains n = (100000 × 0.000100)/(8.314 × 300) = 0.004009 mol, or 4.01 × 10⁻³ mol to three significant figures. A mass of 0.176 g would give M = 0.176/0.004009 = 43.9 g mol⁻¹.

A gas syringe measures volume directly. A leak or delay in sealing loses gas; collection over water can underestimate soluble gases. Measure the actual temperature and pressure. Do not use 24 dm³ mol⁻¹ automatically for heated gases.

Predict the direction of a gas-measurement error

In a molar-mass experiment M = mRT/(pV). Suppose the weighed volatile sample partly escapes before its vapour volume is measured. If the calculation still uses the original full mass, measured V is too small and calculated M is too high. Trace the error through the expression rather than writing “the answer is inaccurate”.

If the apparatus still contains air, the measured total volume includes gas not belonging to the sample, potentially making M too low if all that volume is attributed to the sample. State the experimental arrangement: errors have a direction only once you identify what was measured and what the calculation assumes.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources