2: Concentration Metrics

Molality is m = n_solute/m_solvent, where the solvent mass must be in kilograms. Rearrange the equation to obtain n_solute = m × m_solvent. Substitution gives n = 1 mol kg⁻¹ × 0.5 kg = 0.5 mol. The kilogram units cancel, leaving moles. Therefore option A is correct; using 1 or 2 would ignore the given 0.5 kg mass.

For a dilute aqueous solution, the governing approximation is 1 ppm ≈ 1 mg of solute per kilogram of water or solution. Since the sample contains 3 mg fluoride in 1 kg water, its concentration is approximately 3 ppm, so option A is correct. This is not 0.3, 30, or 300 ppm because the given milligram-to-kilogram ratio is already three parts per million.

The governing concept is the basis used to define a concentration term. Molarity is moles of solute per litre of solution, so it depends on solution volume. Volume percentage is the volume of solute divided by the volume of solution, multiplied by 100. Molality uses solvent mass, mole fraction uses mole ratios, and mass percentage uses mass; therefore option A is the only pair containing two volume-based terms.

A mole fraction is defined as xᵢ = nᵢ / nₜₒₜₐₗ, the moles of one component divided by the total moles of all components. Since both quantities have the unit mol, the units cancel and the result is dimensionless. Molarity has mol L⁻¹, molality has mol kg⁻¹, and mass-volume percentage combines mass and volume; hence A is correct.

For a solution containing exactly two components, the governing relation is x₁ + x₂ = 1, because the component mole amounts add to the total number of moles. Thus, if x₁ = 0.45, then x₂ = 1 − 0.45 = 0.55. Mole fractions must lie between 0 and 1. Choosing 0.45 would give a total of 0.90, while 1.45 is impossible; therefore A is correct.

Mass percentage is defined as (mass of solute / mass of solution) × 100. Rearranging gives mass of solute = (mass percentage / 100) × mass of solution. Substitution gives (5 / 100) × 200 g = 10 g, so A is correct. The denominator is the total solution mass, not the solvent mass; 5 g would correspond to 100 g of solution.

The governing formula is mass percentage = (solute mass / solution mass) × 100. Hence solute mass = (10 / 100) × 300 g = 30 g. This is also clear because 10% means one-tenth of the total solution mass, and one-tenth of 300 g is 30 g. Option B would apply to a 100 g solution, while D represents the entire solution, not just the solute.

Volume percentage is defined as (volume of solute / volume of solution) × 100. Therefore, volume of solute = (20 / 100) × 250 mL = 50 mL. The reference quantity is the total solution volume, not the solvent volume. Option B confuses the percentage with the actual volume, while C and D result from incorrect factors; thus A is correct.

A 2% mass-volume, or m/V, solution contains 2 g of solute in every 100 mL of solution. Since 500 mL is five times 100 mL, the solute mass is 2 × 5 = 10 g. Equivalently, mass = (2 / 100) × 500 = 10 g. This is not mass percentage because the reference quantity is solution volume; therefore A is correct.

Molarity is moles of solute per litre of solution, so the governing relation is n = M × V when volume is in litres. Substituting M = 0.1 mol L⁻¹ and V = 2 L gives n = 0.1 × 2 = 0.2 mol. The units mol L⁻¹ and L cancel to leave mol. Option B is the amount in one litre, while C and D are excessive; hence A is correct.

Molarity relates moles to solution volume in litres: n = M × V. First convert 200 mL to 0.200 L. Then n = 1.5 mol L⁻¹ × 0.200 L = 0.300 mol, so A is correct. The unit litre cancels, leaving moles. Forgetting the millilitre-to-litre conversion would produce an answer 1000 times too large; 1.5 mol would be the amount in one litre.

Molarity is the amount of solute in moles divided by the volume of the final solution in litres: M = n/V. Convert 600 mL to 0.600 L, then calculate M = 0.3 mol ÷ 0.600 L = 0.50 mol L⁻¹, or 0.5 M. Option B ignores the volume, while C and D result from incorrect multiplication or division. The answer is therefore A.

The governing relation is M = moles of solute ÷ litres of solution. First convert 400 mL into 0.4 L. Thus M = 0.8 mol ÷ 0.4 L = 2 mol L⁻¹ = 2 M, so option A is correct. Option B is merely the given amount in moles, option C incorrectly divides by 2, and option D doubles the correct value through an arithmetic error.

Molarity is defined as the number of moles of solute present per litre of the final solution. Therefore, a 1 M solution contains 1 mol solute in 1 L solution, which is exactly option A. Option B defines 1 molal solution because it refers to kilograms of solvent. Option C expresses mass percentage, and option D expresses parts per million, so neither describes molarity.

Mass by mass percentage is calculated as (mass of solute ÷ mass of solution) × 100. Therefore, a 5% solution contains 5 g of solute in every 100 g of the complete solution. Option B uses 100 g solvent, not solution, while C is mass-volume percentage and D describes molarity.

Mass-volume percentage is calculated as (mass of solute in grams ÷ volume of final solution in millilitres) × 100. Consequently, a 5% m/v solution contains 5 g solute in every 100 mL of solution, so A is correct. B is mass by mass percentage, C is volume by volume percentage, and D describes a molality-type quantity because it uses moles per kilogram of solvent.

Volume-by-volume percentage is defined as (volume of solute ÷ volume of final solution) × 100. Therefore, 5% v/v means 5 mL of solute is present in 100 mL of the final solution, making A correct. B is mass percentage, C mixes grams with millilitres and refers to solvent rather than final solution, while D is a molarity statement, not a volume percentage.

ppm means parts per million. Thus, 1 ppm represents one part of solute in one million parts of the sample or solution, using a consistent basis such as mass or volume; option A is correct. One part in 100 or 1000 corresponds to much higher concentrations. Option D gives moles per litre, which is molarity and uses a different concentration scale.

Molarity is defined as moles of solute divided by volume of solution in litres. If the numerator remains constant while the denominator increases on heating, the quotient becomes smaller. Thus molarity decreases. This temperature dependence is why molarity is less convenient than molality for some thermal studies.

Molality is moles of solute per kilogram of solvent. Mass is essentially independent of ordinary temperature changes, whereas solution volume can expand or contract. Therefore molality remains comparatively stable with temperature. Molarity and volume-based percentages can change when volume changes.

Molality is defined as moles of solute per kilogram of solvent, with unit mol kg⁻¹, not moles per litre. Moles per litre of final solution is the definition and unit of molarity. Mole fraction is a ratio and therefore unitless, while ppm means parts per million. Hence option A is the wrong pair; the other three descriptions are correct.

Mole fraction is the ratio of moles of one component to total moles: xᵢ = nᵢ/Σn. Because both numerator and denominator are measured in moles, the units cancel and the result is dimensionless. Molarity uses solution volume, molality uses solvent mass, and percentage formulas are multiplied by 100.

Molarity is the amount of solute in moles divided by the total volume of solution in litres: M = n/V. For A, M = 2/1 = 2 M; for B, it is 1/1 = 1 M; for C, it is 1/2 = 0.5 M; and for D, it is 1/4 = 0.25 M. Since 2 M is the greatest concentration, option A is correct. The distractors have either fewer moles or a larger volume.

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of the complete solution. Thus A gives 0.2/1 = 0.2 m, B gives 1/1 = 1 m, C gives 1/0.5 = 2 m, and D gives 2/2 = 1 m. The smallest value is 0.2 m, so option A is correct. Options B and D are equal, while C is the largest.

Very small amounts are called trace concentrations, and ppm provides a convenient numerical scale for them. Mole fraction cannot exceed 1, molarity remains meaningful even for dilute solutions, and a small solute amount gives a small—not automatically 100%—mass percentage. Therefore A is correct.