2: Concentration Metrics

Mass-volume percentage expresses the grams of solute present in 100 mL of solution. The formula is (mass of solute in g / volume of solution in mL) × 100. Therefore, (1 / 500) × 100 = 0.2%. Option A is correct. The answer is not 1% because the given 1 g is distributed through 500 mL, not 100 mL.

Volume percentage expresses the volume of solute present in every 100 mL of solution. Mathematically, it is (volume of solute ÷ volume of solution) × 100, so option A gives the correct meaning. Option B describes mass percentage, option C describes molality, and option D describes the basis of molarity. The denominator and units distinguish these concentration terms.

For a dilute aqueous solution, ppm by mass is approximately milligrams of solute per kilogram of water. Thus, ppm = 10 mg ÷ 2 kg = 5 mg kg⁻¹, or 5 ppm. This also follows from converting 2 kg water to about 2,000,000 mg and forming the ratio in parts per million. Therefore, option A is correct; the other values use an incorrect mass ratio.

The governing dilution principle is conservation of solute moles: M₁V₁ = M₂V₂. Substituting the data gives M₂ = (0.2 mol L−1 × 250 mL) ÷ 500 mL = 0.10 mol L−1. The volume doubles while the solute amount remains constant, so the concentration becomes half. Thus, option A is correct; option B ignores dilution, while C and D increase the concentration incorrectly.

Dilution follows the conservation relation M₁V₁ = M₂V₂ because the number of solute moles does not change. Therefore, V₂ = (M₁V₁)/M₂ = (1.0 × 100 mL)/0.2 = 500 mL. A lower target concentration requires a larger final volume, so option A is correct. The other options either fail to apply the dilution equation or would not reduce the concentration sufficiently.

For an acid–base reaction, HCl is monoprotic: each mole supplies one acidic H⁺, so its n-factor is 1. Normality is N = molarity × n-factor = 0.5 × 1 = 0.5 N. Hence option A is correct. Option B incorrectly doubles the molarity, option C halves it, and option D uses an unjustified factor of four. The result assumes complete neutralisation of HCl.

Parts per million, abbreviated ppm, expresses the amount of solute as the number of solute parts in one million total parts of solution or mixture. It is used for very dilute compositions. Therefore, option A gives the correct meaning. Option B describes a percentage-type ratio, option C defines molarity, and option D defines molality, so those alternatives represent different concentration measures.

Volume-by-volume percentage is calculated as (volume of liquid solute ÷ volume of final solution) × 100. Thus, a 5% v/v solution contains 5 mL of liquid solute in a final solution volume of 100 mL, making option A correct. Option B is mass percent, option C is mass/volume percent, and option D describes a molarity-like quantity rather than v/v percentage.

The governing concept is mole fraction, which measures the share of a component in the total moles of a solution. For the solute, X_solute = moles of solute ÷ total moles of all components. The total is 2 + 3 = 5 mol, so X_solute = 2 ÷ 5 = 0.4. Mole fraction is dimensionless and, for a component in an ordinary mixture, lies between 0 and 1. Therefore, option A is correct. Option B, 0.6, is the solvent mole fraction because 3 ÷ 5 = 0.6. Option C is an unjustified average of the given numbers, and option D is simply the amount of solute in moles, not its fraction of the mixture. The two mole fractions also add to 1, confirming the result.

During dilution, no solute is added or removed, so the number of solute moles remains constant. The governing relation is M1V1 = M2V2. Hence M2 = (1.2 mol L−1 × 250 mL) ÷ 600 mL = 300 ÷ 600 = 0.5 mol L−1. Therefore, option A is correct. The unchanged value 1.2 ignores dilution, while 2.0 incorrectly suggests concentration increases.

Dilution changes the volume but keeps the solute amount constant, so M1V1 = M2V2. Substitution gives V2 = (M1V1) ÷ M2 = (0.4 × 100 mL) ÷ 0.1 = 400 mL. Thus option A is correct. The final volume is four times the initial volume because the molarity is reduced from 0.4 M to one-fourth, 0.1 M; the other options do not conserve solute moles.

Mass percent is calculated as (mass of solute ÷ mass of solution) × 100. Rearranging, mass of solute = (15 ÷ 100) × 300 g = 45 g. Therefore, option A is correct. The remaining mass is 300 − 45 = 255 g, which is the solvent and explains option D; the percentage must be applied to the total solution mass, not to an arbitrary component.

A 12% by-mass solution contains 12 g solute in every 100 g of solution. For 250 g solution, solute mass = (12 ÷ 100) × 250 = 30 g. Since solution mass equals solute mass plus solvent mass, solvent mass = 250 − 30 = 220 g. Thus option A is correct. Option B is the solute mass, while 250 g and 280 g do not satisfy the stated composition.

The governing concept is volume percentage, which expresses the volume of solute in 100 mL of the final solution. Use volume percentage = (volume of solute ÷ volume of solution) × 100. Therefore, (30 mL ÷ 150 mL) × 100 = 20%. The denominator is the final solution volume, so option A is correct; the other values result from using an incorrect ratio.

Mass-volume percentage means the mass of solute, in grams, present in 100 mL of solution. Apply % m/v = (mass of solute ÷ volume of solution) × 100. Thus, (6 g ÷ 300 mL) × 100 = 2 g per 100 mL, or 2%. Option A is correct. Options B, C, and D do not follow the required mass-to-final-volume ratio.

For a dilute aqueous solution, ppm is numerically equivalent to milligrams of solute per kilogram of water. Therefore, ppm = 25 mg ÷ 5 kg = 5 mg/kg = 5 ppm. Option A is correct. Choosing 25 ignores the water mass, 125 multiplies instead of dividing, and 0.2 reverses the ratio. The approximation is appropriate for dilute water solutions.

In a binary solution there are only two components, so the sum of their mole fractions must equal one: x_solute + x_solvent = 1. Substituting x_solvent = 0.85 gives x_solute = 1 − 0.85 = 0.15. Therefore, option A is correct. Option B repeats the solvent fraction, option C exceeds the possible maximum of one, and option D has no valid calculation basis.

Molality m is defined as moles of solute divided by kilograms of solvent. Rearrange the relation m = n_solute/mass_solvent to get mass_solvent = n_solute/m. Substituting gives 0.5 mol ÷ 2 mol kg−1 = 0.25 kg. Option A is correct. Kilograms are required because molality uses solvent mass, whereas molarity would use solution volume.

Molarity is the number of moles of solute divided by the final volume of solution. During dilution, adding solvent increases the volume while the amount of solute remains essentially constant, so the ratio decreases. Therefore, option A is correct. Adding more solute, reducing volume, or concentrating the solution increases molarity rather than decreasing it.

Molarity is the ratio of moles of solute to the volume of the solution in litres. In a homogeneous solution, every portion has the same composition. Taking 100 mL removes solute and solution in the same proportion, so the ratio n/V remains unchanged. Hence the portion is still 0.5 M. Option B wrongly treats the smaller volume as lower concentration, while C and D do not follow the concentration ratio.

Molarity is defined as moles of solute per litre of solution, so its unit is mol L⁻¹. Multiplying it by volume in litres gives (mol L⁻¹)(L) = mol. If the volume is supplied in millilitres, it must first be divided by 1000. Grams, mass expressed in millilitres, and percentage temperature are not valid quantities for this relation. Therefore, option A is correct.

Molality is defined as the moles of solute present per kilogram of solvent, with unit mol kg⁻¹. Thus, multiplying molality by the solvent mass in kilograms gives (mol kg⁻¹)(kg) = mol. A mass in grams must be converted to kilograms first. Litres of solution belong to molarity, not molality, while the other listed conditions do not supply the required basis.

Mass percentage of solute is calculated as (mass of solute ÷ mass of solution) × 100. The total solution mass is 20 g + 180 g = 200 g. Therefore, mass percentage = (20 ÷ 200) × 100 = 10%. Option B incorrectly uses the solute mass alone, C effectively compares solute with solvent, and D represents the solvent percentage. Hence option A is correct.

A 10% by-mass solution contains 10 g of solute in every 100 g of solution. For a required solution mass of 500 g, solute mass = (10/100) × 500 = 50 g. If the system contains only solute and solvent, the remaining 450 g is solvent, not solute. Therefore, option A gives the required solute mass; D is the complementary solvent mass.

For a solution whose volume is expressed in litres, the amount of solute is calculated using n = M×V. Substituting the given values gives n = (0.2 mol L⁻¹)(1.5 L) = 0.30 mol. The values 0.2 and 1.5 are the molarity and volume separately, not the amount. Dividing them would give an unrelated ratio. Thus, option A is correct.