2: Concentration Metrics

Mass percentage is defined by mass percentage = (mass of solute ÷ mass of solution) × 100. The denominator is the total solution mass, which is usually the sum of solute and solvent masses when no material is lost. Using only solvent mass gives a different ratio, while either volume belongs to volume-based concentration measures rather than mass percentage.

First find the total mass of solution: 5 g salt + 95 g water = 100 g solution. Apply mass percentage = (mass of solute ÷ mass of solution) × 100. Therefore, (5 ÷ 100) × 100 = 5%. Option C is the percentage of water, not salt; 10% and 100% do not follow from the stated masses.

Volume percentage is calculated as (volume of solute ÷ volume of solution) × 100. It is especially convenient for liquid–liquid mixtures, such as alcohol and water, because liquid volumes can be measured readily. Solid–solid, solid–gas, and gas–solid systems are not usually represented most conveniently by this measure; mass, mole, or other bases may be preferred.

The mole fraction of component i is xᵢ = nᵢ ÷ Σn, where nᵢ is its amount and Σn is the total amount of all components. Both quantities are measured in moles, so their units cancel and mole fraction is dimensionless. For an ordinary component it lies between 0 and 1, so it cannot be always greater than 100 or limited to solids.

For a binary solution containing components A and B, xA = nA/(nA + nB) and xB = nB/(nA + nB). Adding them gives xA + xB = (nA + nB)/(nA + nB) = 1. Mole fractions describe the complete composition of the solution, so all component fractions together must equal unity, not 2, 0, or 100.

Normality is an equivalent-based concentration defined as the number of gram-equivalents of solute per litre of solution: N = equivalents/V in litres. The equivalent factor depends on the reaction, such as neutralisation or redox, so normality may be reaction-specific. Moles per kilogram of solvent defines molality, a mole ratio is not normality, and total mass alone is not a concentration definition. Hence A is correct.

Molarity is the amount of solute in moles divided by the final volume of the solution in litres: M = n/V. The denominator must be the total solution volume after mixing, not the mass or volume of solvent alone. Option B is the definition of molality, while C is a mass fraction-like ratio and D is a volume ratio. Therefore option A states the correct molarity formula.

Molality is defined as the number of moles of solute divided by the mass of solvent in kilograms: m = nsolute/kg solvent. It uses solvent mass rather than the final volume of the solution, so it is different from molarity. Option B is molarity, option C is a volume ratio, and option D is a mass-volume concentration expression. Thus option A gives the correct molality formula.

The mole fraction of a component is xᵢ = moles of component i ÷ total moles of all components in the solution. Hence, the denominator requires the sum of solute and solvent moles. Option A is correct. Molality uses kilograms of solvent, molarity uses litres of solution, and mass percentage uses mass values, so those distractors use different concentration bases.

Molality is defined as moles of solute per kilogram of solvent: m = n(solute)/mass of solvent in kg. Therefore, option A is correct. Mass percentage uses the total mass of solution, molarity uses the volume of the final solution, and volume percentage compares volumes. The distinction between solvent mass and solution mass is the key governing concept.

Molarity is the number of moles of solute present in one litre of the final solution. Thus, a molarity of 2 means 2 mol L⁻¹, or 2 moles of solute per 1 litre of solution, so option A is correct. Option B describes molality because it uses kilogram of solvent; options C and D use inappropriate mass or volume statements.

Mass percentage is governed by the formula: mass percentage = [mass of solute / mass of solution] × 100. Substituting the data gives [(15 g)/(150 g)] × 100 = 0.10 × 100 = 10%. Hence option A is correct. Option B incorrectly uses the solute number as the percentage, option C uses an incorrect ratio, and option D is half the correct result. The denominator must be the total mass of solution.

Mass-volume percentage expresses the mass of solute in grams present in 100 mL of solution. Equivalently, % (m/V) = [mass of solute in grams / volume of solution in mL] × 100. Here, (5/100) × 100 = 5%, so option A is correct. Because the volume is already 100 mL, the numerical percentage equals the grams of glucose. Options B, C, and D use incorrect scaling.

Molality is defined as the number of moles of solute divided by the mass of solvent in kilograms: m = n(solute)/mass(solvent in kg). Hence option A is correct. Unlike molarity, molality uses the solvent’s mass rather than the final solution’s volume, so it is less dependent on temperature-related volume changes. Option B defines molarity, while C and D are unrelated expressions.

Molarity is the amount of solute in moles divided by the final volume of solution in litres: M = n/V. Here n = 2 mol and V = 4 L, so M = 2/4 = 0.5 mol L⁻¹. Therefore option A is correct. The value 2 is only the amount of solute, not its molarity; multiplying by volume would also use the formula incorrectly.

Mass percentage is the mass of solute divided by the total mass of solution, multiplied by 100. The solution mass is 25 g + 225 g = 250 g. Therefore, mass percentage = (25/250) × 100 = 10%. Option A is correct. Dividing 25 by 225 gives about 11.1%, but that compares solute with solvent rather than with the complete solution.

Volume percentage is calculated as the volume of solute divided by the final volume of solution, multiplied by 100. The problem gives 40 mL solute and a final solution volume of 200 mL. Thus volume percentage = (40/200) × 100 = 20%. Option A is correct. The denominator is not the solvent volume; using 40 or 80 would confuse a component volume with the total solution volume.

Mass-volume percentage means grams of solute present in 100 mL of solution. The formula is (mass in grams/solution volume in mL) × 100. Substitution gives (2/250) × 100 = 0.8%. Equivalently, scaling 2 g in 250 mL to 100 mL gives 0.8 g per 100 mL. Hence option A is correct; 2% ignores the volume scaling.

In a binary solution there are exactly two components, solute and solvent. Their mole fractions must add to one: xsolute + xsolvent = 1. Therefore, xsolvent = 1 − 0.3 = 0.7, making option A correct. Option B repeats the solute fraction, option C exceeds the maximum possible mole fraction of one, and option D results from an incorrect decimal operation.

During dilution, the amount of solute remains constant while the solution volume increases. Using M1V1 = M2V2, M2 = (1 mol L⁻¹ × 1 L) ÷ 2 L = 0.5 mol L⁻¹. Equivalently, the volume doubles, so the molarity becomes half. Option B ignores dilution, while C and D incorrectly predict an increase. Thus, A is correct.

Use the dilution relation M1V1 = M2V2, with both volumes expressed in the same unit. Substitution gives M2 = (0.5 × 500) ÷ 1000 = 0.25 M. The volume doubles from 500 mL to 1000 mL, so the conserved amount of solute is spread through twice the volume. Therefore, option A is correct; the other choices fail to apply dilution properly.

The governing concept is conservation of mass in a solution: mass of solution equals mass of solute plus mass of solvent. Therefore, mass of solvent = mass of solution − mass of solute = 100 g − 15 g = 85 g. Thus option A is correct. Option B gives only the solute mass, option C gives the total solution mass, and option D incorrectly adds the solute to the already stated total.

Mass percentage means the mass of solute present in every 100 g of solution. Using the formula, mass percent = (mass of solute / mass of solution) × 100, the solute mass is (20/100) × 200 g = 40 g. Therefore option A is correct. Option D, 160 g, is the mass of solvent because it is the remaining 80% of the solution.

Mass percentage is calculated as (mass of solute / mass of solution) × 100. Substituting the given values gives (5 g / 50 g) × 100 = 0.1 × 100 = 10%. Therefore option A is correct. The value 5 g is a mass, not a percentage, and 50% would incorrectly treat the solute as half of the solution instead of one-tenth.

Volume percentage is defined as the volume of solute divided by the total volume of solution, multiplied by 100. Therefore, volume percentage = (75 mL ÷ 300 mL) × 100 = 0.25 × 100 = 25%. Option A is correct. The values 75 and 300 are raw volumes, not percentages, and 30% does not follow from the given ratio.