2: Concentration Metrics

Mole fraction is a dimensionless fraction, whereas mole percent expresses the same fraction per hundred. The conversion is mole percent = mole fraction × 100. Thus, 0.3 × 100 = 30%. In a binary solution, 70% would be the mole percent of the other component because the mole fractions sum to one. Options B and C omit the required scaling.

A 4% mass-by-volume solution contains 4 g of solute in every 100 mL of solution. For 250 mL, scale this proportion directly: mass = (4 g/100 mL) × 250 mL = 10 g. Therefore option C is correct. The values 4 g and 8 g do not apply the 250 mL factor correctly, while 16 g is an excessive fourfold value. The calculation uses final solution volume, not solvent volume.

The mole fraction of a component is defined as its moles divided by the total moles of all components: xsolute = nsolute/ntotal. Rearranging gives nsolute = xsolute × ntotal. Substitution gives nsolute = 0.18 × 5 = 0.90 mol. Therefore option C is correct. The value 0.18 is only a dimensionless fraction, not the number of moles, and 1.80 results from an incorrect multiplication.

For a binary solution containing only components A and B, the mole fractions must sum to unity: xA + xB = 1. Therefore xB = 1 − xA = 1 − 0.75 = 0.25, making option A correct. Choosing 0.75 would incorrectly assign A’s fraction to B as well, and 1.25 cannot be a mole fraction because a mole fraction cannot exceed one. Option B does not satisfy the sum-to-one rule.

Mass percentage expresses the mass of solute as a percentage of the mass of the complete solution: mass percent = (mass of solute/mass of solution) × 100. Substitution gives (2 g/500 g) × 100 = 0.4%. Therefore option B is correct. Using 2 g as the denominator or omitting the multiplication by 100 produces the distractor values; the denominator must be the total solution mass, not merely the solvent mass.

Molality is calculated from the moles of solute and the mass of solvent in kilograms: m = moles of solute ÷ kilograms of solvent. Convert 200 g to 0.200 kg, then m = 0.5 ÷ 0.200 = 2.5 mol kg⁻¹. Thus option C is correct. Option B would result from an incorrect conversion, and using 200 without changing grams to kilograms would produce a wrong numerical value and unit.

For a dilute aqueous solution, the density is taken as approximately 1 kg L−1, so 1 L of solution is approximately 1 kg. Convert the glucose mass: 0.036 g = 36 mg. Thus the concentration is about 36 mg kg−1, and mg per kilogram is numerically equal to ppm. Therefore the answer is 36 ppm, option B; 360 ppm would require ten times more solute.

Molality is defined as moles of solute divided by the mass of solvent in kilograms: m = nsolute/kg solvent. Substituting the data, 0.5 = 0.25/msolvent, so msolvent = 0.25/0.5 = 0.50 kg. Converting kilograms to grams gives 0.50 × 1000 = 500 g. Thus option B is correct. The denominator is solvent mass, not total solution mass.

Mass fraction is the mass of solute divided by the total mass of solution, while mass percent is that fraction multiplied by 100. Total solution mass = 5 + 95 = 100 g. Therefore solute mass fraction = 5/100 = 0.05 and mass percent = 0.05 × 100 = 5%. Hence option A is correct. The value 0.95 is the solvent fraction, not the solute fraction.

Mass fraction is a decimal ratio, whereas mass percentage expresses the same ratio per hundred parts. The conversion is mass percentage = mass fraction × 100. Thus 0.12 × 100 = 12%. Option C is correct. The value 0.12 is the fraction itself, 1.2% results from a factor-of-ten error, and 88% would represent the complementary fraction only after suitable conversion.

A 2% mass-by-volume solution contains 2 g of solute in 100 mL of solution. On scaling to 1 L, the solute mass becomes 20 g. Its amount is n = m/M = 20/40 = 0.50 mol. Because this amount is present in one litre, molarity is 0.50 mol L−1, so option C is correct. Option B would result from failing to scale 100 mL to 1 L.

Mole fraction is defined as the moles of a component divided by the total moles of all components. In a pure solvent, every mole present belongs to the solvent, so nsolvent/ntotal = nsolvent/nsolvent = 1. The solute mole fraction is zero, but the solvent mole fraction is one; a mole fraction cannot exceed one.

Before adding solute, the solvent mole fraction is one. After addition, total moles become n_solvent + n_solute, while solvent moles remain n_solvent. Thus x_solvent = n_solvent/(n_solvent+n_solute), which is less than one. It cannot become two or negative, and it is not still one when solute is present.

A binary solution contains two components, A and B. The mole fraction of a component is its moles divided by the total moles, so the fractions of all components together must add to one: XA + XB = 1. Therefore, option A is correct. Zero would imply no particles, and two is impossible because mole fractions together represent the complete composition.

A binary solution contains only two components, A and B. The mole fractions of all components must add up to one, so X_A + X_B = 1. Substituting X_A = 0.3 gives X_B = 1 − 0.3 = 0.7. The value 0.3 repeats the given fraction, 1.3 exceeds the possible total fraction, and 3.0 cannot be a mole fraction.

In a binary solution, the solute and solvent are the only two components, so their mole fractions obey X_solute + X_solvent = 1. With X_solute = 0.1, rearrange the relation: X_solvent = 1 − 0.1 = 0.9. The answer cannot be 1.1 because a mole fraction cannot exceed one; 0.1 is the solute fraction, and 0.01 results from an incorrect multiplication.

Molality is defined as the amount of solute in moles divided by the mass of solvent in kilograms: m = moles of solute / kg of solvent. It differs from molarity, which uses litres of solution in the denominator. Because molality uses solvent mass, it is largely independent of temperature-related volume changes and is useful in colligative-property calculations.

Molarity is moles of solute per litre of solution. The number of solute moles usually remains fixed, but the solution volume can expand on heating or contract on cooling. Therefore the numerical molarity may change. Molality uses solvent mass, which is essentially temperature-independent, so it is often preferred for colligative calculations.

Molality is defined as the number of moles of solute divided by the mass of solvent in kilograms: m = moles of solute/kg of solvent. It uses the solvent alone, not the total solution. Molarity instead uses the volume of the solution in litres. This mass-based definition is useful because mass changes negligibly with temperature compared with volume.

The mole fraction of a component is defined as its number of moles divided by the total number of moles of all components. Thus, for the solute, \(x_{\text{solute}}=n_{\text{solute}}/n_{\text{total}}=0.2/1.0=0.2\). Therefore option B is correct. The value 0.8 could represent the solvent fraction only if the remaining moles were 0.8, while 1.2 cannot be a mole fraction because mole fractions lie between zero and one.

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. Here m = nsolute/masssolvent = 0.1 mol/1 kg = 0.1 mol kg⁻¹. The fact that the solute is a non-electrolyte is not needed for this concentration calculation; it would matter when selecting the van’t Hoff factor in a colligative-property formula.

Convert the solvent mass into kilograms because molality uses kg of solvent: 500 g = 0.500 kg. Then m = moles of solute/kilograms of solvent = 0.2/0.500 = 0.4 mol kg⁻¹. Using 500 directly without conversion would give an incorrect value and incorrect units, so unit conversion is the essential first step.

The total number of moles is the sum of solute and solvent moles: \(n_{\text{total}}=0.1+0.9=1.0\). The solute mole fraction is therefore \(x_{\text{solute}}=0.1/1.0=0.1\). Option A is correct. The value 0.9 is the solvent mole fraction, 1.0 is the total moles rather than a fraction for this component, and 9.0 results from dividing in the wrong order.

Molality uses kilograms of solvent. Convert 100 g water to 0.100 kg, then calculate m = 0.1 mol/0.100 kg = 1.0 mol kg⁻¹. The non-electrolyte description does not change the concentration calculation; it only indicates that i would be approximately one in a later colligative-property calculation. The answer 10 results from mishandling the unit conversion.

For a binary solution, the solute and solvent are the only two components, so their mole fractions must add up to one. Let x_solvent be the mole fraction of the solvent. Then x_solute + x_solvent = 1, giving x_solvent = 1 - 0.15 = 0.85. Mole fraction is a dimensionless ratio and cannot be 15 in this situation. Therefore, option B, 0.85, is correct.