2: Concentration Metrics

A concentration term is selected according to the quantities and units supplied. Mass percentage uses masses, volume percentage uses volumes, molarity uses moles and solution volume, molality uses moles and solvent mass, and mole fraction uses moles of components. Therefore, checking the given quantities is the logical first step.

The governing relation is M = n/V, where n is the moles of solute and V is the final solution volume in litres. Dilution adds solvent but does not change the amount of solute, so n remains 0.5 mol. With V = 1 L, the new molarity is 0.5/1 = 0.5 M. Thus A is correct. The other choices use an incorrect volume or reverse the concentration relationship.

Mass percentage is defined as (mass of solute/mass of solution) × 100. The total solution mass is 4 g + 96 g = 100 g. Thus mass percentage = (4/100) × 100 = 4%, so option A is correct. Option B is the solvent percentage, option C uses the solvent as denominator, and option D incorrectly doubles the solute percentage.

Volume percentage is calculated as (volume of solute/volume of final solution) × 100. Here the alcohol volume is 30 mL and the final solution volume is 150 mL, so volume percentage = (30/150) × 100 = 20%. Therefore option D is correct. The denominator is the final total volume, not the water volume or the initial alcohol volume; using either would produce a misleading result.

Mass-volume percentage means grams of solute present in 100 mL of solution, calculated by (mass in grams/solution volume in mL) × 100. Here, (12/240) × 100 = 5%. Thus option C is correct. The value 12 g is only the given mass, not a percentage, while 3% and 4% do not result from the stated ratio. The solution volume, not solvent volume, is used.

In a binary solution, the only two components are solute and solvent, so their mole fractions must add to unity: xsolute + xsolvent = 1. Substituting the given value gives xsolvent = 1 − 0.28 = 0.72. Hence option C is correct. A value of 1.28 is impossible because a mole fraction cannot exceed one; 0.28 and 0.56 do not satisfy the complementary relationship.

Mass percentage by mass is defined as (mass of solute / mass of solution) × 100. Thus a 5% m/m solution contains 5 g solute in every 100 g of total solution. For 400 g, solute mass = (5/100) × 400 = 20 g. Therefore option C is correct. The calculation must use total solution mass, not solvent mass; 5 g would apply only to a 100 g sample.

Volume-by-volume percentage is calculated as (volume of solute / volume of solution) × 100. A 10% v/v solution therefore contains solute equal to 10% of the final solution volume. For 350 mL, solute volume = (10/100) × 350 = 35 mL. Hence option B is correct. The remaining solvent volume is irrelevant, while 350 mL would represent pure solute rather than a 10% solution.

A 2% mass-volume solution contains 2 g of solute in every 100 mL of solution. Applying direct proportionality to 750 mL gives mass of solute = (2 g / 100 mL) × 750 mL = 15 g. Thus option B is correct. The denominator refers to the volume of the complete solution, and the answer must be expressed in grams; the other options use an incorrect factor or base amount.

Molarity is defined as the number of moles of solute divided by the volume of the final solution in litres. Adding water changes the volume but does not add or remove solute, so the amount remains 0.2 mol. The new molarity is therefore 0.2 mol ÷ 2 L = 0.1 mol L⁻¹, or 0.1 M. Thus option A is correct; 0.2 M is the original value, not the diluted value.

Molarity is moles of solute per volume of solution. On heating, a liquid solution may expand, so its volume increases while the amount of solute generally remains unchanged. The same moles are then distributed through a larger volume, reducing molarity. Molality is less affected by thermal expansion because it uses solvent mass.

Molality uses kilograms of solvent in its denominator, and mass does not appreciably change with ordinary temperature changes. Molarity uses solution volume, which can expand or contract with temperature. Therefore molality is generally more temperature-independent; it is not related to colour or volume percentage.

Mole fraction is the ratio of the moles of a specified component to the total moles of all components: x_solute = n_solute/n_total. Rearranging gives n_solute = x_solute × n_total = 0.4 × 15 = 6 mol. Therefore option B is correct. In a binary solution, 9 mol would represent the other component because its mole fraction is 0.6, while 15 mol is the total amount.

For any component, mole fraction equals the component’s moles divided by the total moles. Thus n_solvent = x_solvent × n_total = 0.60 × 20 = 12 mol. Option B is therefore correct. Since the solution is binary, the remaining component has 20 − 12 = 8 mol, explaining option A. Option C is the total amount, and option D does not follow from the stated fraction.

Molarity is calculated as moles of solute divided by solution volume in litres. The correct conversion is 500 mL = 0.5 L. If 500 L is used instead, the denominator is 500 ÷ 0.5 = 1000 times too large. For the same number of moles, the calculated molarity will consequently be 1000 times too small. Therefore option B is correct; the quantity still has mol L⁻¹ units.

Molarity is directly proportional to the amount of solute and inversely proportional to solution volume: M = n/V. Let both solutions contain n moles, let the second volume be V, and let the first volume be V/2. Then M₁ = n/(V/2) = 2n/V = 2M₂. Therefore the first solution has twice the molarity, so option B is correct. Equal moles alone do not imply equal molarity.

Molality is moles of solute divided by kilograms of solvent. If the original value is n/m, doubling both quantities gives 2n/2m, which simplifies to n/m. Therefore, molality remains unchanged. It would double only if solute moles alone doubled, and it would halve only if solvent mass alone doubled.

Molarity is moles of solute per litre of the complete solution, so it is volume-based. Molality is moles of solute per kilogram of solvent, so it is mass-based. Neither is unitless: their units are mol L⁻¹ and mol kg⁻¹ respectively. They are also not restricted to percentage notation.

Mole fraction is defined as xᵢ = nᵢ/Σn, the moles of one component divided by the total moles. Both numerator and denominator have the same unit, mol, so the units cancel and mole fraction is dimensionless. Molarity has units mol L⁻¹, molality has units mol kg⁻¹, and mass-volume percentage is expressed as a percentage of mass per volume. Thus option C is correct.

The governing definitions distinguish the denominator used in each concentration term. Molarity is moles of solute per litre of the complete solution, whereas molality is moles of solute per kilogram of solvent. Mole fraction is a ratio of moles and therefore has no unit. Thus option D wrongly assigns molality’s unit to molarity; A, B, and C are correct.

Molarity is the amount of solute in moles divided by the volume of the solution in litres: M = n/V. Therefore, compare the mole-to-litre ratio for every option. For A, M = 0.5/1 = 0.5 M. For B, M = 0.5/0.5 = 1.0 M. For C, M = 1/2 = 0.5 M. For D, M = 0.25/1 = 0.25 M. The largest value is 1.0 M, so option B is correct. Options A and C have equal molarity because both contain 0.5 mole per litre, even though their solute amounts and volumes differ. Option D has the smallest ratio. The governing idea is that concentration depends on the ratio of solute amount to solution volume, not on either quantity considered alone.

Mass percentage directly uses masses of solute and solution. Molality uses moles of solute divided by the mass of solvent in kilograms. Thus both concentration measures require a mass-based quantity and do not depend on solution volume. Option A belongs to volume-based calculations such as molarity.

Molarity is moles of solute per litre of solution, so the denominator is solution volume in litres. Mass-volume percentage is grams of solute per 100 mL of solution, so its denominator is also solution volume, expressed in millilitres. The units differ, but the underlying denominator is the final volume of the solution.

Molarity means moles of solute per litre of solution, so the volume must first be converted from millilitres to litres. Use M = n/V, which can be rearranged as n = M × V. Here M = 3 mol L⁻¹ and V = 100 mL = 100/1000 L = 0.100 L. Hence n = 3 × 0.100 = 0.300 mol. Therefore, option B is correct. Option A would result from treating 100 mL as though it were 1 L. Option C introduces an unnecessary extra factor of ten, while option D treats the smaller sample as if it contained ten litres or otherwise reverses the scale. The governing concept is proportionality: at fixed molarity, the amount of solute is directly proportional to solution volume. Converting units before substitution prevents the common factor-of-1000 error.

Mass percentage is defined as (mass of solute/mass of solution) × 100. Here the solute mass is 10 g and the total solution mass is 200 g, so mass percentage = (10/200) × 100 = 5%. Therefore option B is correct. The denominator must be the entire solution mass, not the solvent mass or solute mass alone.