2: Concentration Metrics

Molarity is the number of moles of solute present in one litre of solution. Use n = M × V, but first convert the volume: 200 mL = 0.200 L. Therefore, n = 0.5 mol L⁻¹ × 0.200 L = 0.100 mol. Hence, option B is correct. The other values result from using an incorrect volume conversion or confusing the given molarity with the required amount of solute.

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. Convert the solvent mass first: 500 g = 0.500 kg. Rearranging m = n_solute/mass_solvent gives n_solute = m × mass_solvent = 3 mol kg⁻¹ × 0.500 kg = 1.5 mol. Thus option C is correct. The value 3.0 mol would require 1 kg of solvent, while the smaller values do not satisfy the stated molality.

Mass percentage is calculated as (mass of solute ÷ mass of solution) × 100. Here the sugar mass is 5% of the total 500 g solution: mass of sugar = (5/100) × 500 g = 25 g. Therefore, option B is correct. The value 50 g represents 10% of 500 g, 100 g represents 20%, and 20 g is less than the required 5% amount.

A 10% by-mass solution contains salt equal to 10/100 of the total solution mass. Salt required = (10/100) × 300 g = 30 g. Since the total mass must be 300 g, water required = 300 − 30 = 270 g. Thus option A is correct. Option D reverses salt and water, option B gives only about 3.33% salt, and option C is not exactly 10% of 300 g.

Mass by volume percentage tells how many grams of solute are present in 100 mL of solution. The formula is % (m/v) = (mass of solute in g ÷ volume of solution in mL) × 100. Substituting the data gives (5 ÷ 250) × 100 = 2%. Therefore, option B is correct. The final solution volume, not the initial water volume, must be used in the denominator.

For a binary solution, the two mole fractions must add to unity: x_solute + x_solvent = 1. With x_solute = 0.2, x_solvent = 1 − 0.2 = 0.8. Hence option D is correct. Option A is simply the given solute fraction, while 0.4 and 0.6 would make the total mole fraction 0.6 and 0.8 respectively, not 1. The result also lies within the valid range from 0 to 1.

First calculate the amount of NaOH using its molar mass, 40 g mol⁻¹: n = 40 g ÷ 40 g mol⁻¹ = 1 mol. Molarity is moles of solute divided by the final solution volume in litres. Thus M = 1 mol ÷ 2 L = 0.50 mol L⁻¹. Option B is correct. The stated 2 L is the final solution volume; using the water volume before dissolving would be inappropriate.

The molar mass of NaCl is 58.5 g mol⁻¹, so the number of moles is n = 5.85 ÷ 58.5 = 0.10 mol. Convert the final volume: 500 mL = 0.500 L. Molarity is n/V, so M = 0.10 ÷ 0.500 = 0.20 mol L⁻¹. Therefore, option B is correct. Ignoring the volume conversion would produce an incorrect concentration.

Molarity relates the amount of solute to the volume of solution in litres. Use n = M × V and convert 200 mL to 0.200 L. Thus n = 0.25 mol L⁻¹ × 0.200 L = 0.050 mol. Therefore, option B is correct. The answer 0.250 mol would require 1 L of solution, while 0.025 mol and 0.125 mol do not follow the stated product of molarity and volume.

The governing principle of dilution is conservation of solute: adding solvent changes the volume but not the amount of solute. Therefore, M₁V₁ = M₂V₂. Substituting the values, M₂ = (0.5 mol L⁻¹ × 0.250 L) ÷ 1.00 L = 0.125 mol L⁻¹. Thus option B is correct. Options C and D do not account for the fourfold increase in volume, while option A is too small.

Molality is defined as the number of moles of solute per kilogram of solvent, not per kilogram of solution. Sodium chloride has molar mass 58.5 g mol⁻¹, so 58.5 g NaCl equals 1 mol. The solvent mass is 2 kg; hence molality = 1 mol ÷ 2 kg = 0.50 mol kg⁻¹. Option B is correct. The other values arise from using an incorrect denominator or arithmetic.

Molarity is the number of moles of solute per litre of solution, so the amount of solute is found from n = M × V. Convert the volume first: 250 mL = 0.250 L. Thus n = 1 mol L⁻¹ × 0.250 L = 0.25 mol. Option A is correct. A value of 1.00 mol would require 1 L of this solution, while 0.50 mol and 2.50 mol result from incorrect scaling.

Molarity is moles of solute per litre of the final solution. H₂SO₄ has molar mass 98 g mol⁻¹, so 98 g corresponds to 1 mol. The final volume is 500 mL = 0.500 L. Therefore, M = 1 mol ÷ 0.500 L = 2.0 mol L⁻¹, so option C is correct. The value 1.0 mol L⁻¹ would incorrectly treat 500 mL as 1 L.

Molality uses moles of solute divided by the mass of solvent in kilograms. For NaCl, n = 11.7 g ÷ 58.5 g mol⁻¹ = 0.2 mol. The water mass is 400 g = 0.4 kg. Thus molality = 0.2 mol ÷ 0.4 kg = 0.50 mol kg⁻¹, making option B correct. Using grams directly or the total solution mass would give an incorrect result.

Mole fraction is the amount of the selected component divided by the total amount of all components. From the ratio, take solute = 1 mol and solvent = 4 mol, so total moles = 1 + 4 = 5 mol. Hence x_solute = 1/5 = 0.20, making option A correct. Option D is the solvent mole fraction, and option B incorrectly uses only the solvent amount as a denominator.

Molarity is defined as moles of solute per litre of solution, so use n = M × V with volume in litres. Convert 750 mL to 0.750 L. Thus n = 0.4 mol L−1 × 0.750 L = 0.300 mol. Therefore, option C is correct. Options A, B, and D result from using an incorrect volume or an incorrect multiplication.

Molarity is the number of moles of solute divided by the volume of the final solution in litres. Convert 600 mL to 0.600 L, then calculate M = n/V = 0.30 mol ÷ 0.600 L = 0.50 mol L−1. Therefore, option C is correct. Using 600 without conversion or using the solvent volume instead of final solution volume would be incorrect.

Molality is defined as moles of solute per kilogram of solvent. Convert the solvent mass: 250 g = 0.250 kg. Therefore, moles of solute = molality × solvent mass in kilograms = 2 mol kg−1 × 0.250 kg = 0.50 mol. Option B is correct. Failing to convert grams to kilograms would produce an answer that is 1000 times too large.

Mass percentage is calculated from the mass of solute and the total mass of solution: mass percent = (mass of solute/mass of solution) × 100. Substitution gives (120 g/1200 g) × 100 = 10%. Therefore, option B is correct. The one-litre volume is not needed because this concentration unit is based entirely on mass; 12% would incorrectly use the solute-to-volume figures.

A 20 mass percent solution contains 20 g solute in every 100 g of solution. For 150 g solution, solute mass = (20/100) × 150 = 30 g. Since total solution mass = solute mass + solvent mass, solvent mass = 150 − 30 = 120 g. Therefore, option C is correct. Option A gives the solute mass, not the requested solvent mass.

For a dilute aqueous solution, the density is taken as approximately 1 kg L−1. Consequently, 1 mg of solute in 1 L of water is approximately 1 mg kg−1, and mg kg−1 is equivalent to ppm. Therefore, 40 mg L−1 is approximately 40 ppm. Option A is correct. The other choices result from multiplying or dividing by an extra factor of ten or one hundred.

Mass percentage is calculated as (mass of solute ÷ mass of solution) × 100. Assume 100 g of solution, so it contains 12 g solute and 100 − 12 = 88 g solvent. Therefore, the solute-to-solvent mass ratio is 12:88, which can also be simplified to 3:22. Option A gives the direct masses corresponding to 100 g solution; 12:100 incorrectly uses solution mass as solvent mass.

The governing idea is that mass percentage uses the total solution mass as the denominator, not the solvent mass alone. Let the solute mass be 1 part and the solvent mass be 4 parts; the total solution mass is therefore 5 parts. Solute percentage = (1 ÷ 5) × 100 = 20%. Option B would divide by 4, while 80% is the solvent percentage.

A 4% mass-by-volume solution contains 4 g of solute in every 100 mL of solution. Convert 1.5 L to 1500 mL. Since 1500 mL contains 15 groups of 100 mL, the solute mass is 15 × 4 = 60 g. The percentage is based on final solution volume, not solvent volume. Options C and D reflect incorrect scaling, while B does not apply the stated concentration.

Volume-by-volume percentage is defined as (volume of solute ÷ volume of solution) × 100. Therefore, solute volume = 15/100 × 2 L = 0.30 L. Converting 0.30 L to millilitres gives 300 mL. Option B uses only 1 L as a reference, and option C loses a factor of ten; the percentage must be applied to the full 2 L solution volume.