2: Concentration Metrics

Molarity is the number of moles of solute per litre of final solution. First find the moles of glucose using n = mass/molar mass: n = 9 g ÷ 180 g mol⁻¹ = 0.05 mol. The solution volume is half a litre, so V = 0.5 L. Applying M = n/V gives M = 0.05 mol ÷ 0.5 L = 0.10 mol L⁻¹, or 0.1 M. Thus option B is correct. The value 0.05 M confuses the number of moles with molarity. The values 0.5 M and 1 M would require incorrect multiplication, an incorrect volume, or an incorrect conversion of the given mass. The final volume, not the solvent volume, is used in the molarity formula.

Molarity is the number of moles of solute per litre of final solution. Urea has molar mass 60 g mol−1, so the amount present is n = 6/60 = 0.10 mol. The final volume is 500 mL = 0.500 L. Therefore, M = n/V = 0.10/0.500 = 0.20 mol L−1. Hence option B is correct; using 500 as litres or using the initial water volume would be incorrect.

Mole fraction requires amounts in moles, not the original masses. Ethanol has molar mass 46 g mol−1, so 46 g ethanol equals 1 mol. Water has molar mass 18 g mol−1, so 90 g water equals 5 mol. The total amount is therefore 1 + 5 = 6 mol, and the ethanol mole fraction is 1/6. Thus option A is correct; using the mass ratio would not calculate mole fraction.

During dilution, no solute is removed, so the number of solute moles remains constant. The dilution relation is M1V1 = M2V2. Substitution gives M2 = (2 mol L−1 × 50 mL) / 500 mL = 0.20 mol L−1. The same result follows because the volume becomes ten times larger and the concentration becomes one-tenth of 2 M. Therefore option B is correct.

Dilution changes the volume but preserves the moles of solute. First calculate the initial amount: n = M1V1 = 0.4 mol L−1 × 0.250 L = 0.100 mol. The final volume is 1.00 L, so the final molarity is M2 = 0.100 mol / 1.00 L = 0.10 mol L−1. Therefore option B is correct. The lower value reflects distribution through a larger volume.

Use the molarity relation n = M × V, with volume expressed in litres. The required moles of KCl are n = 0.1 mol L−1 × 0.500 L = 0.050 mol. The molar mass of KCl is approximately 39.1 + 35.5 = 74.6 g mol−1. Hence the required mass is 0.050 × 74.6 = 3.73 g approximately. Therefore option B is correct.

Molality is defined as moles of solute per kilogram of solvent. Thus, a 1 molal aqueous solution contains 1 mol solute in 1 kg water. The water amount is 1000/18 ≈ 55.5 mol. Total moles = 1 + 55.5 = 56.5 mol, so the solute mole fraction is 1/56.5 ≈ 0.0177. Therefore, option B is correct. Option D is close to the solvent mole fraction, while A incorrectly ignores the solute in the total-mole denominator.

A 2 molal solution contains 2 mol NaOH in 1 kg water. Using the molar mass of NaOH as 40 g mol−1, solute mass = 2 × 40 = 80 g. The total solution mass is 1000 + 80 = 1080 g. Therefore, mass percent = (80/1080) × 100 = 7.407%, or about 7.41%. Option B is correct. The value 8% results from rough or incorrect handling of the denominator, while 16% doubles the solute contribution.

For a dilute aqueous solution, the density is approximately 1 kg L−1, so 1 ppm is approximately 1 mg of solute per litre of solution. Thus 50 ppm corresponds to about 50 mg L−1. In 2 L, solute mass = 50 mg L−1 × 2 L = 100 mg. Therefore, option C is correct. The 50 mg choice applies to only 1 L, whereas 25 mg and 200 mg use incorrect volume factors.

A 2 molal solution contains 2 mol sucrose in 1 kg water. The water amount is 1000/18 ≈ 55.5 mol. Total moles = 55.5 + 2 = 57.5 mol. Hence the mole fraction of water is 55.5/57.5 ≈ 0.965, so option A is correct. The value 0.035 is approximately the sucrose mole fraction, while 0.982 would correspond to a lower solute amount and is not consistent with 2 molal concentration.

Molality is moles of solute divided by kilograms of solvent. Glucose has molar mass 180 g mol⁻¹, so 36 g glucose represents 36 ÷ 180 = 0.2 mol. Rearranging m = n/kg solvent gives solvent mass = n/m = 0.2 ÷ 0.2 = 1.0 kg. Therefore option B is correct. The calculation uses the mass of water only, not the total mass or volume of the final solution.

For a mass-based concentration, 1 ppm means one part of solute per million parts of solution, equivalent to 1 mg solute per kg of solution. Therefore, a concentration of 3 ppm means 3 mg per kilogram. For 5 kg of solution, solute mass = 3 mg kg⁻¹ × 5 kg = 15 mg. Option C is correct. The value 3 mg applies only to 1 kg, not to the stated 5 kg sample.

Molality is defined as moles of solute per kilogram of solvent. Rearranging the relation m = n/msolvent gives msolvent = n/m. Substituting n = 0.2 mol and m = 0.4 mol kg⁻¹, the solvent mass is 0.2 ÷ 0.4 = 0.5 kg. Converting to grams gives 500 g, so option B is correct. Using 250 g would produce 0.8 mol kg⁻¹, twice the required molality.

The governing dilution principle is conservation of solute: M1V1 = M2V2, because adding solvent does not change the solute moles. Substituting values gives 1.2 × 100 = M2 × 300, so M2 = 120/300 = 0.4 mol L−1. Thus, option B is correct. The final volume is three times larger, so the concentration becomes one-third, not higher.

Molality is moles of solute per kilogram of solvent. With 1 kg of water and a molality of 0.5 mol kg−1, the required glucose amount is 0.5 mol. Using the molar mass of glucose, 180 g mol−1, mass = 0.5 × 180 = 90 g. Hence option B is correct. The 45 g choice corresponds to 0.25 mol, while 180 g corresponds to one mole.

Volume percentage is calculated as (volume of solute ÷ volume of solution) × 100. Dilution adds solvent but does not change the amount or volume of ethanol, so ethanol remains 50 mL while the final solution volume is 400 mL. Therefore, volume percentage = (50 ÷ 400) × 100 = 12.5%. Option B is the original percentage before dilution, not the final value.

For a dilute aqueous solution with density 1 g mL−1, one litre of solution has a mass of approximately one kilogram. Thus, concentration in mg L−1 is numerically equal to concentration in ppm. Here, concentration = 10 mg ÷ 2 L = 5 mg L−1, so the answer is approximately 5 ppm. The other options result from using the wrong volume or failing to divide by 2.

During dilution, the number of moles of solute remains constant, so the dilution relation M1V1 = M2V2 applies. Substituting the values gives M2 = (2 M × 200 mL) ÷ 500 mL = 0.8 M. Millilitres can be used because the same volume unit appears in both terms. The result must be below the initial 2 M because the solution volume increases.

The governing principle of dilution is conservation of solute moles: adding water changes the volume but not the amount of dissolved solute. Therefore, M₁V₁ = M₂V₂. Substitution gives M₂ = (0.4 × 150)/600 = 60/600 = 0.10 M. Since the volume becomes four times larger, the molarity becomes one-fourth. Thus option A is correct; option C ignores dilution, while D reverses the concentration change.

In dilute water, 1 L has approximately the mass of 1 kg, so concentration in mg L−1 is numerically close to concentration in ppm. The nitrate concentration is 12 mg divided by 2 L, giving 6 mg L−1. Hence the approximate concentration is 6 ppm, option A. The value 12 ppm incorrectly uses the total nitrate mass without dividing by the water volume.

Molality is defined as moles of solute divided by the mass of solvent in kilograms, not the mass of the complete solution. The solution mass equals solvent mass plus solute mass and is therefore larger than the solvent mass. Keeping the same numerator but using this larger denominator produces a smaller quotient. Hence the calculated value is less than the true molality, so option A is correct.

Molality is calculated using moles of solute per kilogram of solvent. The solute amount is 10/50 = 0.20 mol, and the water mass is 90 g = 0.090 kg. Therefore, molality = 0.20/0.090 = 2.222... mol kg−1, or approximately 2.22 m. Option A is correct. Using grams without conversion or using total solution mass would produce the distractor values.

Molarity is the number of moles of solute per litre of solution. First, moles of AlCl₃ = 27 ÷ 133.5 ≈ 0.202 mol. Dividing by the solution volume, 2 L, gives M = 0.202 ÷ 2 ≈ 0.101 M, which rounds to 0.10 M. Therefore, option A is correct. Option B results from forgetting the two-litre volume, while C and D are much larger calculation errors.

Molality is defined as moles of solute per kilogram of solvent. The solvent mass is 500 g = 0.5 kg. For a 1 m solution, glucose moles = 1 × 0.5 = 0.5 mol. Its mass is therefore 0.5 × 180 = 90 g, so option A is correct. Option B would represent one full mole and ignores the 0.5 kg solvent amount; the other options use incorrect factors.

Use the definition of molality: m = moles of solute ÷ kilograms of solvent. The 250 g water equals 0.25 kg. Therefore, urea moles = 2 × 0.25 = 0.5 mol. Multiplying by its molar mass gives mass = 0.5 × 60 = 30 g, so option A is correct. Option C corresponds to one mole, while B and D apply incorrect numerical factors.