2: Concentration Metrics

Take 1 L of solution as the basis. Its mass is density × volume = 1.11 g mL−1 × 1000 mL = 1110 g. Since the solution is 10% NaOH by mass, it contains 0.10 × 1110 = 111 g NaOH. With molar mass 40 g mol−1, the amount is 111/40 = 2.775 mol. Dividing by 1 L gives approximately 2.78 mol L−1, so B is correct.

Take 1 L of solution. A 2 M solution contains 2 mol NaCl, whose mass is 2 × 58.5 = 117 g. The total mass of 1 L solution is 1.08 g mL−1 × 1000 mL = 1080 g. Therefore, solvent mass = 1080 − 117 = 963 g = 0.963 kg. Molality is moles of solute per kilogram of solvent: 2/0.963 = 2.08 mol kg−1, so C is correct.

Take 1 kg of solvent as the basis. A 5 molal solution contains 5 mol glucose. With glucose molar mass 180 g mol−1, its mass is 5 × 180 = 900 g, so the solution mass is 1900 g. Using density 1.2 g mL−1, solution volume = 1900/1.2 = 1583.3 mL = 1.5833 L. Hence molarity = 5/1.5833 ≈ 3.16 mol L−1, making B correct.

Choose 1 L of solution. Its mass from the density is 1.18 × 1000 = 1180 g. Since the solution is 36.5% HCl by mass, HCl mass = 0.365 × 1180 = 430.7 g. HCl has molar mass 36.5 g mol−1, so moles of HCl = 430.7/36.5 = 11.8 mol. Because this amount is present in 1 L, the molarity is 11.8 mol L−1; C is correct.

Choose 1 L of solution as the basis because molarity is moles of solute per litre of solution. One litre equals 1000 mL, so the mass of this solution is density × volume = 1.2 g mL⁻¹ × 1000 mL = 1200 g. Since the solution is 20% solute by mass, solute mass = 0.20 × 1200 = 240 g. Convert this mass to moles using the molar mass: n = 240 g/(60 g mol⁻¹) = 4 mol. These 4 mol are present in 1 L of solution, so the molarity is 4 mol L⁻¹, or 4 M. Option A is correct. Option B would reflect an incorrect mass or percentage calculation; option C confuses density with molarity; and option D results from a factor-of-ten error. The essential sequence is volume basis, solution mass, solute mass, moles, then molarity.

Choose 1 L of solution. Its mass is density × volume = 1.08 g mL−1 × 1000 mL = 1080 g. A 10% mass-by-mass solution therefore contains 108 g urea per litre. The number of moles is 108 ÷ 60 = 1.8 mol. Since this amount is present in 1 L of solution, the molarity is approximately 1.8 M. The smaller value 0.18 M misses a factor of ten.

A 25% mass-by-mass solution contains 25 g glucose and 75 g solvent in every 100 g solution. Thus the solute-to-solvent mass ratio is 25:75 = 1:3. For 100 g solvent, glucose mass is (25/75) × 100 = 33.33 g, or 33.33/180 = 0.185 mol. The solvent mass is 0.100 kg, so molality = 0.185/0.100 ≈ 1.85 m. Molality uses solvent mass, not solution mass.

A 20% solution contains 20 g solute and 80 g solvent per 100 g solution, so the solute-to-solvent mass ratio is 1:4. For 200 g solvent, sodium chloride mass is 50 g. Its amount is 50/58.5 ≈ 0.8547 mol. Converting solvent mass to kilograms gives 0.200 kg. Therefore molality = 0.8547/0.200 ≈ 4.27 mol kg−1, or 4.27 m. Molality is based on solvent mass.

For mixing solutions of the same solute, total solute moles are added and the stated volumes are added. The first portion contains 1 × 0.100 = 0.100 mol, and the second contains 3 × 0.200 = 0.600 mol. Total moles = 0.700 mol; total volume = 0.300 L. Hence M = 0.700/0.300 = 2.33 M, so option A is correct. A simple average would be wrong because the volumes differ.

Molarity after mixing is found from total moles divided by total volume. The first solution contributes 0.5 × 0.200 = 0.100 mol, while the second contributes 1.5 × 0.300 = 0.450 mol. Thus total moles are 0.550 mol and total volume is 0.200 + 0.300 = 0.500 L. Final molarity = 0.550/0.500 = 1.1 M, making option A correct.

First determine the solution volume from density: volume = mass ÷ density = 200 ÷ 1.25 = 160 mL. Mass-volume percentage is grams of solute per 100 mL solution, so it equals (25 ÷ 160) × 100 = 15.625%. Hence option A is correct. Option C is the solute mass percentage by mass, not by volume; options B and D result from incorrect volume or ratio calculations.

Take 1 L of solution. Its mass from the density is 1.08 × 1000 = 1080 g. It contains 2 mol NaOH, whose mass is 2 × 40 = 80 g. Therefore, solvent mass = 1080 − 80 = 1000 g = 1.000 kg. Molality = 2 mol/1.000 kg = 2.00 mol kg⁻¹, so option B is correct. Dividing by solution mass would incorrectly calculate a different quantity.

First calculate the amount of calcium chloride: n = M × V = 0.25 mol L−1 × 0.400 L = 0.10 mol. Complete ionisation follows CaCl2 → Ca2+ + 2Cl−, so every formula unit produces three ionic particles. Therefore total ionic-particle amount is 0.10 × 3 = 0.30 mol. Option A counts formula units only, while the other larger values overcount the ions.

Initially, the solution contains 0.2 × 1.0 = 0.20 mol solute. The removed 0.100 L portion contains 0.2 × 0.100 = 0.020 mol, so 0.180 mol remains. Adding 100 mL water restores the total volume to 1.0 L without adding solute. Therefore the new molarity is 0.180 mol L−1, or 0.18 mol L−1, making option B correct.

Choose 1 L of solution. At 1.5 mol L−1, it contains 1.5 mol solute, whose mass is 1.5 × 100 = 150 g. The mass of 1 L solution, using density 1.10 g mL−1, is 1.10 × 1000 = 1100 g. Therefore mass percentage is (150/1100) × 100 = 13.64%, approximately 13.6%. Option C is correct.