2: Concentration Metrics

Mass-volume percentage means grams of solute present in 100 mL of solution. Using density = mass ÷ volume, the volume of 200 g solution is 200 ÷ 1 = 200 mL. Therefore, mass-volume percentage = (10 ÷ 200) × 100 = 5%, so option A is correct. Option B incorrectly treats 100 g as the volume basis, while C and D use other incorrect ratios.

Molarity means moles of solute per litre of solution. First convert 300 mL to 0.300 L. The amount of solute is n = M × V = 1.5 mol L−1 × 0.300 L = 0.45 mol. Its mass is n × molar mass = 0.45 × 40 = 18 g. The litre conversion is essential; the other options arise from incorrect volume or arithmetic handling.

Use the molarity relation n = M × V, with volume expressed in litres. Here, 250 mL = 0.250 L, so n = 0.8 × 0.250 = 0.20 mol. The solute mass is m = n × molar mass = 0.20 × 98 = 19.6 g. Option B represents half the required amount, while the larger values result from using the wrong volume or multiplying without the proper mole calculation.

Choose 1.00 L of solution as the basis. With density 1.0 g mL−1, this volume has a mass of 1000 g. A 5.85% by-mass solution therefore contains 0.0585 × 1000 = 58.5 g NaCl. Taking the molar mass of NaCl as approximately 58.5 g mol−1 gives 1.00 mol in 1.00 L. Thus the molarity is about 1.00 M, so option B is correct.

Use the definition xsolute = nsolute/(nsolute + nsolvent). Substituting the data gives 0.2 = 2/(2 + nsolvent). The total moles must therefore be 2/0.2 = 10 mol. Subtracting the 2 mol of solute gives nsolvent = 10 − 2 = 8 mol. Thus option D is correct. Substitution back gives 2/(2+8) = 0.2, confirming the result.

Take 100 g of solution as the basis. A 5% by-mass solution contains 5 g solute and 95 g water. The solute amount is 5/50 = 0.10 mol, while the solvent mass is 95 g = 0.095 kg. Molality uses kilograms of solvent, so m = 0.10/0.095 = 1.0526 ≈ 1.05 mol kg−1. Hence option B is correct; using 100 g solution as the denominator would incorrectly calculate molarity-like concentration.

Take 1 L of solution. A 1 M solution contains 1 mol glucose; using its molar mass of about 180 g mol−1, the solute mass is 180 g. The solution mass is 1.06 g mL−1 × 1000 mL = 1060 g, so the water mass is 1060 − 180 = 880 g = 0.880 kg. Thus molality = 1/0.880 ≈ 1.14 mol kg−1, so option C is correct.

Molality is defined as moles of solute per kilogram of solvent. Taking 1 kg of water, a 2 molal solution contains 2 mol NaCl. Its mass is 2 × 58.5 = 117 g, while the total solution mass is 1000 + 117 = 1117 g. Therefore, mass percent = (117/1117) × 100 = 10.46% approximately. Option A ignores the solvent contribution, and option C uses an incorrect denominator.

Molarity is moles of solute per litre of solution. First convert 500 mL to 0.500 L. Required moles of Na₂CO₃ = M × V = 0.2 × 0.500 = 0.100 mol. The molar mass of anhydrous Na₂CO₃ is 106 g mol⁻¹, so mass = 0.100 × 106 = 10.60 g. Option A results from using half the required amount, while the larger values use an incorrect volume or factor.

Use a 1 L basis because molarity is moles per litre of solution. At density 1.2 g mL⁻¹, 1 L of solution has mass 1200 g. A 20% by-mass solution therefore contains 0.20 × 1200 = 240 g glucose. Its amount is 240/180 = 1.333 mol. Since this is present in 1 L, the molarity is 1.33 mol L⁻¹, so option B is correct.

A 1 molal solution contains 1 mol urea in 1 kg of solvent. The urea mass is therefore 1 × 60 = 60 g, and the total solution mass is 1000 + 60 = 1060 g. Mass percentage = (60/1060) × 100 = 5.66% approximately. Thus option A is correct. Using 1000 g as the denominator would give 6%, but that is the percentage relative only to solvent, not to the complete solution.

Choose 1 L of solution. Its mass is density × volume = 1.07 g mL⁻¹ × 1000 mL = 1070 g. Since the solution is 10% NaCl by mass, it contains 107 g NaCl. Number of moles = 107/58.5 = 1.829 mol. These moles occur in 1 L, so molarity is approximately 1.83 mol L⁻¹ and option C is correct.

Use 1 kg of water as the reference because molality is based on kilograms of solvent. A 0.2 molal solution contains 0.2 mol glucose. The water amount is 1000/18 = 55.56 mol. Therefore, glucose mole fraction = 0.2/(0.2 + 55.56) = 0.00359, which rounds to 0.0036. The denominator must include glucose and water; omitting glucose gives a slightly different value. Thus option A is correct.

The governing definition is ppm by mass = (mass of solute ÷ mass of solution) × 1,000,000. The solution mass is 250 mL × 1 g mL−1 = 250 g. Hence ppm = (0.005 g ÷ 250 g) × 1,000,000 = 20 ppm. Option B is correct. The other values arise from using the wrong solution mass or an arithmetic error.

Molarity after mixing is total solute moles divided by the stated final volume. The first portion contains 5 × 0.050 = 0.250 mol, and the second contains 2 × 0.100 = 0.200 mol. Total moles are 0.450 mol; final volume is 0.150 L. Thus M = 0.450 ÷ 0.150 = 3.00 mol L−1, so B is correct.

The governing concept is conservation of solute during mixing. Let V₁ be the volume of the 1 mol L⁻¹ solution and V₄ the volume of the 4 mol L⁻¹ solution. The amount of solute contributed is 1V₁ + 4V₄, while the total volume after mixing is V₁ + V₄. Therefore, (V₁ + 4V₄)/(V₁ + V₄) = 2. On rearranging, V₁ + 4V₄ = 2V₁ + 2V₄, so 2V₄ = V₁. Hence V₁:V₄ = 2:1 and option B is correct. This also agrees with the idea that the target concentration lies closer to 1 M than to 4 M, so the dilute solution must be used in the larger volume. Equal volumes would give 2.5 M, whereas 1:2 would give 3 M.

Choose 1 L of solution as the basis. Its mass is 1.18 g mL−1 × 1000 mL = 1180 g. At 30% by mass, HNO3 contributes 0.30 × 1180 = 354 g. Its moles are 354 ÷ 63 = 5.62 mol. Since these moles are present in 1 L, the molarity is 5.62 mol L−1, so C is correct.

Take 100 g of solution, containing 40 g ethanol and 60 g water. Ethanol moles = 40 ÷ 46 = 0.869 mol, while water moles = 60 ÷ 18 = 3.333 mol. Thus ethanol:water = 0.869:3.333 = 0.261:1, approximately 0.26:1. Option A is correct. Mass percentages cannot be used directly as mole ratios because the molar masses differ.

Molality is the number of moles of solute per kilogram of solvent, not per kilogram of solution. Assume 100 g of solution: it contains 40 g methanol and 60 g solvent. Methanol moles = 40/32 = 1.25 mol, and solvent mass = 0.060 kg. Therefore molality = 1.25/0.060 = 20.83 mol kg−1, which rounds to 20.8 mol kg−1. Hence option C is correct.

For mixing solutions of the same solute, add the solute moles and divide by the combined volume. The first portion contains 0.1 × 0.200 = 0.020 mol, and the second contains 0.3 × 0.300 = 0.090 mol. Total moles = 0.110 mol and total volume = 0.500 L. Hence M = 0.110/0.500 = 0.22 mol L−1, option C. A simple average is invalid because the volumes differ.

The aliquot, not the original 100 mL sample, is diluted. Its solute amount is n = M1V1 = 0.5 mol L−1 × 0.020 L = 0.010 mol. After adding water, this amount is present in 0.100 L. Therefore M2 = n/V2 = 0.010/0.100 = 0.10 mol L−1, so option B is correct. Water changes volume but not solute moles.

Dilution conserves the amount of solute. The original solute amount is 0.5 mol L−1 × 0.100 L = 0.050 mol in 0.500 L, giving final concentration 0.050/0.500 = 0.10 mol L−1. A 50 mL portion is 0.050 L, so its amount is 0.10 × 0.050 = 0.0050 mol. Therefore option B is correct; using 0.5 M after dilution would overestimate it.

The governing definition is that a 12% mass-by-volume solution contains 12 g of NaCl in every 100 mL of solution. Therefore, 250 mL contains 12 × 250/100 = 30 g NaCl. Using n = m/M, the amount is 30/58.5 = 0.5128 mol, which rounds to 0.513 mol. Thus option B is correct; option A results from an incorrect scaling or division, while the larger values overestimate the amount.

For volumes V1 and V2, the mixture concentration is (0.5V1 + 1.5V2)/(V1 + V2). Setting this equal to 1.0 gives 0.5V1 + 1.5V2 = V1 + V2, so 0.5V2 = 0.5V1 and V1:V2 = 1:1. Equivalently, 1.0 M lies halfway between the two concentrations. Thus option A is correct.

Molality is based on moles of solute per kilogram of solvent, and the mass of solvent remains essentially constant when temperature changes. Molarity is based on moles per litre of solution, and the solution volume can expand or contract with temperature. Therefore molality gives a concentration measure that is less temperature-dependent and is preferred in boiling and freezing point relations.