Mole concept and molar mass

For a binary solution, take 1 mol of total solution. The solute amount is 0.2 mol and water is 0.8 mol. The solvent mass is 0.8 × 18 = 14.4 g = 0.0144 kg. Molality equals moles of solute divided by kilograms of solvent, so m = 0.2/0.0144 = 13.89 mol kg−1, approximately 13.9 mol kg−1. Hence B is correct; using total solution mass or 1 kg water without conversion produces the distractors.

Assume 1 mol of total solution. A solute mole fraction of 0.1 means 0.1 mol solute and 0.9 mol water. The water mass is 0.9 × 18 = 16.2 g = 0.0162 kg. Molality is moles of solute per kilogram of solvent: 0.1/0.0162 = 6.1728 mol kg−1, approximately 6.17 mol kg−1. Therefore B is correct. The denominator must be solvent mass, not total solution mass.

Choose a convenient basis of 1 mol total solution. Then solute moles = 0.05 mol and water moles = 0.95 mol. Using the molar mass of water, 18 g mol−1, the solvent mass is 0.95 × 18 = 17.1 g = 0.0171 kg. Molality = moles of solute ÷ kilograms of solvent = 0.05/0.0171 ≈ 2.92 mol kg−1. Hence option B is correct; total solution mass must not be used.

Molality is moles of solute per kilogram of solvent, whereas molarity is moles per litre of solution. Take 1 kg water, so 3 mol glucose are present. Glucose mass = 3 × 180 = 540 g; solution mass = 1540 g. Its volume is 1540/1.15 = 1339 mL = 1.339 L. Therefore molarity = 3/1.339 ≈ 2.24 mol L−1, so option B is correct. Options A, C and D result from using an incorrect solution mass or volume.

Molarity refers to moles per litre of solution, while molality refers to moles per kilogram of solvent. Take 1 L solution: it contains 4 mol NaOH, whose mass is 4 × 40 = 160 g. The solution mass is 1.16 × 1000 = 1160 g, so water mass is 1160 − 160 = 1000 g = 1 kg. Thus molality = 4/1 = 4.00 mol kg−1, making option C correct. The density is essential for finding solvent mass.

Molarity is the number of solute moles in one litre of solution. For 1 L solution, density gives mass = 1.42 × 1000 = 1420 g. Since the solution is 68% HNO3 by mass, acid mass = 0.68 × 1420 = 965.6 g. Using molar mass HNO3 = 63 g mol−1, moles = 965.6/63 ≈ 15.3 mol. Therefore the molarity is about 15.3 mol L−1, so option B is correct; the other values use an incorrect percentage or volume.

Molality uses kilograms of solvent, not kilograms of the entire solution. Choose 100 g solution: 68 g is HNO3 and 32 g is water. Moles of HNO3 = 68/63 ≈ 1.079 mol, while solvent mass = 32 g = 0.032 kg. Hence molality = 1.079/0.032 ≈ 33.7 mol kg−1, so option C is correct. Dividing by 0.100 kg would incorrectly use solution mass and produce a different concentration.

Mole fraction is based on the ratio of solute moles to total moles. Assume 1 mol total solution: solute = 0.25 mol and water = 0.75 mol. Solute mass = 0.25 × 60 = 15 g; water mass = 0.75 × 18 = 13.5 g. Total mass = 28.5 g. Therefore mass percent = (15/28.5) × 100 ≈ 52.6%, so option C is correct. Treating mole fraction as mass fraction would incorrectly suggest 25%.

The mole fractions of solute and water sum to one. Thus, if xwater = 0.90, xsolute = 0.10. Take 1 mol total mixture: water = 0.90 mol and solute = 0.10 mol. Water mass = 0.90 × 18 = 16.2 g = 0.0162 kg. Molality = 0.10/0.0162 ≈ 6.17 mol kg−1, so option B is correct. The denominator must be solvent mass, not total solution mass.

Use a 1 L sample because molarity directly gives the solute moles in that volume. Solute moles = 2.5 mol, so solute mass = 2.5 × 100 = 250 g. Density gives solution mass = 1.25 × 1000 = 1250 g. Therefore mass percent = (250/1250) × 100 = 20%, making option C correct. Molarity alone would not provide solution mass; the density is needed for the denominator.

Take exactly 1 L of solution, because molarity is moles of solute per litre of solution. Its mass is density × volume = 1.2 g mL−1 × 1000 mL = 1200 g. A 20% mass fraction gives 0.20 × 1200 = 240 g solute. The solute amount is 240 ÷ 40 = 6 mol, so molarity = 6 mol ÷ 1 L = 6 M. Option D is correct; ignoring density would give an incorrect value.

Consider 1 L of solution. A 3 M solution contains 3 mol solute in this volume, so solute mass = 3 × 40 = 120 g. Since this is 12% of the total solution mass, total mass = 120 ÷ 0.12 = 1000 g. Density is mass divided by volume: 1000 g ÷ 1000 mL = 1.00 g mL−1. Therefore option B is correct; confusing solute mass with solution mass would be incorrect.