Mole concept and molar mass

Take 100 g of solution. It contains 36.5 g HCl, equal to 36.5/36.5 = 1 mol, and 63.5 g water, equal to 63.5/18 ≈ 3.528 mol. Total moles are 1 + 3.528 = 4.528 mol. Therefore, the HCl mole fraction is 1/4.528 ≈ 0.221, so B is correct. The value 0.365 is the mass fraction, not the mole fraction, and 0.635 is the water mass fraction.

Let y represent the moles of solvent. Because the solution is binary, its only components are solvent and solute, so the total moles are y + 3. The solvent mole fraction is therefore y/(y + 3). Set this equal to the given value: y/(y + 3) = 0.75. Solving gives y = 0.75y + 2.25, so 0.25y = 2.25 and y = 9 mol. Hence option A is correct. Verification gives 9/(9 + 3) = 9/12 = 0.75. The binary condition matters because no third component is added to the denominator. The other choices produce different solvent fractions and therefore cannot satisfy the stated information.

For a binary solution containing only solute and solvent, their mole fractions must add to one: xsolute + xsolvent = 1. Therefore, xsolvent = 1 − 0.6 = 0.4. Mole percent is obtained by multiplying the mole fraction by 100, so solvent mole percent = 0.4 × 100 = 40%. Option A is correct; 60% is the solute mole percent, not the solvent value.

Mass percent means that the solute mass is 8/100 of the total solution mass. Thus, solute mass = (8/100) × 250 g = 20 g. A solution contains solute plus solvent, so solvent mass = total mass − solute mass = 250 − 20 = 230 g. Therefore option A is correct. Option B is only the solute mass, while option C ignores the solute and option D comes from subtracting 8 g instead of 20 g.

For mixing the same solute, add the moles contributed by each solution and divide by the stated final volume. First solution: n = 3 × 0.200 = 0.600 mol. Second solution: n = 1.5 × 0.400 = 0.600 mol. Total moles = 1.200 mol; final volume = 600 mL = 0.600 L. Therefore M = 1.200/0.600 = 2.0 M, so option B is correct. Options based on averaging concentrations without volume weighting are incorrect.

Mole fraction requires moles of each component, not their masses directly. Sucrose moles = 34.2/342 = 0.100 mol. Water moles = 500/18 ≈ 27.78 mol. Total moles = 27.78 + 0.100 = 27.88 mol. Thus sucrose mole fraction = 0.100/27.88 ≈ 0.0036, so option A is correct. Using the mass ratio or forgetting water in the denominator would give the larger distractor values.

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of the entire solution. Moles of solute = 10 g ÷ 50 g mol−1 = 0.20 mol. The water mass is 190 g = 0.190 kg. Thus, molality = 0.20 ÷ 0.190 = 1.0526, approximately 1.05 mol kg−1. Therefore option B is correct; using 10 g or 200 g as the solvent mass would produce distractor values.

A 5 molal aqueous solution contains 5 mol of solute in 1 kg of water. Taking the molar mass of water as 18 g mol−1, 1 kg water contains 1000 ÷ 18 ≈ 55.5 mol water. Total moles are therefore 5 + 55.5 = 60.5 mol. The solute mole fraction is 5 ÷ 60.5 ≈ 0.083. Thus option B is correct; molality itself is not a mole fraction.

Choose a convenient 100 g sample of solution. It contains 30 g urea and 70 g water. Using molar masses urea = 60 g mol−1 and water = 18 g mol−1, the amounts are 30 ÷ 60 = 0.50 mol urea and 70 ÷ 18 ≈ 3.89 mol water. Hence the urea mole fraction is 0.50 ÷ (0.50 + 3.89) ≈ 0.114. Option B is correct; 0.300 is the mass fraction, not the mole fraction.

Assume 100 g solution. It contains 15 g glucose and 85 g water. Moles of glucose = 15/180 = 0.0833 mol, and solvent mass = 0.085 kg. Hence molality = 0.0833/0.085 ≈ 0.980 mol kg−1. The denominator is the mass of water only, which is why using 100 g would produce a wrong result.

The initial solution contains 10% of 200 g = 20 g solute. Let x g pure solute be added. The final solute mass is 20 + x, while the final solution mass is 200 + x because the added material is also part of the solution. Set the required mass percentage: (20 + x)/(200 + x) = 0.20. Solving gives 20 + x = 40 + 0.2x, so 0.8x = 20 and x = 25 g. Option C is correct.

The original solution contains 20% of 300 g = 60 g solute. Adding water changes the total mass but does not change the solute mass. If x g water is added, the final mass is 300 + x, and the required condition is 60/(300 + x) = 0.12. Thus 300 + x = 500, giving x = 200 g. Therefore option C is correct; 500 g is the final solution mass, not the amount of water added.

Dilution preserves the moles of solute, so use M1V1 = M2V2. Initially, n = 2 mol L−1 × 1 L = 2 mol. To obtain a final concentration of 0.8 M, the required final volume is V2 = 2 mol ÷ 0.8 mol L−1 = 2.5 L. The original volume was 1.0 L, so water added = 2.5 − 1.0 = 1.5 L. Hence option B is correct; 2.5 L is the final volume, not the added amount.

During dilution, the amount of solute remains constant, so M1V1 = M2V2. Substitution gives 5 × 100 mL = 0.5 × V2, hence V2 = 1000 mL. This is the final volume of the diluted solution. The initial volume was 100 mL, therefore the volume of water added is 1000 − 100 = 900 mL. Option C is correct; choosing 1000 mL confuses final volume with added water.

Choose a convenient 100 g solution basis: it contains 40 g solute and 60 g water. Solute moles = 40/80 = 0.50 mol, while water moles = 60/18 ≈ 3.33 mol. Total moles are about 3.83 mol, so the solute mole fraction is 0.50/3.83 ≈ 0.130. Option B is correct; 0.400 is the mass fraction, not the mole fraction.

Take 100 g of solution, containing 60 g ammonia and 40 g water. With molar masses 17 g mol⁻¹ for NH₃ and 18 g mol⁻¹ for H₂O, ammonia moles = 60/17 ≈ 3.53 and water moles = 40/18 ≈ 2.22. Total moles ≈ 5.75, so x(NH₃) = 3.53/5.75 ≈ 0.61. Option C is correct.

Take 1 mol of total solution: solute moles are 0.01 and water moles are 0.99. The solvent mass is 0.99 × 18 = 17.82 g = 0.01782 kg. Molality is moles of solute per kilogram of solvent, so m = 0.01/0.01782 ≈ 0.561 mol kg⁻¹, which rounds to 0.56 mol kg⁻¹. Therefore option B is correct.

A 1 molal aqueous solution contains 1 mol of solute in 1 kg of water. Using the molar mass of water, 18 g mol−1, the solvent contains 1000/18 ≈ 55.5 mol water. Therefore, total moles are 1 + 55.5 = 56.5 mol, and the solute mole fraction is x = 1/56.5 ≈ 0.0177. Hence option A is correct; the other statements confuse volume, mass, or concentration units.

Mole fraction of water is the ratio of water moles to total moles. Let the solute amount be x mol. Then 9/(9 + x) = 0.9. Solving gives 9 = 8.1 + 0.9x, so 0.9 = 0.9x and x = 1.0 mol. Equivalently, 9 mol is 90% of the total, so total moles are 10 and solute moles are 1. Option B is correct.

Association means that two or more solute molecules combine to form a larger associated particle, such as a dimer. Consequently, the number of independent particles in solution becomes smaller than the number expected from the original molecules. Since the van’t Hoff factor compares actual particles with ideal particles, association gives i < 1. Acetic acid dimerisation in benzene is a standard example.