Mole concept and molar mass

Correct answer: B. 0.221

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.

Correct answer: A. 9 mol

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.

Correct answer: A. 40 percent

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.

Correct answer: A. 230 g

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.

Correct answer: B. 2.0 M

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.

Correct answer: A. 0.0036

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.

Correct answer: B. 1.05 mol kg−1

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.

Correct answer: B. 0.083

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.

Correct answer: B. 0.114

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.

Correct answer: B. 0.98 mol kg−1

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.

Correct answer: C. 25 g

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.

Correct answer: C. 200 g

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.

Correct answer: B. 1.5 L

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.

Correct answer: C. 900 mL

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.

Correct answer: B. 0.130

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.

Correct answer: C. 0.61

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.

Correct answer: B. 0.56 mol kg−1

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.

Correct answer: A. Because 1 kg water has about 55.5 mol water

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.

Correct answer: B. 1.0 mol

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.

Correct answer: A. Less than 1

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.