Mole concept and molar mass

Molarity is moles of solute per litre of solution, so the volume must be expressed in litres. Convert 100 mL to 0.100 L, then use n = M × V = 0.1 mol L⁻¹ × 0.100 L = 0.010 mol. Hence, option A is correct. Options B, C, and D result from failing to convert millilitres or applying the formula incorrectly.

Molality is the number of moles of solute per kilogram of solvent, not per litre of solution. Convert 400 g solvent to 0.400 kg. Required moles = molality × solvent mass = 2.5 mol kg−1 × 0.400 kg = 1.0 mol. Therefore, option A is correct. The value 2.5 would apply to 1 kg solvent, while 0.4 is only the converted solvent mass.

The number of moles is calculated by dividing the given mass by the molar mass: n = m/M. Therefore, n = 18 g ÷ 180 g mol⁻¹ = 0.1 mol. The gram units cancel appropriately, leaving the unit mol. Thus, the sample contains 0.1 mole of glucose. This is a direct application of the mole concept and molar mass relationship.

Mole percent is the mole fraction multiplied by 100. Therefore, mole fraction = 35/100 = 0.35. A mole fraction is a ratio of moles, so it has no unit and must lie between 0 and 1. Thus, 3.5 and 35 cannot be valid mole fractions; 0.65 would represent the other component only in a binary solution.

For mass concentration units, 1 ppm means one part in 1,000,000, while 1% means one part in 100. Consequently, 1% = 10,000 ppm. Therefore, 500 ppm expressed as a percentage is 500 ÷ 10,000 = 0.05%. Equivalently, 500 × 10−6 × 100 = 0.05%. The other options shift the decimal point by factors of ten or one hundred.

The conversion factor between percentage and ppm is 1% = 10,000 ppm. Therefore, multiply the percentage value by 10,000: 0.075 × 10,000 = 750 ppm. The same result follows by writing 0.075% as 0.00075 of the whole and multiplying by 1,000,000. Options B and C are smaller by factors of ten and one hundred; D is ten times too large.

For mass-based concentration, parts per million is calculated as ppm = (mass of solute ÷ mass of solution) × 1,000,000. Convert the solution mass to the same unit: 2 kg = 2,000,000 mg. Therefore, ppm = (50 ÷ 2,000,000) × 1,000,000 = 25 ppm. Option A is correct; using 2 instead of 2,000,000 would mix units and give an incorrect result.

Molarity is defined as the amount of solute in moles per litre of solution: M = n/V. Rearranging the equation gives V = n/M. Substituting the data, V = 0.4 mol ÷ 0.8 mol L−1 = 0.5 L. Thus option A is correct. The value 0.32 L results from multiplying n and M, while 2 L incorrectly reverses the required ratio.

Molarity follows M = n/V, where V must initially be expressed in litres. Hence V = n/M = 0.6 mol ÷ 2 mol L−1 = 0.3 L. Converting to millilitres gives 0.3 × 1000 = 300 mL, so option A is correct. The 600 mL distractor comes from multiplying instead of dividing, and 200 mL reverses the numerical relationship.

Molarity is the ratio M = n/V. If the volume is doubled while the amount of solute remains unchanged, the molarity becomes one-half of its original value. To keep M constant, the numerator must increase in the same proportion as the denominator. Therefore, when the solution volume doubles, the moles of solute must also double. Option A is correct; halving or leaving the moles unchanged would reduce molarity.

Molality is defined as moles of solute per kilogram of solvent: m = n divided by the solvent mass. If the solvent mass becomes three times larger and solute moles stay fixed, the molality becomes one-third. To preserve the original ratio, the solute amount must increase by the same factor. Thus, tripling the solvent mass requires tripling the solute moles, making option A correct.

Molarity is defined by M = n/V. In this situation the number of solute moles, n, remains constant, while heating increases the solution volume V by 2%. Since the denominator becomes larger and the numerator is unchanged, the ratio decreases slightly. More precisely, the new value is M/1.02, about 98.04% of the original. Therefore option A is correct; molarity is temperature dependent.

Temperature can change the volume of a solution, so molarity, volume percentage and mass-volume percentage may change even when composition is unchanged. Molality uses moles of solute per kilogram of solvent. Mass is essentially independent of ordinary temperature changes, so molality is the least affected among these choices.

Molarity is moles of solute per litre of solution, whereas molality is moles of solute per kilogram of solvent. These denominators are the key distinction. Mole fraction is a ratio of moles, mass percent uses masses, and volume percent uses volumes; therefore the mixed definitions in the other options are incorrect.

The mole fraction of component i is defined as xᵢ = nᵢ/Σn, meaning the moles of that component divided by the total moles of every component in the mixture. It is a dimensionless quantity, and all component mole fractions add to one. Thus option A is correct. Option B describes a mass-per-volume concentration, while C and D are not definitions of mole fraction.

Mass-volume percentage is defined as grams of solute present in 100 mL of solution. Therefore, data stated as solute mass in grams and solution volume in millilitres can be used directly. Moles would suggest mole fraction or molarity, kilograms of solvent suggest molality, and temperature alone cannot determine a concentration.

Molarity is defined as moles of solute per litre of final solution. Sodium hydroxide has molar mass 40 g mol−1, so 8 g contains 8/40 = 0.20 mol. The final volume is 200 mL = 0.200 L. Therefore M = 0.20/0.200 = 1.00 mol L−1, or 1.00 M, so option C is correct. The volume of water before dissolution must not be used; the stated final solution volume is required.

For a very dilute aqueous solution, 1 L of water is approximately 1 kg because water has density close to 1 g mL−1. The lead-ion mass is 0.05 g, which equals 50 mg. Therefore, concentration = 50 mg ÷ 1 kg = 50 mg kg−1, and mg kg−1 is numerically equal to ppm. Hence option B is correct; the other choices arise from decimal or unit-conversion errors.

Volume percent is calculated using the volume of the component divided by the final volume of the solution, multiplied by 100. Thus ethanol volume percent = (20 mL/95 mL) × 100 = 21.05%, which rounds to 21.1%. The initial total of 100 mL is not used because the final volume is 95 mL after contraction.

Mass by volume percentage is the mass of solute, in grams, present in 100 mL of solution. Since 1 cm³ equals 1 mL, the concentration is 0.001 g per mL. Therefore, in 100 mL the solute mass is 0.001 × 100 = 0.100 g. Hence, mass by volume percentage is 0.10%, so option B is correct; the other values result from decimal or unit-conversion errors.

For a dilute aqueous solution, 1 ppm is approximately 1 mg per litre because 1 L of water has a mass close to 1 kg. Thus, 40 ppm means about 40 mg of calcium ions per litre. For 5 L, mass = 40 mg L⁻¹ × 5 L = 200 mg. Therefore option C is correct; 40 mg ignores the volume multiplier and the larger values use an incorrect conversion.

Mole fraction is defined as the moles of the chosen component divided by the total moles of all components. The ratio permits us to take 2 mol solute and 3 mol solvent, giving 5 mol total. Thus, solute mole fraction = 2/(2 + 3) = 2/5 = 0.40. Option C is correct; 0.60 is the solvent fraction, while the smaller values do not use the total correctly.

In a binary solution, the mole fractions of solute and solvent add to one. Therefore, solvent mole fraction = 1 − 0.4 = 0.6. The required solute-to-solvent ratio is 0.4:0.6 = 4:6 = 2:3 after simplification. Hence option A is correct. Option B reverses the requested order, and options C and D do not represent the given fractions.

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. Convert the solvent mass: 250 g = 0.250 kg. Using moles of solute = molality × mass of solvent in kg gives n = 2 mol kg⁻¹ × 0.250 kg = 0.50 mol. Therefore option B is correct; using 250 directly would ignore the required kilogram conversion.

Mass percent means solute mass divided by total solution mass, multiplied by 100. Solute mass = 10/100 × 500 g = 50 g. The solvent mass is 500 − 50 = 450 g. Hence the solute-to-solvent ratio is 50:450, which simplifies by 50 to 1:9. Option A is correct; 9:1 reverses the requested order.