2: Concentration Metrics

Molarity is the amount of solute in moles divided by the solution volume in litres, so its unit is mol L⁻¹, commonly written as M. Mole fraction is the ratio of moles of one component to the total moles of all components. Since it is a ratio of like units, the units cancel and it is dimensionless. Therefore, option A is correct.

The mole fraction of a component equals the moles of that component divided by the total moles of all components. Here, total moles = 1 mol solute + 9 mol solvent = 10 mol. Thus, mole fraction of solvent = 9/10 = 0.9. The mole fraction has no unit and must lie between 0 and 1, so option A is the only valid answer.

A mole fraction of 0.05 means that the solute contributes 0.05 of the total number of moles. Expressed as a percentage, 0.05 × 100 = 5%. Thus, the solute is present in a relatively small amount compared with the solvent or the total mixture. It is a valid mole fraction because mole fractions range from 0 to 1. Hence, option A is correct.

Molarity is defined as moles of solute per litre of solution. Therefore, when the solute is given by mass, its amount must first be converted into moles using the relation: moles = mass in grams ÷ molar mass in grams per mole. After finding the moles, divide by the solution volume in litres. Thus, option A is correct.

First calculate the amount of glucose: n = mass/molar mass = 18/180 = 0.1 mol. Molarity is the number of moles of solute per litre of solution, so M = n/V = 0.1 mol/1 L = 0.1 mol L⁻¹, or 0.1 M. The volume used must be the final volume of the solution, not merely the solvent volume.

The number of moles of NaOH is n = m/M = 40 g ÷ 40 g mol⁻¹ = 1 mol. Molarity equals moles of solute divided by the volume of the solution in litres. Therefore, molarity = 1 mol ÷ 1 L = 1 M. The equality of the mass and molar mass indicates exactly one mole of NaOH.

A concentrated solution contains a relatively large amount of solute in a given amount of solution or solvent, compared with a dilute solution. The word “concentrated” is generally a qualitative description unless a numerical concentration is supplied. It does not mean that the solution has no solvent or that its volume is zero.

A dilute solution contains a relatively small quantity of solute compared with the amount of solvent or solution. “Dilute” indicates comparatively low concentration, but it does not require the solute to be a solid; solutes may be solids, liquids, or gases. Mole fraction cannot be 2 because mole fractions lie between 0 and 1.

Volume percentage is given by (volume of solute ÷ volume of solution) × 100. Here, the solute volume is 10 mL and the final solution volume is 50 mL. Thus, volume percentage = (10/50) × 100 = 20%. The final solution volume, rather than the amount of solvent alone, must be used in the denominator.

Mass-volume percentage is defined as the mass of solute in grams per 100 mL of solution. Using the formula (mass of solute ÷ volume of solution) × 100, we get (5 g ÷ 250 mL) × 100 = 2 g per 100 mL, or 2%. Therefore, option A is correct. The volume must be the total solution volume.

Molarity is the number of moles of solute per litre of solution. Convert the volume first: 250 mL = 0.250 L. Then calculate M = n/V = 0.25 mol ÷ 0.250 L = 1 mol L⁻¹, or 1 M. Correct conversion from millilitres to litres is essential; using 250 directly would give an incorrect result.

Molality is defined as the number of moles of solute per kilogram of solvent, not per kilogram of solution. Therefore, m = n divided by the mass of solvent = 0.1 mol ÷ 0.5 kg = 0.2 mol kg⁻¹. The unit is written as m, or molal. Unlike molarity, molality is based on solvent mass and is not dependent on solution volume.

Mole fraction expresses the proportion of one component in the total amount of substance. It is calculated as the moles of that component divided by the total moles of all components: Xᵢ = nᵢ/Σn. Because it is a ratio of part to whole, it is dimensionless and is naturally understood as a fraction of the complete solution. Molarity and molality use volume or solvent mass, while molar mass is not a concentration term.

Mass percentage is calculated by dividing the mass of the solute by the mass of the entire solution and multiplying by 100: mass percentage = (mass of solute/mass of solution) × 100. Substituting the given values gives (15 g/150 g) × 100 = 0.1 × 100 = 10%. The masses must be expressed in the same unit, and the denominator is the mass of the solution, not the solvent alone.

Volume percentage is calculated as the volume of the solute divided by the total volume of the solution, multiplied by 100: volume percentage = (volume of solute/volume of solution) × 100. Therefore, the value is (30 mL/150 mL) × 100 = 0.2 × 100 = 20%. Since both measurements are in millilitres, no unit conversion is necessary.

Mass-volume percentage, written as % (m/V), is calculated from the mass of solute and the final volume of solution: % (m/V) = (mass of solute in grams ÷ volume of solution in millilitres) × 100. Option A has the correct order and quantities. B reverses them, C is a mole ratio, and D uses the wrong substances.

Molarity is the amount of solute per litre of solution and is defined by M = n/V. Here n is the number of moles of solute and V is the volume of the solution in litres. Initially, M = n/V. If only the solute amount changes from n to 2n while the volume remains V, the new molarity is Mnew = 2n/V = 2M. Thus the molarity becomes twice its original value, so option A is correct. It would become half if the volume doubled while the amount of solute stayed fixed. It would remain unchanged only if both the relevant ratio and its numerator-denominator changes compensated each other. It cannot become zero when solute is added.

Molality is moles of solute per kilogram of solvent: m = n/S. Keeping n constant while changing the solvent mass from S to 2S gives m′ = n/(2S) = one-half of the original value n/S. Thus option A is correct. Molality depends on solvent mass, not solution volume, so the unchanged solute amount cannot compensate for the doubled denominator.

Mass percentage expresses the mass of solute as a percentage of the total mass of solution. Use mass percentage = (mass of solute/mass of solution) × 100. Therefore, (4 g/200 g) × 100 = 0.02 × 100 = 2%. The denominator is the solution mass, not the solvent mass, so option A is correct.

Volume percentage is calculated as (volume of solute/volume of solution) × 100. Substituting the given values gives (8 mL/400 mL) × 100 = 0.02 × 100 = 2%. The two volumes already use the same unit, so no conversion is needed. The ratio 8/400 alone is 0.02, not 2%, until it is multiplied by 100.

The symbol M denotes molarity, defined as moles of solute per litre of solution. Thus a 0.1 M solution contains 0.1 mol of solute in every 1 L of the final solution. Option B describes molality because it uses kilograms of solvent. Option C represents a mass-based percentage idea, while option D refers to parts per million, so option A is correct.

Molarity is M = n/V, where n is the number of moles and V is volume in litres. Rearranging gives n = M × V. For this solution, n = 0.5 mol L⁻¹ × 2 L = 1 mol. The litre units cancel, leaving moles. Therefore option A is correct; dividing by volume or ignoring the concentration would produce the distractor values.

Use n = M × V, but first express volume in litres because molarity is measured per litre. Thus 250 mL = 0.250 L, and n = 2 mol L⁻¹ × 0.250 L = 0.500 mol. Failing to convert millilitres would make the units inconsistent and lead to an incorrect result. Hence option A is correct.

The governing relation for molarity is M = n/V, where M is in mol L−1, n is the number of moles, and V is the solution volume in litres. Rearranging gives n = M × V. The stated volume is 500 mL, which must be converted to litres: 500 mL = 0.500 L. Substitution gives n = 0.2 mol L−1 × 0.500 L = 0.100 mol. Therefore option A is correct. Option B would incorrectly use one litre or ignore the half-litre volume. Option C confuses millilitres with litres, and option D is four times the calculated amount. Correct unit conversion is essential in this calculation.

Molality is the number of moles of solute divided by the mass of solvent in kilograms: m = nsolute/kg solvent. Unlike molarity, molality uses the mass of solvent rather than the volume of the complete solution. Rearranging the equation gives nsolute = m × mass of solvent. For a 2 m solution prepared with 1 kg of solvent, nsolute = 2 mol kg−1 × 1 kg = 2 mol. Hence option A is correct. Option B would produce a 1 m solution under the stated conditions. Option C would produce 0.5 m, and option D would produce 4 m. The fact that the solvent mass, rather than solution volume, is specified confirms that molality is the relevant concept.