2: Concentration Metrics

For a dilute aqueous solution, 1 litre of water has a mass of approximately 1 kg, because water density is close to 1 kg L⁻¹. Thus the mixture contains about 2 mg solute per 1 kg water. Since 1 kg equals 1,000,000 mg, the ratio is approximately 2 parts per million, or 2 ppm. Therefore A is correct; the other choices misplace the decimal factor.

Molarity is defined by M = n/V, where n is the amount of solute and V is the volume of the final solution in litres. If n stays constant and the volume changes from V to 2V, the new value is M′ = n/(2V) = M/2. Thus option A is correct: doubling the volume halves the molarity. Option B reverses the inverse relationship, C ignores the volume change, and D incorrectly implies that the solute disappears.

The governing relation is M = n/V. Because the solution volume V remains fixed, changing the solute amount from n to 2n changes the molarity to M′ = 2n/V = 2M. Therefore option A is correct: molarity doubles. It would become half only if the amount stayed constant and the volume doubled. It cannot become zero, and option D would apply only when neither the solute amount nor the volume changes.

In a complete acid–base neutralisation, one mole of H₂SO₄ can furnish two acidic protons, so its acid–base n-factor is 2. Normality is calculated as N = molarity × n-factor. Therefore, N = 1 × 2 = 2 N, making option A correct. Option B would incorrectly use n-factor 1; C and D apply incorrect factors. The specified reaction context is important because n-factor depends on the reaction.

Molarity is defined by M = n/V, where n is the moles of solute and V is the final solution volume in litres. If the same amount of solute is present but the final volume is accidentally lower, the denominator decreases while n remains constant. Consequently, the actual molarity becomes greater than the intended 0.1 M. Option B reverses the ratio, and C and D ignore the unchanged solute amount.

The concentration principle is M = moles/volume in litres. Convert 250 mL to 0.250 L, then calculate moles: n = 2 mol L⁻¹ × 0.250 L = 0.500 mol. The solute mass is n × molar mass = 0.500 × 58.5 = 29.25 g, so A is correct. B uses one mole, C doubles the required amount, and D halves the correct mass.

For a complete acid–base neutralisation, normality equals molarity multiplied by the n-factor. Each mole of Ca(OH)₂ provides two moles of OH⁻, so its acid–base n-factor is 2. Therefore N = 0.2 × 2 = 0.4 N, making A correct. B ignores the two hydroxide ions, C divides instead of multiplying, and D applies an incorrect factor.

Molarity requires the number of moles and the final solution volume in litres. The glucose amount is n = mass/molar mass = 90 g ÷ 180 g mol⁻¹ = 0.5 mol. The volume is 500 mL = 0.5 L. Therefore M = 0.5 mol ÷ 0.5 L = 1 mol L⁻¹, so option A is correct. Option B is only the mole amount, not the molarity.

Use the definition M = n/V, where n is moles of solute and V is litres of final solution. For NaOH, n = 20 g ÷ 40 g mol⁻¹ = 0.5 mol. Also, 500 mL = 0.5 L. Hence M = 0.5 mol ÷ 0.5 L = 1 M, making option A correct. Option B gives only the calculated moles, while C and D use incorrect arithmetic or units.

The mole fraction of solute is the moles of solute divided by the total moles of all components: Xsolute = nsolute/(nsolute + nsolvent). A gives 4/(4+1) = 0.80, B gives 1/5 = 0.20, C gives 2/10 = 0.20, and D gives 3/10 = 0.30. Therefore, option A has the highest mole fraction. The comparison must use total moles, not just solute moles.

Molarity uses the moles of solute and the final volume of solution in litres. First, moles of glucose = mass/molar mass = 36/180 = 0.2 mol. Next, 200 mL = 0.200 L. Therefore M = n/V = 0.2/0.200 = 1 mol L⁻¹, so option A is correct. Option B is only the calculated amount in moles; C and D arise from incorrect division or volume handling.

Molarity is calculated from moles of solute divided by litres of final solution. The glucose amount is n = 18 g/(180 g mol⁻¹) = 0.1 mol. The final volume is 200 mL = 0.2 L. Hence M = 0.1/0.2 = 0.5 mol L⁻¹, making option A correct. Option C confuses moles with molarity, while B and D do not follow from the correct volume conversion.

Use M = n/V. The number of NaOH moles is n = 20 g/(40 g mol⁻¹) = 0.5 mol. Convert the final volume: 250 mL = 0.250 L. Therefore M = 0.5/0.250 = 2 mol L⁻¹, so option B is correct. Option A reports the moles rather than molarity; C and D result from incorrect division or failure to use the litre conversion.

The governing concept is mole fraction: the mole fraction of one component equals its moles divided by the total moles of all components. Total moles = 3 + 12 = 15 mol, so the solute mole fraction is xsolute = 3/15 = 0.20. Therefore, option B is correct. The value 0.80 is the solvent mole fraction, while 0.25 and 0.75 do not result from the required total-mole calculation.

Molality is defined as the number of moles of solute per kilogram of solvent, so the solvent mass must be converted from grams to kilograms. Here, 400 g = 0.400 kg. Therefore, molality = 0.8 mol / 0.400 kg = 2 mol kg⁻¹, or 2 m. Thus option D is correct. The other values arise from failing to convert units or using an unrelated given quantity.

Molarity is moles of solute per litre of solution, so the relationship is n = M × V when volume is expressed in litres. Convert 800 mL to 0.800 L, then calculate n = 0.25 mol L⁻¹ × 0.800 L = 0.20 mol. Therefore, option A is correct. The values 0.25 and 0.8 are the given molarity and converted volume, not the number of moles.

Molality is moles of solute divided by kilograms of solvent. Rearrange this definition as moles of solute = molality × kilograms of solvent. Since 250 g = 0.250 kg, the required amount is 2 mol kg⁻¹ × 0.250 kg = 0.500 mol. Thus option B is correct. Option C merely repeats the molality, whereas the other choices ignore the proper relationship or unit conversion.

For a dilute aqueous sample, concentration in ppm is approximately expressed as milligrams of solute per kilogram of water. Divide the impurity mass by the water mass: 15 mg / 3 kg = 5 mg kg⁻¹, which is approximately 5 ppm. Therefore option A is correct. Taking 15 directly ignores the denominator, multiplying gives an incorrect value, and 0.5 results from reversing or mishandling the division.

The conversion follows the definitions of the two units: 1% equals 1 part in 100, whereas 1 ppm equals 1 part in 1,000,000. Consequently, 1% = 10,000 ppm. To convert 50 ppm to percent, divide by 10,000: 50 / 10,000 = 0.005%. Therefore option C is correct. The other values result from using an incorrect conversion factor or shifting the decimal point improperly.

The governing conversion is 1 percent = 1/100 of the whole, while 1 ppm = 1/1,000,000 of the whole. Therefore, 1% equals 10,000 ppm. For the given impurity, 0.02% × 10,000 = 200 ppm. Equivalently, 0.02/100 = 0.0002 and 0.0002 × 1,000,000 = 200. Hence option B is correct; 20 and 2000 result from incorrect conversion factors.

During dilution, the number of solute moles remains constant, so the relation M₁V₁ = M₂V₂ applies. Substituting the data gives M₂ = (0.5 M × 400 mL) ÷ 800 mL = 0.25 M. The volume has doubled, so the concentration becomes half its original value. Option A ignores the dilution, option C incorrectly increases concentration, and option D halves it twice.

Mass percentage is defined as (mass of solute ÷ mass of solution) × 100. Let the total mass of solution be M. Then 10 = (20/M) × 100. Dividing by 100 and solving gives M = (20 × 100)/10 = 200 g. Thus option B is correct. The total mass must exceed 20 g because it includes both solute and solvent; 100 g would give 20% instead.

Volume percentage is defined as (volume of solute ÷ volume of solution) × 100. If the total solution volume is V, then 25 = (40/V) × 100. Solving gives V = (40 × 100)/25 = 160 mL. The same result follows because 25% is one-fourth: 40 mL is one-fourth of the total, so the total is 4 × 40 = 160 mL. Hence option B is correct.

Molality is defined as the moles of solute divided by the mass of solvent in kilograms: m = n_solute/mass of solvent. The denominator is the mass of solvent, not the volume or total mass of the solution. Calculate each value carefully. A gives 0.5/1 = 0.5 m. B gives 1/1 = 1.0 m. C gives 1/2 = 0.5 m. D gives 0.25/2 = 0.125 m. Since 0.125 m is smaller than both 0.5 m and 1.0 m, option D is the correct answer. A and C are equal, so neither is uniquely the lowest. The unit kilogram is important: using grams without conversion would produce an incorrect numerical value. This calculation also illustrates that lowering the solute-to-solvent ratio lowers molality.

The solute mole fraction is calculated by Xsolute = nsolute/(nsolute + nsolvent). The denominator must include both solute and solvent moles. For A, X = 1/(1+9) = 1/10 = 0.10. For B, X = 2/(2+3) = 2/5 = 0.40. For C, X = 3/(3+12) = 3/15 = 0.20. For D, X = 4/(4+16) = 4/20 = 0.20. The greatest value is 0.40, so option B is correct. C and D have equal mole fractions, despite their different absolute amounts, because their solute-to-total ratios are the same. Comparing solute moles alone would give a wrong conclusion.