2: Concentration Metrics

Correct answer: A. 2 parts per million / 2 ppm

Explanation: 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.

Correct answer: A. Molarity becomes half.

Explanation: 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.

Correct answer: A. Molarity doubles.

Explanation: 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.

Correct answer: A. 2 normal

Explanation: 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.

Correct answer: A. Actual molarity becomes higher

Explanation: 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.

Correct answer: A. 29.25 g

Explanation: 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.

Correct answer: A. 0.4 N

Explanation: 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.

Correct answer: A. 1 M

Explanation: 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.

Correct answer: A. 1 M

Explanation: 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.

Correct answer: A. 4 moles solute and 1 mole solvent

Explanation: 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.

Correct answer: A. 1 M

Explanation: 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.

Correct answer: A. 0.5 M

Explanation: 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.

Correct answer: B. 2 M

Explanation: 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.

Correct answer: B. 0.20

Explanation: 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.

Correct answer: D. 2 m

Explanation: 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.

Correct answer: A. 0.2 mol

Explanation: 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.

Correct answer: B. 0.5 mol

Explanation: 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.

Correct answer: A. 5 ppm

Explanation: 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.

Correct answer: C. 0.005 percent

Explanation: 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.

Correct answer: B. 200 ppm

Explanation: 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.

Correct answer: B. 0.25 M

Explanation: 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.

Correct answer: B. 200 g

Explanation: 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.

Correct answer: B. 160 mL

Explanation: 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.

Correct answer: D. 0.25 mol solute in 2 kg solvent

Explanation: 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.

Correct answer: B. 2 moles solute and 3 moles solvent

Explanation: 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.