Parts Per Million And Concentration
A sample of drinking water was found to contain 0.032 g of fluoride ions dissolved in 16 kg of water. Express the fluoride ion concentration in parts per million (ppm) by mass.
Select the correct option:
Solution
2.0 ppm
Parts per million (ppm) by mass is defined as the mass of solute per million parts ((10^6)) of solution, expressed as: ppm (= (\text{mass of solute}/\text{mass of solution}) \times 10^6). Approximating the solution mass as the solvent mass (since solute is negligible): ppm (= (0.032 , \text{g} / 16000 , \text{g}) \times 10^6 = (0.032/16000) \times 10^6 = 2 \times 10^{-6} \times 10^6 = 2.0 , \text{ppm}). Option 0.2 ppm would result from a unit conversion error, perhaps using 160,000 g instead of 16,000 g, or missing a factor of 10. Option 20 ppm would result if the mass of solvent were taken as 1,600 g (1.6 kg) rather than 16 kg. Option 200 ppm would require the solvent to be only 160 g, which contradicts the problem. The WHO permissible limit for fluoride in drinking water is 1.5 mg/L ≈ 1.5 ppm, so 2.0 ppm is slightly above the limit — a realistic JEE practical-context question. Plausibility check: 0.032 g in 16,000 g (= 2 \times 10^{-6}) g/g, and (2 \times 10^{-6} \times 10^6 = 2.0) ppm.
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- parts per million and concentration
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2.0 ppm
Parts per million (ppm) by mass is defined as the mass of solute per million parts ((10^6)) of solution, expressed as: ppm (= (\text{mass of solute}/\text{mass of solution}) \times 10^6). Approximating the solution mass as the solvent mass (since solute is negligible): ppm (= (0.032 , \text{g} / 16000 , \text{g}) \times 10^6 = (0.032/16000) \times 10^6 = 2 \times 10^{-6} \times 10^6 = 2.0 , \text{ppm}). Option 0.2 ppm would result from a unit conversion error, perhaps using 160,000 g instead of 16,000 g, or missing a factor of 10. Option 20 ppm would result if the mass of solvent were taken as 1,600 g (1.6 kg) rather than 16 kg. Option 200 ppm would require the solvent to be only 160 g, which contradicts the problem. The WHO permissible limit for fluoride in drinking water is 1.5 mg/L ≈ 1.5 ppm, so 2.0 ppm is slightly above the limit — a realistic JEE practical-context question. Plausibility check: 0.032 g in 16,000 g (= 2 \times 10^{-6}) g/g, and (2 \times 10^{-6} \times 10^6 = 2.0) ppm.
This medium difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of parts per million and concentration. It appeared in the 2025 exam.
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