Problem 15 - Chemistry: An Atoms-Focused Approach

Question

The concentrations of very dilute solutions are sometimes expressed as parts per million. Express the concentration of each of the following trace and ultratrace essential elements in parts per million: a. Fluorine, \(110 \mathrm{mg}\) in \(70 \mathrm{kg}\) b. Silicon, \(525 \mathrm{mg} / \mathrm{kg}\) c. Iodine, \(0.043 \mathrm{g}\) in \(100 \mathrm{kg}\)

Step-by-Step Solution

Answer

Answer: The concentrations of the trace elements in parts per million are: a. Fluorine: ≈ 1.57 ppm b. Silicon: 525 ppm c. Iodine: ≈ 0.43 ppm.

Step 1: 1. Convert Fluorine concentration to ppm

To find the concentration of Fluorine in ppm, we need to divide the amount of Fluorine (\(110 \mathrm{mg}\)) by the total amount of the solution (\(70 \mathrm{kg}\)) and then multiply by \(10^6\). First, we convert the total amount to milligrams so the units match. \(70 \mathrm{kg}\) is equal to \(70 \times 10^6 \mathrm{mg}\). So the calculation is: \(\frac{110 \mathrm{mg}}{70 \times 10^6 \mathrm{mg}} \times 10^6 = \frac{110}{70} \approx 1.57\) ppm.

Step 2: 2. Convert Silicon concentration to ppm

The concentration of Silicon is given in \(\mathrm{mg/kg}\). To convert it to ppm, we simply multiply by \(10^6 \frac{\mathrm{mg}}{\mathrm{kg}}\): \(525 \mathrm{mg/kg} \times 10^6 \frac{\mathrm{mg}}{\mathrm{kg}} = 525 \times 10^6 \mathrm{ppm} = 525 \mathrm{ppm}\).

Step 3: 3. Convert Iodine concentration to ppm

To find the concentration of Iodine in ppm, we need to divide the amount of Iodine (\(0.043 \mathrm{g}\)) by the total amount of the solution (\(100 \mathrm{kg}\)) and then multiply by \(10^6\). First, we convert both Iodine and the total amount to milligrams, so the units match. \(0.043 \mathrm{g}\) is equal to \(0.043 \times 10^3 \mathrm{mg}\) and \(100 \mathrm{kg}\) is equal to \(100 \times 10^6 \mathrm{mg}\). So the calculation is: \(\frac{0.043 \times 10^3 \mathrm{mg}}{100 \times 10^6 \mathrm{mg}} \times 10^6 = \frac{43}{100} \approx 0.43\) ppm. In conclusion, the concentrations of the given trace elements in parts per million are: a. Fluorine: \(\approx 1.57 \mathrm{ppm}\) b. Silicon: \(525 \mathrm{ppm}\) c. Iodine: \(\approx 0.43 \mathrm{ppm}.\)