A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentration exceeds the recommended 4.00 ppm. A 5.00-g sample of this water is found to have 0.152 mg of fluoride. Should the water be declared safe for drinking?

3.04 ppm < 4.00 ppm, safe to drink

7.60 ppm > 4 ppm, safe to drink

30.4 ppm > 4 ppm, unsafe to drink

30,400 ppm > 4 ppm, unsafe to drink

"30.4 ppm > 4 ppm, unsafe to drink" is the one among the following choices given in the question that shows that the water should be declared unsafe for drinking. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has helped you.

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RomeliaThurston RomeliaThurston · Answer 2 · 2018-12-18T01:17:49-05:00

Answer: The water is unsafe to drink because 30.4 ppm > 4 ppm.

Explanation: The recommended concentration of fluoride in the water is 4.00 ppm. If the concentration of fluoride increases the recommended limit, the water is unsafe to drink and vice-versa.

Parts Per Million (ppm) is the measurement of the concentration of the solution. It is expressed in mg/kg, which means:

\(ppm=\frac{\text{Weight of solute (in mg)}}{\text{Weight of the sample ( in kg)}}\)

We are given a sample of 5.00 grams and the solute is fluoride which has a weight of 0.152 mg.

Weight of the sample = 5.00 grams = 0.005 kg (Conversion factor: 1 kg = 1000g)

Putting values in ppm equation, we get:

\(ppm=\frac{0.152mg}{0.005kg}=30.4mg/kg=30.4ppm\)

As the ppm value is more than the recommended value. Hence, the water is unsafe to drink.