Question

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?

Answer 1

Answer: The given sample of water is not safe for drinking.

Explanation:

We are given:

Concentration of fluorine in water recommended = 4.00 ppm

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of fluorine in water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

Mass of fluorine = $0.152mg=0.152\times 10^{-3}g$    (Conversion factor:  1 g = 1000 mg)

Mass of water = 5.00 g

Putting values in above equation, we get:

$\text{ppm of fluorine in water}=\frac{0.152\times 10^{-3}}{5}\times 10^6\\\\\text{ppm of fluorine in water}=30.4$

As, the calculated concentration is greater than the recommended concentration. So, the given sample of water is not safe for drinking.

Hence, the given sample of water is not safe for drinking.