Question

The maximum acceptable concentration of fluoride in tap water is 1.5mg/L. Express this concentration in ppm. Then figure what volume of tap water contains 1.0 g of fluroide

Answer 1

Given that the maximum acceptable concentration of fluoride in tap water = 1.5 mg/L.

1 mg/L is equivalent to 1 parts per million(ppm).

Converting 1.5 mg/L to ppm:

$1.5\frac{mg}{L} *\frac{1 ppm}{1\frac{mg}{L} } =1.5 ppm$

So the maximum acceptable concentration of fluoride in tap water in ppm is 1.5 ppm.

Finding out the volume of tap water that would contain 1.0 g fluoride:

Converting 1.0 g fluoride to mg:$1.0g*\frac{1000mg}{1g}=1000mg$

Taking the concentration of fluoride in tap water to be 1.5 mg/L,

Volume of tap water that contains 1000 mg fluoride

                              =$1000mg*\frac{1mL}{1.5mg}=666.67 mL$

Rounding the volume to three significant figures, 667 mL of tap water.