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:
[tex]1.5\frac{mg}{L} *\frac{1 ppm}{1\frac{mg}{L} } =1.5 ppm[/tex]
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:[tex]1.0g*\frac{1000mg}{1g}=1000mg[/tex]
Taking the concentration of fluoride in tap water to be 1.5 mg/L,
Volume of tap water that contains 1000 mg fluoride
=[tex]1000mg*\frac{1mL}{1.5mg}=666.67 mL[/tex]
Rounding the volume to three significant figures, 667 mL of tap water.
