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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

User Emilee
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1 Answer

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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(mg)/(L) *(1 ppm)/(1(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*(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*(1mL)/(1.5mg)=666.67 mL

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



User NCA
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