Question

A mom is concerned about fluoridation in drinking water. She has read that the maximum contaminant concentration level goal for fluoride in drinking water is 4 ppm. The mom has discovered that her drinking water supply contains 0.8 ppm of sodium fluoride. How many milligrams of sodium fluoride are contained in one liter of drinking water containing 0.8 ppm of sodium fluoride? What is the ratio strength concentration of sodium fluoride in drinking water that contains 0.8 ppm 1I. of sodium fluoride? What is the ppm of fluoride in the mom's supply of drinking water? (molecular weight of NaF 42, I molecular weight of F = 19)

Answer 1

- (I.) As, 1 ppm = 1 mg / liter (when density of water is taken o as 1 kg / LTD So 0.8 ppm of Naf means 0.8 mg Nat/liter of water. E (I) 0.8 ppm of Nat means 0.8 mg Naf/ 1 of water And 0.8 mg Naf = 0.8 g Naf 0.0008 g Naf 1 lite of water Ratio Strength = 72 = 0.00089 = 10000 1000 nr. 8 10000 Ratio strength = 0.0008 .

(1) As given in question. ppm of Naf in water supply = 0.8. - ie 0.8 mg Naf in 1 l of water. Now, W. Naf molecular weight: 42. and f's molecular weight = 19. for converting fronde compound into flusside ion from sodium fluoride (Naf) divide (6 42 = 2.21. Thus, 1 mg of Nat contains 0.45 ng Fion. 0.8 mg of Nat contains 0.45 x0.8 mg Fion - 360 1000 = 0.36 mg of fion. is, 0.36 mg /L of fluoride in water supply So, 0.36 ppm is there inmonds water supply