Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of bo
ID: 733670 • Letter: F
Question
Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of body weight can cause death. Nevertheless, in order to prevent tooth decay, F- ions are added to drinking water at a concentration of 1 mg of F- ion per L of water. How many liters of fluoridated drinking water would a 70-kg person have to consume in one day to reach this toxic level? times 10 L How many kilograms of sodium fluoride would be needed to treat a 5.06 times 105-gal reservoir? times 10 kg NaF Details of last check answer:Explanation / Answer
The lethal dose for a 70 Kg body is 70 x 0.20 = 14g Since 1mg = 0.001g (or 0.000001 Kg) The lethal dose is contained in 14/0.001 = 14 000 litres. 2) NaF contains: (19 / 23 + 19) x 100 = 45.2 % fluoride (ie 100 Kg of NaF contain only 45.2 Kg of fluoride, F-) And since 1 gallon = 3.79 litres 5.06x 10^5 gallons x 3.79 = 1917740 litres at 0.000001 Kg per litre and NaF containing 45.2% F- we need: 0.000001 x 1917740 x 100/45.2 = 0.4243Kg of NaF