Question
How many gram ofsodiun fluoride should be added to 100000liter of drinking water containing 0.6 ppm of sodium fluoride to provide a recommended concentration of 1.75 ppm
Answers
look at the amounts of NaF involved:
(.6/10^6)*100000 + 1.0*x = (1.75/10^6)(100000+x)
x = 0.115
check:
.6 ppm in 10^5L = 0.06g
add 0.115 g and you have 0.175g, or 1.75ppm
(.6/10^6)*100000 + 1.0*x = (1.75/10^6)(100000+x)
x = 0.115
check:
.6 ppm in 10^5L = 0.06g
add 0.115 g and you have 0.175g, or 1.75ppm
Thank you .. but the right answer in my book is 115g
Hmm. Obviously I misplaced a decimal point. I expect you can find the flaw in my reasoning.
1.75-0.6ppm=1.15ppm
Which means 1.15g in 10^6mls
So how many grams in 10litres.
Change litres into mls and calculate
Answer will be 115g
Which means 1.15g in 10^6mls
So how many grams in 10litres.
Change litres into mls and calculate
Answer will be 115g
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