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Sagot :
Let's start with calculating the concentration of the fluoride ion with the provided data:
[tex]\%m/v\text{ F=}\frac{20.8\text{ g}}{1.37*10^4L}=1.5182\text{ *10}^{-3}\text{ g/L}[/tex]To convert to ppm, we have to remember the formula:
[tex]ppm=\frac{mass\text{ of substrate \lparen mg\rparen}}{mass\text{ of the solution \lparen kg\rparen}}[/tex]But, as the dissolvent is water and we can assume that the density is 1 g/ml, the same as 1 kg/L, we can calculate the ppm by dividing the mg of F- into the provided volume of water:
[tex]ppm=\frac{(20.8*1000)\text{ mg F}}{1.37*10^4\text{ L sln}}=1.5182\text{ ppm F}^-\text{ }[/tex]So, the answer is 1.5182 ppm F-.
For the second part, we only have to notice that ppm is the same (in this case) that mg/L. So, it means that the level is acceptable as it is 1.5182 mg/L, which is lower than 2 mg/L.
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