Molar mass of fluoride is 19g/mol
Given concentration =0.0400 M
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= 0.0400 mol /L
We know that number of moles,n = mass/molarmass
So mass of fluoride,m = 0.0400 mol x 19 g/mol
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= 0.76 g
So 0.76 g of F- Present in 1L=1000 mL of solution
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= 1000 mL x 1.00 g/mL
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= 1000 g
So mass percentage = mass of F- / mass of
solution
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= 0.76 g / 1000 g
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= 0.00076 %
0.76 g present in 1000 mL = 0.76 g/1L
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= 760 mg/1L
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= 760
ppm                          Â
since 1ppm = mg/L