Given that
% of FeF2 = 0.0398 %
That means 100 g of solution contains 0.0398 g of FeF2
The mass of F- present in 0.0398 g of FeF2 = 0.0398g*(38g/mol)/(93.845g/mol)
= 0.0161 g
∴ The mass of F- present in million g of solution = (0.0161 g/100g) *106 g
= 161 g F- / million g of solution
=
161 ppm
Hence the concentration of F- = 161 ppm