5 kg of water sample contains 0.2 g of fluoride. Calculate the concentration of fluoride ions in terms of ppm.

The correct option is D 40
Given, mass of $F^{-}$ ion= 0.2 g
We know, 1 kg =1000g so 5 kg = 5000g
Concentration of flouride ion in ppm $=\frac{\text{weight of fluoride ion}\left(g\right)}{\text{total weight of water (g)}}\times {10}^6$
$=\frac{0.2}{5000}\times {10}^6=40$ ppm