Question

how many grams of sodium fluoride should be added to 100000 liters of drinking water containing 0.6 ppm of sodium fluoride to provide a recommended concentration of 1.75 ppm

Answer 1

Answer:

ok I figured it out

Key was 1 ppm is 1mg/L.

1.15 mg/L * 100,000 L = 115,000 mg (115g)

Answer 2

The sodium fluoride should be added to 100000 liters as $115g$.

Explanation:

Never mind. I figured it out.

Key was $1 ppm is 1mg/L$ .

$1.15 mg/L*100.000 L=115.000 mg(115g)$

Therefore, the sodium fluoride added as 115g.