Question

Be sure to answer all parts. Fluoride ion is poisonous in relatively low amounts: 0.2 g of F− per 70 kg of body weight can cause death. Nevertheless, in order to prevent tooth decay, F− ions are added to drinking water at a concentration of 1 mg of F− ion per L of water. How many liters of fluoridated drinking water would a 70−kg person have to consume in one day to reach this toxic level?

Answer 1

To reach the toxic level of 0.2 g of F−, a 70-kg person would need to consume approximately 285.71 liters of fluoridated drinking water in one day.

Step-by-step explanation:

To determine how many liters of fluoridated drinking water a 70-kg person would have to consume in one day to reach the toxic level of 0.2 g of F−, we need to find the amount of F− ions ingested from 1 liter of water and then divide the toxic level by this amount.

Given that the concentration of F− ions in fluoridated water is 1 mg/L, or 0.001 g/L, and the person's body weight is 70 kg, we can calculate the number of liters of water needed using the equation:

Number of liters = Toxic level / (Concentration of F− ions per liter × Body weight)

Plugging in the values, we get:

Number of liters = 0.2 g / (0.001 g/L × 70 kg) = 285.71 L

Therefore, a 70-kg person would need to consume approximately 285.71 liters of fluoridated drinking water in one day to reach the toxic level of 0.2 g of F−.

Answer 2

200 liters of fluoridated drinking water would have to be consumed by as person to reach the toxic level.

Step-by-step explanation:

Toxic level of fluoride ions = 0.2 g per 70 kg body weight

Concentration of fluoride ions in drinking water = 1 mg/ L = 0.001 g/L

1 mg = 0.001 g

Let the volume of drinking water consumed by person to reach the toxic level of fluoride concentration be V.

For 70 kg body weight 0.2 grams of fluoride ions will be toxic.

`V* 0.001 g/L=0.2 g`

`V=(0.2 g)/(0.001 g/L)=200 L`

200 liters of fluoridated drinking water would have to be consumed by as person to reach the toxic level.