Question

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 in Scientific Notation

Answer 1

Answer:

$2.0 \times 10^2 L$ liters of fluoridated drinking water would a 70−kg person have to consume in one day to reach this toxic level

Explanation:

1g =1000mg

So,

$0.2g \times \frac {1000mg}{1g}=200mg$ is toxic for 70kg person

Fluoride present in drinking water is 1mg per L.

So 200mg is present in 200L

200L of fluoridated drinking water is toxic to a 70kg person which  can cause death.

Using dimensional analysis,

$0.2 g F^{-} i o n s \times \frac{1000 m g}{1 g} \times \frac{1 L \text {drinking water}}{1 m g F^{-}ions}=200 L \text { drinking water }$

$2.0 \times 10^2 L$ drinking water (Answer)

Answer 2

Answer:

200 L per day

Explanation:

If there is 1 mg of F- per L of water, we can calculate the volume for 2 g like this:

1) Converse the toxic mass of fluoride to miligrams

1 g ------- 1000 mg

0.2 g ----  X

X = 200 mg

2) Calculate the liters of fluoridated drinking water that a person can consume to reach 200 mg

1 mg -------- 1 L

200 mg ---- X

x = 200 L