A sample of water is analyzed and found to contain 18.6 ppb of arsenic. If a person were to drink 2.0 L of this water per day for 1.4 years, what is the maximum amount of arsenic they could accumulate in their body?
Assume that 1 year = 365 days
anwser________ grams or arsenic
1 ppb = 1 ug/L
18.6 ppb = 18.6 ug/L x 2.0L/day x 365 days/yr x 1.4 yrs = ? ug As/yr.
To calculate the maximum amount of arsenic a person could accumulate in their body, we need to first convert the concentration of arsenic in the water from ppb (parts per billion) to grams. Here's how you can do it:
1. Convert PPB to grams per liter (g/L):
Since 1 ppb is equivalent to 1 microgram per liter (μg/L), we can multiply the concentration in ppb by 0.001 to convert it to grams per liter.
Concentration in g/L = 18.6 ppb × 0.001 = 0.0186 g/L
2. Calculate the total amount of arsenic consumed per day:
Multiply the concentration in grams per liter by the volume of water consumed per day. In this case, the person drinks 2.0 liters of water per day.
Amount consumed per day = 0.0186 g/L × 2.0 L = 0.0372 grams of arsenic
3. Calculate the total amount of arsenic consumed in 1.4 years:
Multiply the amount consumed per day by the number of days in 1.4 years (1.4 years × 365 days/year).
Amount consumed in 1.4 years = 0.0372 grams/day × (1.4 years × 365 days/year) = 18.6 grams of arsenic
Therefore, the maximum amount of arsenic the person could accumulate in their body after drinking 2.0 L of water containing 18.6 ppb arsenic for 1.4 years is 18.6 grams.