Suppose a group of volunteers is planning on building a park near a local lake.
ID: 535735 • Letter: S
Question
Suppose a group of volunteers is planning on building a park near a local lake. The lake is known to contain low levels of arsenic (As). Therefore, prior to starting construction, the group decides to measure the current level of arsenic in the lake. a) If a 12.7 cm^3 sample of lake water is found to have 163.3 ng As, what is the concentration of arsenic in the sample in parts per billion (ppb), assuming that the density of the lake water is 1.00 g/cm^3? One of the volunteers suggests hiring an on-site water treatment company to remove the arsenic from the lake. The company claims their process takes 2.37 days to remove 51.40 kg of As from a water source. b) Calculate the total mass (in kg) of arsenic in the lake that the company will have to remove if the total volume of water in the lake is 0.610 km^3. c) Based on the company's claim and the concentration of arsenic in the lake, how many years will it take to remove all of the arsenic from the lake, assuming that there are always 365 days in a year?Explanation / Answer
Ans. #A. As content = 163.3 ng / 12.7 cm3 [1 cm3 = 1 mL]
As content per 1000 mL water = 1000 mL x (163.3 ng / 12.7 mL)
= 12858.27 ng
So, [As] = 12858.27 ng/ L
= 12.85827 ug/ L ; [1 ug = 1000 ng]
= 12.85 ug/ L
= 12.85 ppb ; [1 ppb = 1 µg/ L ]
# B. Total volume of water in lake = 0.610 km3
= 0.610 x 1012 L ; [1 km3= 1012 L]
= 0.610 x 1012 L
Total As content in lake = Total vol. of water x [As] in ppb
= 0.610 x 1012 L x (12.85 µg/ L)
= 7.8385 x 1012 µg
= 7.8385 x 103 g ; [1 µg = 10-9 kg]
#C. Time (days) taken for As removal = total As content / rate of As removal
= 7.8385 x 103 kg / (51.40 kg / 2.37 days)
= 361.425 days
= 361.425 / 365 year [1 day = 1/365 year]
= 0.9902 year
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.