Suppose a group of volunteers is planning on building a park near a local lake.

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. If a 12.9 cm^3 sample of lake water is found to have 161.5 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 2.61 days to remove 54.40 kg of As from a water source. 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 0.790 km^3. 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 m a year?

Explanation / Answer

Ans. A. 1 ppb = 1 µg solute per kg solution

Given,

            Mass of water sample = Volume x density = 12.9 cm3 x (1.0 g cm-3)

                                                = 12.9 g

                                                = 0.0129 kg                                       ; [ 1 kg = 103 g]

            Mass of As = 161.5 ng = 161.5 x 10-3 µg = 0.1615 µg          ; [1 ng = 10-3 µg]

Concentration in ppb = Mass (µg) of As / Mass of water

                                    = 0.1615 µg / 0.0129 kg

                                    = 12.52 µg/kg

                                    = 12.52 ppb  

Ans. B. Total volume of water in lake = 0.790 km3

                                                = 0.790 x 1012 L                                   ; [1 km3= 1012 L]

Total mass of water = Volume x density

= 0.790 x 1012 L x (1.0 kg L-1)

= 0.790 x 1012 kg

Total As content in lake = Total mass of water x [As] in ppb

                                    = 0.790 x 1012 kg x (12.52 µg/kg)

                                    = 9.89 x 1012 µg

                                    = 9.89 x 1012 x 10-6 g                                       ; [1 µg = 10-6 g]

                                    = 9.89 x 106 g

                                    = 9.89 x 106 x 10-3 kg                                      ; [1 g = 10-3 kg]

                                    = 9.89 x 103

Ans. C. Time (days) taken for As removal = total As content / rate of As removal

                                                = 9.89 x 103 kg / (54.40 kg / 2.61 days)

                                                = 471.52 days

                                                = 471.52 / 365 year                             [1 day = 1/365 year]

                                                = 1.3 year                              

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