Question 4 [15 points]. Chromium was found sorbed to the sediment in a sandy sat

Question

Question 4 [15 points]. Chromium was found sorbed to the sediment in a sandy saturated aquifer that is 550 m long, 210 m wide, and 4.5 m thick, with a porosity of 0.36. A sediment core with a volume of 0.155 L was dried, and the sediment grains in the core were found to have a mass of 226 g and a chromium mass fraction of 27.07 ppb. Chromium was also dissolved in the groundwater in the aquifer at a concentration of 21 nmol/L. a) Accounting for both the mass dissolved in the groundwater and the mass sorbed to sediment, what was the total mass of chromium in the aquifer? b) You have enough resources to precisely re-measure one (and only one) of the listed values. If you want to most improve your estimate of total chromium, which value do you re-measure, and why? (Multiple answers are possible. I am more interested in your justification rather than your choice.)

Explanation / Answer

a) We know that ,

Volume of sediment core = 0.155 L

Mass of sediment graint grains = 226 g

i.e. mass fraction of chromium =27.07 ppb or 27.07 microgram/L

Mass of chromium present in the unknown solution = 27.07 (microgram/L)*0.155 L

= 4.1959 microgram

For known solution,

Concentration of chromium = 21 nmol/L

Volume of solution should be the same as volume of unknown solution for comparision.

i.e. Volume of known solution = 0.155 L

Mass of chromium can be calculate from concentration formula,

Mass = Required concentration (M) * Molecular weight of solute * required volume (L)

Therefore, Mass of Chromium = 21*10-9 (M) * 51.9961 (g/mol) * 0.155 L

Mass of Chromium = 169.2473*10-9

Mass of Chromiumin known solution = 0.1692 microgram

Total mass of Chromium = 0.1692 microgram + 4.1959 microgram

Total mass of Chromium = 4.3651 microgram

b) To achieve the precision, we need to change the concentration of known solution which is using for comparision instead of unknown solution, from this we can improve the estimation of chromium present in the sediment.

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