A 100 mL water sample is collected from the activated sludge process of municipa

Question

A 100 mL water sample is collected from the activated sludge process of municipal wastewater treatment. The sample is placed in a drying dish (weight 0.5000g before the sample is added) and then placed in an oven at 104 degree C until all moisture is evaporated. The weight of the dried dish is recorded as 0.5625 g. A similar 100 mL sample is filtered, and the 100 mL liquid sample that passes through the filter is collected and placed in another drying dish (weight 0.5000 g before the sample is added). This sample is dried at 104 degree C, and the dried dish's weight is recorded as 0.5325 g. Determine the concentration (in mg/L) of total solids, total suspended solids, total dissolved solids, and volatile suspended solids. (Assume VSS = 0.7TSS).

Explanation / Answer

a) in 100 ml sample , mass of total solids = 0.5625 - 0.5 = 0.0625 g = 62.5 mg

so concentration (in mg/L) of total solids = 62.5 / 0.1 = 625 mg/L

b) in 100 ml sample... mass of dissolved solids = 0.5325 - 0.5 = 0.0325 g = 32.5 mg

so concentration of dissolved solids = 32.5 / 0.1 = 325 mg/L

c) concentraion of suspended solids = conc. of total solids - conc. of dissolved solids

= 625 - 325 = 300 mg/L

d) conc. of volatile suspended solids = 0.7 * 300 = 210 mg'L

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