(5) The following is from a laboratory test to determine the various solids conc

Question

(5) The following is from a laboratory test to determine the various solids concentration of a sample of untreated wastewater. A 100 mL sample is filtered through a filter pad that removes all the suspended solids. The dry and cool weight of the pad and crucible before filtration is 48.3152 g. After filtration, drying at 105C, and cooling, the weight of the crucible, filter pad, and dried solids is 48.6402 g. The dish placed in a furnace at 600C, then cooled. Weight is measured as 48.6315. What is the concentration of total, suspended and volatile solids in the wastewater sample, expressed in mg/L?

Explanation / Answer

In 100 ml of water;

Given, Mass of Crucible+filter = 48.3152 g

Mass of suspended particles = 48.6402 - 48.3152 = 0.325 g

Concentration of suspended particles = (0.325*1000 mg)*(1000 ml/100 ml) = 3.25*10^3 mg/L

Mass of particles remaining after removal of volatile substances = 48.6315 - 48.3152 = 0.3163 g

Mass of Volatile substances = 0.325-0.3163 = 0.0087 g

Concentration of suspended particles = (0.0087*1000 mg)*(1000 ml/100 ml) = 87 mg/L

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