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14 September, 04:00

As a quality control check, a sample of acetone is taken from a process to determine the concentration of suspended particulate matter. An 850 mL sample was placed in a beaker and evaporated. The remaining suspended solids were determined to have a mass of 0.001 g. The specific gravity of acetone is 0.79 g/cm.

(a) Determine the concentration of the sample as mg/L.

(b) Determine the concentration of the sample as ppm (problem from EPA Air Pollution Training Institute)

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  1. 14 September, 04:10
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    The volume of the sample given is 850 ml, the density given is 0.79 gram per cm. Now the weight of the sample will be,

    Weight = volume * density = 850 * 0.79

    = 671.5 grams

    Weight of the suspended solids given is 0.001 gram

    The concentration of the sample can be determined by using the formula,

    Concentration = wt. of sample/volume

    = [671.5 - 0.001) 10³ mg / 0.85 L

    = 789998.82 mg/L or 789998.82 ppm

    Now the concentration of suspended solids is.

    Css = 0.001 * 10³ mg / 0.85 L = 1.1764 mg per L or 1.1764 ppm
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