Question

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)

Answer 1

Answer:

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