Homework Answers
Given data
Dry mass in 200 mL of water sample or suspended solids = 48.2 mg
Answer
Based on given data, dry mass in 1000 mL of water sample or suspended solids = 48.2 mg * 5 = 241 mg
Since 1000 ml = 1 L
Dry mass in 1 L of water sample or suspended solids = 241 mg
Dry mass in 1 L of water sample or suspended solids = 241 mg * (1*10-3) g /mg = 241 * 10-3 g
Since 1 L = 1000 cc
Dry mass in 1000 cc of water sample or suspended solids = 241 * 10-3 g
hence dry mass in 1 cc of water sample or suspended solids = 241 * 10-3 g / 1 * 103 cc = 241 * 10-6 g / cc ... eqn (1)
Now parts per million is the gram of suspended solids per million parts of water. Hence ppm = 1 * 10-6 g / cc... eqn (2)
From eqn (1) & eqn (2) the concentration of suspended solids = 241 ppm
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