Consider the following data : Weight of a dish = 48·6212 g 100 mL of sample is placed in the dish and evaporated. The new weight of the dish and dry solids = 48·6432 g The dish is placed in a 550 °C furnace, then cooled. New weight = 48·6300 g In this case, the total volatile solids are
- A132 mg/LCorrect
- B220 mg/L
- C88 mg/L
- D308 mg/L
Solution
Concept
Total, fixed and volatile solids
Formula
VS = TS − FS
Step-by-step solution
1. Total solids = 48.6432 − 48.6212 = 0.0220 g in 100 mL = 220 mg/L. 2. Fixed solids (residue after ignition) = 48.6300 − 48.6212 = 0.0088 g = 88 mg/L. 3. Volatile solids = 220 − 88 = 132 mg/L.
Explanation
Volatile solids are the part of total solids lost on ignition at 550 °C (mostly organic matter). They equal total solids minus fixed solids.
Why the other options are wrong
220 mg/L is the total solids, 88 mg/L is the fixed solids, and 308 mg/L is their sum.
Shortcut
VS = TS − FS; weigh in grams, convert to mg/L (× 10 000 for 100 mL).
Common mistake
Reporting total or fixed solids.
Correct answer: A) 132 mg/L