Degree of saturation from masses, volume, and Gs A moist soil sample has volume 60 cm³, wet mass 108 g, and oven-dry mass 86.4 g. The specific gravity of solids is 2.52. Determine the degree of saturation (in %).
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A54%
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B64%
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C74%
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D84%
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E92%
Answer
Correct Answer: 84%
Explanation
Introduction / Context:Degree of saturation S quantifies how much of the void space is filled with water. Using basic phase relationships, S can be obtained from measured masses, total volume, and Gs without directly measuring void volume or water volume.
Given Data / Assumptions:
- Total volume V = 60 cm³.
- Wet mass Mw+d = 108 g; dry mass Md = 86.4 g ⇒ water mass Mw = 21.6 g.
- Water content w = Mw / Md = 21.6 / 86.4 = 0.25 (25%).
- Specific gravity of solids Gs = 2.52; ρw = 1 g/cm³.
Concept / Approach:
Two convenient formulas: ρd = Md / V ρd = (Gs * ρw) / (1 + e) ⇒ e = (Gs * ρw / ρd) − 1 S = (w * Gs) / e (with w as a fraction) These allow solving for e first and then S.
Step-by-Step Solution:
Compute dry density: ρd = 86.4 / 60 = 1.44 g/cm³.Find void ratio: e = (2.52 / 1.44) − 1 = 1.75 − 1 = 0.75.Compute degree of saturation: S = (0.25 * 2.52) / 0.75 = 0.84 = 84%.Verification / Alternative check:
Bulk density ρ = 108/60 = 1.80 g/cm³ leads to the same phase relationships and confirms plausibility of the result.
Why Other Options Are Wrong:
54%, 64%, and 74% underestimate S; 92% overestimates it. The calculated value is 84%.
Common Pitfalls:
Using wet mass in the dry density formula; forgetting to convert water content to a fraction before substituting; arithmetic slips when computing e.
Final Answer:
84%