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 %).

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:

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%