Pycnometer calculation — a pycnometer containing 400 g of sand and filled to the top with water weighs 2150 g. The same pycnometer filled only with clean water weighs 1950 g. If the specific gravity of soil solids Gs is 2.5, estimate the water content w of the sand sample.

Introduction / Context:
The pycnometer test is often used to determine specific gravity and, with appropriate mass balances, to back-calculate moisture content of a granular sample. Here, Gs is given, and two weighed conditions allow us to solve for the mass of water within the sand before dilution, hence its water content w.

Given Data / Assumptions:

• Mass of dry sand solids Ms is unknown; total initial sand mass is 400 g (includes water).
• Mass of pycnometer + sand + water to top = 2150 g.
• Mass of pycnometer + water to top (no sand) = 1950 g.
• Specific gravity of solids Gs = 2.5; density of water rho_w is the reference.

Concept / Approach:

Let M_p be the mass of empty pycnometer. In the water-only case, mass of water is 1950 - M_p. In the sand case, the solids displace an equivalent volume of water equal to Vs * rho_w, where Vs = Ms / (Gs * rho_w). Balancing masses in the filled-to-top condition yields a simple relation independent of M_p and rho_w, allowing direct solution for Ms and then w.

Step-by-Step Solution:

Verification / Alternative check:

Cross-check by computing solid volume Vs = Ms / (Gs * rho_w) and confirming that the difference in water mass between the two pycnometer fillings equals Ms - Ms/Gs. The numbers are internally consistent.

Why Other Options Are Wrong:

5%, 10%, 15% are inconsistent with the derived Ms and Mw values from the displacement relation. 25% would require a larger mass difference than observed.

Common Pitfalls:

Forgetting that mass of displaced water equals Ms/Gs when using grams with water density as 1 g/cc; mixing up wet mass (400 g) with dry mass Ms; attempting to use pycnometer volume explicitly (not needed here).

Final Answer:

20%