Potato Cores, Sucrose, Salt, and Boiling Water
Students cut 24 potato cylinders of equal length and mass (2.0 g each) and blot them dry. Six cylinders go into each of four beakers containing 0.0 M, 0.2 M, 0.4 M, and 0.6 M sucrose solution. After 45 minutes, the cylinders are blotted and reweighed. Mean mass changes are +0.22 g, +0.06 g, −0.09 g, and −0.21 g. A second group repeats the 0.4 M treatment but uses 0.4 M NaCl instead of sucrose and finds a mean mass change of −0.19 g. A third group boils potato cylinders for five minutes before placing them in 0.6 M sucrose and finds a mean mass change of −0.02 g. All cylinders were blotted the same way and weighed on the same balance. The teacher asks the class to estimate the osmolarity of potato cell contents and to explain the NaCl and boiling results.
1. Estimate the osmolarity of the potato cell contents and justify your estimate with the water-potential reasoning behind the mass changes.
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Water moves by osmosis from higher water potential (lower solute concentration) toward lower water potential (higher solute concentration). Mass increased at 0.0 M (+0.22 g) and 0.2 M (+0.06 g), so those solutions were hypotonic and the cells took up water. Mass decreased at 0.4 M (−0.09 g) and 0.6 M (−0.21 g), so those were hypertonic.
The zero-change point lies between 0.2 and 0.4 M. The change in mass is roughly 0.15 g per 0.2 M step, so the crossing is near 0.25 M. Potato cell contents are therefore about 0.25 Osm, or 0.25 M sucrose equivalent. If the tissue had no functioning membrane, this consistent trend would not appear.
2. Explain the boiled-potato result. Why is the mass change so small compared with the living tissue in 0.6 M sucrose?
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Boiling denatures membrane proteins, destroys the phospholipid bilayer, and kills the cells. Without an intact selectively permeable membrane there is no osmosis: water-potential differences no longer drive net water movement across a membrane. Solute and water can move freely into and out of the tissue, and some soluble cell contents leak out.
The −0.02 g change is near zero and reflects leaching of cell material, not osmotic shrinkage. The comparison is important: it acts as a control showing that the mass changes in living tissue depend on a functioning membrane, not on the cell wall or on sucrose itself.
3. Explain why 0.4 M NaCl produced a different result from 0.4 M sucrose, and state which treatment the NaCl result most closely resembles.
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NaCl dissolves into two particles, Na+ and Cl−. A 0.4 M NaCl solution therefore has an osmolarity of about 0.8 Osm, roughly twice that of 0.4 M sucrose (0.4 Osm). The cells lose more water to 0.4 M NaCl than to 0.4 M sucrose, which matches the data: −0.19 g versus −0.09 g.
The NaCl value is close to the 0.6 M sucrose value (−0.21 g), consistent with roughly 0.8 Osm versus 0.6 Osm; the small gap could reflect some ion uptake or a shrinking gradient as cells lose water. The point is that molarity is not osmolarity: comparing solutions requires counting dissolved particles.