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Chapter 14: Problem 96

Snails are very sensitive to the concentration of salt in their environment. Because their "skin" is a semipermeable membrane, water can easily flow through this membrane. If you spill salt onto the "skin" of a snail, water from the inside of the body will flow out very quickly. Explain this phenomenon in terms of the chemistry described in this chapter.

Short Answer

Expert verified
Osmosis causes water to move from the inside of the snail to the salty environment outside, leading to rapid dehydration.

Step by step solution

01

Understand Osmosis

Osmosis is the process by which water molecules move from an area of lower solute concentration to an area of higher solute concentration through a semi-permeable membrane.
02

Snail's Semi-permeable Membrane

The 'skin' of the snail acts as a semi-permeable membrane, allowing only certain substances like water to pass through it while blocking most solutes like salt.
03

Effect of Salt on the Membrane

When salt is spilled onto the snail's 'skin', it creates an area of high solute (salt) concentration outside the snail's body.
04

Water Movement

Due to osmosis, water inside the snail's body will move out through the semi-permeable membrane to balance the solute concentration on both sides of the membrane.
05

Rapid Water Loss

This movement of water happens very quickly, causing the snail to lose water and potentially dehydrate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Semi-permeable Membrane
The concept of a semi-permeable membrane is crucial in understanding the process of osmosis. A semi-permeable membrane is a type of barrier that allows certain molecules or ions to pass through while blocking others. In biological systems, this selective permeability is vital for maintaining cellular balance.
The snail's 'skin', like many biological membranes, is semi-permeable. This means it permits the passage of water molecules but restricts the movement of larger molecules such as salts.
By controlling what enters and leaves the cell, semi-permeable membranes help maintain an optimal internal environment, which is essential for cells and organisms to function properly.
Solute Concentration
Solute concentration refers to the amount of dissolved substances, or solutes, in a solution. In the context of osmosis, it’s important to understand how solute concentration affects water movement.
When salt is applied to a snail's 'skin', it leads to a higher solute concentration outside the snail. Inside the snail's body, the solute concentration is lower in comparison.
This discrepancy in solute concentrations is what drives the process of osmosis. The snail’s body tries to equalize the solute concentrations on either side of the membrane by moving water to the area with higher solute concentration, which is outside in this scenario.
Water Movement
In osmosis, water moves from an area of lower solute concentration to one of higher solute concentration to achieve equilibrium. This is a passive process, meaning it doesn’t require energy.
With the snail's 'skin' exposed to salt, the outside of the snail now has a higher solute concentration than the inside. Due to osmosis, water inside the snail’s body moves out through its semi-permeable membrane to the higher solute concentration area.
This movement of water is continuous until the solute concentrations are balanced or until the organism can no longer sustain the imbalance.
Dehydration
Dehydration occurs when an organism loses more water than it takes in, and osmosis can be a significant cause of this in certain scenarios. In the case of the snail, spilling salt on its 'skin' leads to rapid water loss through its semi-permeable membrane.
The quick movement of water from inside the snail to the outside environment can deplete the snail’s internal water reserves. If the snail loses water too quickly and cannot replace it, it leads to dehydration.
Understanding how osmosis and semi-permeable membranes work helps explain why organisms like snails are so sensitive to changes in their environment, particularly with substances like salt.

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Most popular questions from this chapter

Use the equation to calculate the following: \(2 \mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (a) the moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) produced from \(3.6 \mathrm{~mol} \mathrm{H} \mathrm{H}_{2} \mathrm{SO}_{4}\) (b) the moles of \(\mathrm{H}_{2} \mathrm{O}\) produced from \(0.025 \mathrm{~mol} \mathrm{NaOH}\) (c) the moles of \(\mathrm{NaOH}\) required to react with \(2.50 \mathrm{~L}\) of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (d) the grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) that can be obtained from \(25 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{NaOH}\) (e) the volume of \(0.250 \mathrm{MH}_{2} \mathrm{SO}_{4}\) needed to react with \(25.5 \mathrm{~mL}\) of \(0.750 \mathrm{M} \mathrm{NaOH}\) (f) the molarity \((M)\) of the \(\mathrm{NaOH}\) solution when \(48.20 \mathrm{~mL}\) react with \(35.72 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\)

A solution of \(5.36 \mathrm{~g}\) of a molecular compound dissolved in \(76.8 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) has a boiling point of \(81.48^{\circ} \mathrm{C}\). What is the molar mass of the compound? Boiling point for benzene \(=80.1^{\circ} \mathrm{C}\) \(K_{b}\) for benzene \(=2.53^{\circ} \mathrm{C} \mathrm{kg}\) solvent \(/ \mathrm{mol}\) solute

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