Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

Step by step solution

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a. Assumption to use freezing-point depression to calculate osmotic pressure

In addition to assuming ideal behavior of the solutions, the assumption that must be made to use freezing-point depression to calculate osmotic pressure is that the plant cell's contents and the surrounding salt solution have the same molality (also referred to as molal concentration).

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b. Conditions for the assumption to be reasonable

The assumption in part a is reasonable when the solute concentration is low, allowing the plant cells to maintain internal solute concentrations in equilibrium with the surrounding solution. In such conditions, the plant cells can effectively absorb water from the surrounding medium to maintain osmotic balance.

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c. Calculation of osmotic pressure at 25°C

To calculate the osmotic pressure at 25°C, we can use the formula for freezing point depression: ΔT_f = K_f * m where ΔT_f is the freezing point depression, K_f is the cryoscopic constant of the solvent (for water, K_f = 1.86 K kg/mol), and m is the molality of the solution. We are given ΔT_f as 0.621°C, so we can calculate the molality: m = ΔT_f / K_f m = 0.621 / 1.86 = 0.33384 mol/kg Now, we can use the molality to calculate the osmotic pressure (π) using the formula: π = n * R * T where n is the number of moles of the solute, R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin (298.15 K for 25°C). Since molality (mol/kg) is the number of moles of solute per kilogram of solvent, we can rewrite the formula as: π = (m * M) * R * T where M is the molar mass of the solvent (for water, M = 18.015 g/mol). Now we can calculate the osmotic pressure: π = (0.33384 * 18.015) * 8.314 * 298.15 π = 134850 Pa Thus, the osmotic pressure of the solution in the plant cell at 25°C is approximately 134.85 kPa.

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d. Effect of placing plant cells in a solution with a higher boiling point

When the plant leaf is placed in an aqueous salt solution at 25°C with a boiling point of 102.0°C, the solution has a higher concentration of dissolved salts than the original solution the plant thrived in. This means the cells will experience a higher osmotic pressure from the surrounding solution, causing water to move out of the cells and into the surrounding solution to achieve osmotic balance. As a result, the plant cells will likely shrink and may undergo plasmolysis, which could be harmful to the plant.

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