Chapter 11, Problem 120CP

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°C) that has a freezing point equal to −0.62l°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°C) of the solution in the plant cell.

d. The plant leaf is placed in an aqueous salt solution (at 25°C) that has a boiling point of l02.0°C. What will happen to the plant cells in the leaf?

Interpretation Introduction

Interpretation: The conditions and assumption for the freezing point calculation, the calculation of osmotic pressure and state of leaf after boiling at certain temperature has to be explained.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The depression in freezing point can be given by the equation,

${\text{ΔT=K}}_{\text{f}}{\text{m}}_{\text{solute}}$

Where, $\text{ΔT}$=change in freezing point depression

                         ${\text{K}}_{\text{f}}$= molal freezing point depression constant

                          ${\text{m}}_{\text{solute}}$ = molality of solute

Answer to Problem 120CP

The molality of a solution for concentration units is used to determine the freezing point depression and the osmotic pressure is calculated from the molarity units. An assumption can be made that the molarity and molality of solution are equal.

Explanation of Solution

To explain the assumption made for ideal solution where the freezing point is used to calculate the osmotic pressure.

The depression in freezing point can be given by the equation,

${\text{ΔT=K}}_{\text{f}}{\text{m}}_{\text{solute}}$

Where, $\text{ΔT}$=change in freezing point depression

                          ${\text{K}}_{\text{f}}$= molal freezing point depression constant

                          ${\text{m}}_{\text{solute}}$ = molality of solute

The osmotic pressure can be given by the equation,

$\text{Π=MRT}$

$\begin{array}{l}\text{Here,Π=Osmotic}\text{pressure}\\ \text{M=Molarity}\text{of}\text{solution}\\ \text{R=}\text{Gas}\text{law}\text{constant}\\ \text{T=Temperature}\end{array}$

The assumption that can be made for the calculation of osmotic pressure from the boiling point is that the molarity and molality of solution remains the same.

The molality of a solution is used to determine the freezing point while the molarity units are used to calculate the osmotic pressure.

Interpretation Introduction

Interpretation: The conditions and assumption for the freezing point calculation, the calculation of osmotic pressure and state of leaf after boiling at certain temperature has to be explained.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The depression in freezing point can be given by the equation,

${\text{ΔT=K}}_{\text{f}}{\text{m}}_{\text{solute}}$

Where, $\text{ΔT}$=change in freezing point depression

                         ${\text{K}}_{\text{f}}$= molal freezing point depression constant

                          ${\text{m}}_{\text{solute}}$ = molality of solute

Answer to Problem 120CP

The condition where the molarity and molality remains the same is when the litres of the solution are same as the kilograms of the solvent present in the solution.

Explanation of Solution

To explain the reasonable condition for the assumption made in part a

The molality and molarity of solution remains equal under the condition when the litres of the solution are same as the kilograms of the solvent present in the solution.

$\text{Molarity=}\frac{\text{Moles}\text{of}\text{solvent}}{\text{Litres}\text{of}\text{solution}}\text{;Molality=}\frac{\text{Moles}\text{of}\text{solvent}}{\text{Kilograms}\text{of}\text{solvent}}$

The litres of solution are equal to the kilograms of solvent present in the solution, thus making the molarity and molality equal. This is seen in aqueous solution where the water’s density will be equal to the density of the solution.  The density of the solution will be equal to one, when there isn’t lot of solute dissolved in the solution. Therefore, for dilute solutions, the values of molarity and molality will be approximately equal to each other o

Conclusion

The molality of a solution for concentration units is used to determine the freezing point depression and the osmotic pressure is calculated from the molarity units. An assumption can be made that the molarity and molality of solution are equal. The condition where the molarity and molality remains the same is when the litres of the solution are same as the kilograms of the solvent present in the solution.

Interpretation Introduction

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$\text{Π=MRT}$

$\begin{array}{l}\text{Here,Π=Osmotic}\text{pressure}\\ \text{M=Molarity}\text{of}\text{solution}\\ \text{R=}\text{Gas}\text{law}\text{constant}\\ \text{T=Temperature}\end{array}$

Answer to Problem 120CP

The osmotic pressure is found to be $\text{8}\text{.17}\text{atm}$

Explanation of Solution

Record the given data

Freezing point = $\text{-0}\text{.621°C}$

To calculate the molality of solution

Molal freezing point depression of water = $\text{1}\text{.86°C}\text{kg}{\text{mol}}^{\text{-1}}$

From the freezing point equation

$\begin{array}{l}{\text{ΔT=K}}_{\text{f}}\text{m}\\ \\ \text{m=}\frac{\text{ΔT}}{{\text{K}}_{\text{f}}}\\ \\ \text{Molality}\text{=}\frac{\text{0}\text{.621°C}}{\text{1}\text{.86°C}\text{kg/mol}}\\ \\ \text{=0}\text{.334}\text{mol}{\text{kg}}^{\text{-1}}\end{array}$

Molality of solution = $\text{0}\text{.334}\text{mol}{\text{kg}}^{\text{-1}}$

To calculate the osmotic pressure of solution

Assume that the molality = molarity

R= $\text{0}\text{.08206}\text{L}\text{atm}$

Temperature = $\text{298}\text{K}$

$\begin{array}{l}\text{Π=MRT}\\ \\ \text{=}\frac{\text{0}\text{.334}\text{mol}}{\text{L}}\text{×}\frac{\text{0}\text{.08206}\text{L}\text{atm}}{\text{K}\text{mol}}\text{×298}\text{K}\\ \\ \text{=8}\text{.17}\text{atm}\end{array}$

Osmotic pressure of solution = $\text{8}\text{.17}\text{atm}$

Conclusion

The osmotic pressure of solution was calculated by using the values of molarity, gas law constant and temperature. The osmotic pressure of the solution is found to be $\text{8}\text{.17}\text{atm}$

Interpretation Introduction

Concept Introduction:

The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties.

The elevation in boiling point changed can be given by the equation,

${\text{ΔT=K}}_{\text{b}}{\text{m}}_{\text{solute}}$

Where, $\text{ΔT}$= change in boiling point elevation

                   ${\text{K}}_{\text{b}}$ = molal boiling point elevation constant

                ${\text{m}}_{\text{solute}}$ = molality of solute

Answer to Problem 120CP

The plant will shrink and die since the water is exited from the plant cells.

Explanation of Solution

Record the given data

Boiling temperature = $\text{102°C}$

To calculate the moles of solute

Molal boiling point constant = $\text{0}\text{.51°C}\text{kg}{\text{mol}}^{\text{-1}}$

From the elevation of boiling point equation,

$\begin{array}{l}{\text{ΔT=K}}_{\text{b}}\text{m}\\ \\ \text{m=}\frac{\text{ΔT}}{{\text{K}}_{\text{b}}}\\ \\ \text{=}\frac{\text{2}\text{.0°C}}{\text{0}\text{.51°C}\text{kg/mol}}\\ \\ \text{=3}\text{.92}\text{mol}{\text{kg}}^{\text{-1}}\end{array}$

Conclusion

The moles of solute were calculated using the values of change in boiling temperature to the molal boiling point elevation constant. The moles of solute were found to be $\text{3}\text{.92}\text{mol}{\text{kg}}^{\text{-1}}$ The isotonic solution in part c is much less concentrated that solution calculated in this part. In order to try to equilibrate the ion concentration both inside and outside the cell, the water would leave the plant cell. Since there is more concentration inconsistency, the water would leave the plant cells making them to shrink and expire