
Concept explainers
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?
a)

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,
Where,
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,
Where,
The osmotic pressure can be given by the equation,
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.
b)

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,
Where,
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.
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
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.
c)

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,
Answer to Problem 120CP
The osmotic pressure is found to be
Explanation of Solution
Record the given data
Freezing point =
To calculate the molality of solution
Molal freezing point depression of water =
From the freezing point equation
Molality of solution =
To calculate the osmotic pressure of solution
Assume that the molality = molarity
R=
Temperature =
Osmotic pressure of solution =
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
d)

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,
Where,
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 =
To calculate the moles of solute
Molal boiling point constant =
From the elevation of boiling point equation,
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
Want to see more full solutions like this?
Chapter 11 Solutions
OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl's Chemistry, 9th
Additional Science Textbook Solutions
Chemistry: Structure and Properties (2nd Edition)
SEELEY'S ANATOMY+PHYSIOLOGY
Campbell Biology (11th Edition)
General, Organic, and Biological Chemistry - 4th edition
Microbiology with Diseases by Body System (5th Edition)
- 8. (16 pts) Provide the stepwise mechanism for the synthesis of the following compound via an enaminearrow_forwardDraw the titration curve of (i) weak acid vs. strong base; (ii) weak acid vs. weakbase; (iii) diprotic acid with strong base (iii) triprotic acid with strong base.arrow_forwardComplete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. More... No reaction. my ㄖˋ + 1. Na O Me Click and drag to start drawing a structure. 2. H +arrow_forward
- Predict the intermediate 1 and final product 2 of this organic reaction: NaOMe H+ + 1 2 H H work up You can draw 1 and 2 in any arrangement you like. Note: if either 1 or 2 consists of a pair of enantiomers, just draw one structure using line bonds instead of 3D (dash and wedge) bonds at the chiral center. Click and drag to start drawing a structure. X $ dmarrow_forwardPredict the major products of this organic reaction: 1. NaH (20°C) 2. CH3Br ? Some notes: • Draw only the major product, or products. You can draw them in any arrangement you like. • Be sure to use wedge and dash bonds where necessary, for example to distinguish between major products that are enantiomers. • If there are no products, just check the box under the drawing area. No reaction. Click and drag to start drawing a structure. G Crarrow_forwardPredict the major products of this organic reaction: 1. LDA (-78°C) ? 2. Br Some notes: • Draw only the major product, or products. You can draw them in any arrangement you like. . • Be sure to use wedge and dash bonds where necessary, for example to distinguish between major products that are enantiomers. • If there are no products, just check the box under the drawing area. No reaction. Click and drag to start drawing a structure. Xarrow_forward
- Please draw the structuresarrow_forwardDraw the missing intermediates 1 and 2, plus the final product 3, of this synthesis: 0 1. Eto 1. Eto- 1 2 2. MeBr 2. EtBr H3O+ A 3 You can draw the three structures in any arrangement you like. Explanation Check Click and drag to start drawing a structure.arrow_forwardDraw the missing intermediate 1 and final product 2 of this synthesis: 1. MeO- H3O+ 1 2 2. PrBr Δ You can draw the two structures in any arrangement you like. Click and drag to start drawing a structure.arrow_forward
- What is the differences between: Glyceride and phosphoglyceride Wax and Fat Soap and Fatty acid HDL and LDL cholesterol Phospho lipids and sphingosine What are the types of lipids? What are the main lipid components of membrane structures? How could lipids play important rules as signaling molecules and building units? The structure variety of lipids makes them to play significant rules in our body, conclude breifly on this statement.arrow_forwardWhat is the differences between DNA and RNA for the following: - structure - function - type What is the meaning of: - replication - transcription - translation show the base pair connection(hydrogen bond) in DNA and RNAarrow_forwardWhat is the IP for a amino acid- give an example what are the types of amino acids What are the structures of proteins The N-Terminal analysis by the Edman method shows saralasin contains sarcosine at the N-terminus. Partial hydrolysis of saralasin with dilute hydrochloric acid yields the following fragments: Try-Val-His Sar-Arg-Val His-Pro-Ala Val- Tyr- Val Arg-Val-Tyr What is the structure of saralasin?arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningGeneral, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning




