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Chapter 5: Problem 8

A \(1 \mathrm{M}\) solution of a glucose gives a pressure more than 25 times greater than that of the atmosphere. A \(1 \mathrm{M}\) solution of a salt gives an even larger osmotic pressure. Explain.

Short Answer

Expert verified
Salt solutions have more dissociated particles, increasing osmotic pressure.

Step by step solution

01

Understanding Osmotic Pressure

Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. It depends on the concentration of solute particles in the solution.
02

Glucose in Solution

Glucose is a molecular compound. In a solution, each glucose molecule contributes individually to the number of solute particles. For a 1 M glucose solution, the number of solute particles is equal to the concentration, so it contributes to the osmotic pressure as a single unit.
03

Salt in Solution

Salt, such as sodium chloride (NaCl), is an ionic compound that dissociates into ions in solution. For NaCl, each molecule dissociates into two ions: Na鈦� and Cl鈦�. Therefore, a 1 M NaCl solution has twice the number of solute particles compared to a 1 M glucose solution.
04

Comparing Osmotic Pressures

Because the osmotic pressure is proportional to the number of solute particles, a salt solution, which produces more ions, will have a higher osmotic pressure. Thus, a 1 M salt solution has a larger osmotic pressure than a 1 M glucose solution.

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Key Concepts

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

Solute Particles
In solutions, solute particles play a vital role in determining the properties of the solution, including osmotic pressure. Solute particles can either be ions or molecules, depending on the nature of the dissolved substance.
Understanding the distinction between different types of solute particles is essential.
  • **Ionic Compounds:** When dissolved, they break into individual ions. This increases the number of solute particles significantly.
  • **Molecular Compounds:** They remain intact as whole molecules in a solution, thus contributing fewer solute particles.
The number of solute particles directly affects properties like boiling point elevation, freezing point depression, and most importantly, osmotic pressure.
More solute particles in a solution lead to a higher osmotic pressure because there is more solute concentration for the solvent to balance.
Concentration of Solutions
The concentration of solutions refers to the amount of solute that is dissolved in a given quantity of solvent. This concept is fundamental as it determines many of the physical properties of a solution.
More concentrated solutions have more solute particles per unit volume, affecting osmotic pressure.
Concentration is commonly expressed in molarity (M), which is moles of solute per liter of solution. In the context of the original exercise, we referenced a 1 M solution, indicating a moderate concentration level.
  • **Higher Concentration:** Leads to increased osmotic pressure due to higher solute particle count.
  • **Lower Concentration:** Results in lower osmotic pressure.
The relationship is linear, meaning that doubling the concentration doubles the osmotic pressure, providing other factors remain constant.
Ionic Compounds
Ionic compounds, like sodium chloride (NaCl), are characterized by their ability to dissociate into ions when dissolved in water.
This dissociation increases the number of solute particles in the solution, thereby influencing properties such as osmotic pressure.
  • **Dissociation of NaCl:** One unit of NaCl separates into two ions: Na鈦� and Cl鈦�.
  • **Effect on Osmotic Pressure:** A 1 M NaCl solution results in a 2 M concentration of solute particles because each formula unit of NaCl provides two ions.
This doubling of solute particles results in a significantly higher osmotic pressure. Thus, ionic compounds are key in processes that depend on particle interactions in a solution, like ion exchange or salting out processes.
Molecular Compounds
Molecular compounds, such as glucose, do not dissociate into ions in a solution. They exist as individual molecules when dissolved.
This impacts the overall count of solute particles within a solution.
  • **Molecular Behavior:** Each molecule acts as a single solute particle, not increasing in number when in solution.
  • **Molecular Example -- Glucose:** In a 1 M glucose solution, each glucose molecule contributes one solute particle.
Because these compounds do not increase the number of solute particles, the osmotic pressure of molecular compounds like glucose solutions is lower compared to that of ionic solutions at the same molarity.
This distinction is crucial for understanding the behaviors of various substances in pressure-sensitive applications.

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Most popular questions from this chapter

Use the following osmotic pressure data for horse hemoglobin in \(0.2 \mathrm{M}\) phosphate and at \(3{ }^{\circ} \mathrm{C}\) to determine the molecular mass of the protein. $$ \begin{array}{lc} \hline \begin{array}{c} \text { Concentration of } \\ \text { hemoglobin }(\mathrm{g} / \mathrm{I} 00 \mathrm{ml}) \end{array} & \text { Osmotic pressure }\left(\mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right) \\ \hline 0.65 & 3.84 \\ 0.81 & 3.82 \\ 1.11 & 3.51 \\ 1.24 & 3.79 \\ 1.65 & 3.46 \\ 1.78 & 3.82 \\ 2.17 & 3.82 \\ 2.54 & 3.40 \\ 2.98 & 3.76 \\ 3.52 & 3.80 \\ 3.90 & 3.74 \\ 4.89 & 4.00 \\ 6.06 & 3.94 \\ 8.01 & 4.27 \\ 8.89 & 4.36 \\ \hline \end{array} $$

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Analysis of gene regulation involves study of structural and thermodynamic aspects of how proteins bind nucleic acid. One area of such research is the recognition of DNA operator sites by repressor molecules. Suppose protein P binds a single specific sequence on a molecule of DNA D. This is a common mechanism for the regulation of gene expression. At equilibrium, \(\mathrm{P}+\mathrm{D}\) \(\Leftrightarrow \mathrm{P} \bullet \mathrm{D} .\) A bacterial cell contains one molecule of DNA. Assume that cell is cylindrical, and that its diameter and length are \(1 \mu \mathrm{m}\) and \(2 \mu \mathrm{m}\), respectively. Calculate the total concentration of \(\mathrm{D}\). Assume that \(K_{\mathrm{eq}}=10^{-10} \mathrm{M}\). Calculate the \([\mathrm{P} \bullet \mathrm{D}]\), assuming that \([\mathrm{P}]=[\mathrm{D}]\). The concentration of bound \(\mathrm{D}\) is just \([\mathrm{P} \bullet \mathrm{D}]\). Calculate the concentration of unbound D. Calculate \([\mathrm{P} \bullet \mathrm{D}] /[\mathrm{P}]\). Give an interpretation of this quantity. The subject of binding will be discussed in detail in Chapter \(7 .\)

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