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From the following: pure water solution of \(\mathbf{C}_{12} \mathbf{H}_{22} \mathbf{O}_{11}(m=0.01)\) in water solution of \(\mathrm{NaCl}(m=0.01)\) in water solution of \(\mathrm{CaCl}_{2}(m=0.01)\) in water Choose the one with the a. highest freezing point. b. lowest freezing point. c. highest boiling point. d. lowest boiling point. e. highest osmotic pressure.

Short Answer

Expert verified
a. Highest freezing point: C12H22O11 solution b. Lowest freezing point: CaCl2 solution c. Highest boiling point: CaCl2 solution d. Lowest boiling point: C12H22O11 solution e. Highest osmotic pressure: CaCl2 solution

Step by step solution

01

Identify the solutions given

The given solutions are: 1. Pure water solution of C12H22O11 (sucrose) with a molality (m) of 0.01 2. Solution of NaCl with a molality (m) of 0.01 3. Solution of CaCl2 with a molality (m) of 0.01
02

Determine the Van't Hoff factor for each solute

The Van't Hoff factor, i, is the number of ions that a solute dissociates into when dissolved in solution. We have: 1. For sucrose (C12H22O11), i = 1, as it doesn't dissociate into ions. 2. For NaCl, i = 2, as it dissociates into Na+ and Cl-. 3. For CaCl2, i = 3, as it dissociates into Ca2+ and two Cl- ions.
03

Apply the formulas for colligative properties

We'll use the formulas for freezing point depression, boiling point elevation, and osmotic pressure. Here, we assume that all three solution properties are related to the total number of solute particles. 1. Freezing point depression: \(\Delta T_f = K_f \cdot m \cdot i\) 2. Boiling point elevation: \(\Delta T_b = K_b \cdot m \cdot i\) 3. Osmotic pressure: \(\Pi = i \cdot M \cdot R \cdot T\) where K_f and K_b are the molal freezing-point depression and boiling-point elevation constants, respectively; m is the molality; i is the Van't Hoff factor; M is the molar concentration, which can be considered equivalent to molality in this case; R is the ideal gas constant; and T is the temperature in Kelvin.
04

Compare freezing points

The solution with the highest freezing point will have the lowest freezing point depression. Using the formula and the data we have for all three solutions, we can order them as follows: C12H22O11 - Lowest Freezing Point Depression NaCl - Intermediate Freezing Point Depression CaCl2 - Highest Freezing Point Depression Hence, the solution of C12H22O11 in water has the highest freezing point. The solution with the lowest freezing point is CaCl2. a. Highest freezing point: C12H22O11 solution b. Lowest freezing point: CaCl2 solution
05

Compare boiling points

The solution with the highest boiling point will have the highest boiling point elevation. Using the formula and the available data for all three solutions, we can order them as follows: C12H22O11 - Lowest Boiling Point Elevation NaCl - Intermediate Boiling Point Elevation CaCl2 - Highest Boiling Point Elevation Hence, the solution of CaCl2 in water has the highest boiling point. The solution with the lowest boiling point is C12H22O11. c. Highest boiling point: CaCl2 solution d. Lowest boiling point: C12H22O11 solution
06

Compare osmotic pressure

The solution with the highest osmotic pressure will have the highest concentration of solute particles. Using the formula and the available data for all three solutions, we can order them as follows: C12H22O11 - Lowest Number of Solute Particles NaCl - Intermediate Number of Solute Particles CaCl2 - Highest Number of Solute Particles Hence, the solution of CaCl2 in water has the highest osmotic pressure. e. Highest osmotic pressure: CaCl2 solution

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Key Concepts

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

Van't Hoff factor
The Van't Hoff factor, denoted as \( i \), is an important concept in understanding colligative properties of solutions. It represents the number of particles into which a substance dissociates in solution. Knowing this factor is crucial because it affects how a solute will change the boiling point, freezing point, and osmotic pressure of a solvent.
  • For non-electrolytes, like sucrose (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)), the Van't Hoff factor is 1 because it does not dissociate into ions.
  • In contrast, sodium chloride (NaCl) dissociates into two ions: \( \text{Na}^{+} \) and \( \text{Cl}^{-} \), giving it a Van't Hoff factor of 2.
  • For calcium chloride (CaCl2), it dissociates into three ions: one \( \text{Ca}^{2+} \) and two \( \text{Cl}^{-} \) ions, resulting in a Van't Hoff factor of 3.
Understanding the Van’t Hoff factor helps predict how much a solute will affect the properties of a solution.
Freezing point depression
Freezing point depression is a colligative property, which means it depends on the number of solute particles in a solution, not the identity of the solute. This property can be described by the formula:\[\Delta T_f = K_f \cdot m \cdot i\]where:
  • \( \Delta T_f \) is the change in freezing point.
  • \( K_f \) is the freezing-point depression constant of the solvent.
  • \( m \) is the molality of the solution.
  • \( i \) is the Van’t Hoff factor.
When a solute is added to a solvent, the freezing point is lowered because the solute particles disrupt the formation of a solid structure of the solvent. In the exercise, the sucrose solution has the highest freezing point because it has the lowest \( i \) value, leading to the smallest freezing-point depression. In contrast, the calcium chloride solution has the lowest freezing point because it has the greatest number of solute particles affecting the solvent.
Boiling point elevation
Boiling point elevation is another colligative property that describes the increase in the boiling point of a solvent when a solute is added. The formula for boiling point elevation is:\[\Delta T_b = K_b \cdot m \cdot i\]Here:
  • \( \Delta T_b \) represents the elevation in boiling point.
  • \( K_b \) is the boiling-point elevation constant of the solvent.
  • \( m \) is the molality of the solution.
  • \( i \) is the Van't Hoff factor.
The addition of solute particles leads to a higher boiling point because they hinder the escape of solvent molecules into the gas phase. In our context, calcium chloride exhibits the highest boiling point, as it dissociates into the most ions, increasing the boiling point considerably. Conversely, the sucrose solution has the lowest boiling point elevation due to its minimal change in the number of particles.
Osmotic pressure
Osmotic pressure is an important colligative property relevant in biological and chemical processes. It refers to the pressure required to stop the flow of solvent molecules through a semipermeable membrane from a dilute solution into a concentrated one. The osmotic pressure equation is:\[\Pi = i \cdot M \cdot R \cdot T\]where:
  • \( \Pi \) is the osmotic pressure.
  • \( i \) is the Van't Hoff factor.
  • \( M \) is the molar concentration of the solution.
  • \( R \) is the ideal gas constant.
  • \( T \) is the temperature in Kelvin.
The higher the number of solute particles, the higher the osmotic pressure. In the given exercise, the calcium chloride solution has the highest osmotic pressure due to its large Van’t Hoff factor, which indicates a high concentration of dissolved ions contributing to osmotic pressure. This property is significant in scenarios such as the maintenance of cell integrity and function in biological systems.

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Most popular questions from this chapter

You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: "The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil." What do you say to your friend?

a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing \(1.0 \mathrm{g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{g} / \mathrm{mol}\) ) if the density of the solution is \(1.0 \mathrm{g} / \mathrm{cm}^{3} .\) b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

A solution of sodium chloride in water has a vapor pressure of 19.6 torr at \(25^{\circ} \mathrm{C} .\) What is the mole fraction of solute particles in this solution? What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C} ?\) The vapor pressure of pure water is 23.8 torr at \(25^{\circ} \mathrm{C}\) and 71.9 torr at \(45^{\circ} \mathrm{C},\) and assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

Rationalize the trend in water solubility for the following simple alcohols: $$\begin{array}{lc} \text { Alcohol } & \begin{array}{c} \text { Solubility } \\ \left(\mathrm{g} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { at } 20^{\circ} \mathrm{C}\right) \end{array} \\ \hline \text { Methanol, } \mathrm{CH}_{3} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Ethanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Propanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & \text { Soluble in all proportions } \\ \text { Butanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{2} \mathrm{OH} & 8.14 \\ \text { Pentanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{OH} & 2.64 \\ \text { Hexanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{OH} & 0.59 \\ \text { Heptanol, } \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CH}_{2} \mathrm{OH} & 0.09 \\ \hline \end{array}$$

Liquid A has vapor pressure \(x,\) and liquid B has vapor pressure y. What is the mole fraction of the liquid mixture if the vapor above the solution is \(30 . \%\) A by moles? \(50 . \%\) A? \(80 . \%\) A? (Calculate in terms of \(x\) and \(y .\) ) Liquid A has vapor pressure \(x,\) liquid B has vapor pressure y. What is the mole fraction of the vapor above the solution if the liquid mixture is \(30 . \%\) A by moles? \(50 . \%\) A? \(80 . \%\) A? (Calculate in terms of \(x\) and \(y .\) )

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