Problem 47
Question
Challenge \(A 0.045 m\) solution (consisting of a nonvolatile, nonelectrolyte solute) is experimentally found to have a freezing pointdepression of \(0.08^{\circ} \mathrm{C}\) . What is the freezing point depression constant\(\left(K_{\mathrm{f}}\right) .\) Which is most likely to be the solvent: water, ethanol, or chloroform?
Step-by-Step Solution
Verified Answer
The freezing point depression constant, \( K_f \), is approximately 1.78 °C/m; the solvent is most likely water.
1Step 1: Understanding Freezing Point Depression
When a solute is dissolved in a solvent, the freezing point of the solution decreases compared to that of the pure solvent. This phenomenon is called freezing point depression. It is represented by the formula:\[ \Delta T_f = i \cdot K_f \cdot m \]where \( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor (which is 1 for nonelectrolytes), \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution.
2Step 2: Substitute Known Values into the Equation
In this problem, we are given that \( \Delta T_f = 0.08^{\circ} \mathrm{C} \) and the molality \( m = 0.045 \). The solute is nonelectrolyte (\( i = 1 \)). We need to find \( K_f \). Substitute the known values into the formula:\[ 0.08 = 1 \times K_f \times 0.045 \]
3Step 3: Solve for Freezing Point Depression Constant
To find \( K_f \), solve the equation:\[ K_f = \frac{0.08}{0.045} \]Calculate the value:\[ K_f \approx 1.78^{\circ} \mathrm{C/m} \]
4Step 4: Identify the Most Likely Solvent
Compare the calculated \( K_f \) value with the typical \( K_f \) values for each solvent:- Water: \( K_f = 1.86^{\circ} \mathrm{C/m} \)- Ethanol: \( K_f = 1.99^{\circ} \mathrm{C/m} \)- Chloroform: \( K_f = 4.68^{\circ} \mathrm{C/m} \)The calculated value of \( K_f \approx 1.78^{\circ} \mathrm{C/m} \) is closest to that of water.
Key Concepts
Colligative PropertiesVan't Hoff FactorMolalitySolution Chemistry
Colligative Properties
Colligative properties are fascinating because they depend solely on the number of solute particles in a solution, not the identity of the solute. This means that it doesn't matter what kind of solute you dissolve, just how much of it is dissolved.
Freezing point depression is an example of a colligative property, where adding a solute lowers the freezing point of the solvent. When you add any solute that doesn't evaporate easily (a nonvolatile solute) or carry electric charge (nonelectrolyte), the freezing point of the solution becomes lower.
Freezing point depression is an example of a colligative property, where adding a solute lowers the freezing point of the solvent. When you add any solute that doesn't evaporate easily (a nonvolatile solute) or carry electric charge (nonelectrolyte), the freezing point of the solution becomes lower.
- Other colligative properties include boiling point elevation, vapor pressure lowering, and osmotic pressure.
- These properties are vital for various applications, like antifreeze in car engines or even understanding how organisms survive in cold climates.
Van't Hoff Factor
The van't Hoff factor, denoted as \(i\), is crucial for calculating colligative properties like freezing point depression. It tells us how many particles a solute splits into once dissolved in a solution.
For nonelectrolytes, like in our exercise, \(i\) is usually 1 because they do not dissociate into ions. For electrolytes, which can split into ions, \(i\) can be greater than 1. For instance, sodium chloride (NaCl) in water would have \(i\) around 2 because it dissociates into two separate ions: Na+ and Cl-.
This factor is used to adjust the calculations for colligative properties to account for the number of particles contributing to the effect.
For nonelectrolytes, like in our exercise, \(i\) is usually 1 because they do not dissociate into ions. For electrolytes, which can split into ions, \(i\) can be greater than 1. For instance, sodium chloride (NaCl) in water would have \(i\) around 2 because it dissociates into two separate ions: Na+ and Cl-.
This factor is used to adjust the calculations for colligative properties to account for the number of particles contributing to the effect.
- The calculation for freezing point depression becomes different when \(i\) is greater than 1, leading to a more significant effect per mole of solute added.
- Always identify whether your solute impacts \(i\) to ensure accurate calculations.
Molality
Molality is a measure of concentration, much like molarity, but it works a bit differently, especially useful for colligative properties. It is defined as the number of moles of solute per kilogram of solvent.
In the context of solution chemistry, molality is highly beneficial because it does not change with temperature. Unlike molarity, which changes because it's based on volume (which can expand or contract with heat), molality is grounded on mass.
In the context of solution chemistry, molality is highly beneficial because it does not change with temperature. Unlike molarity, which changes because it's based on volume (which can expand or contract with heat), molality is grounded on mass.
- Since colligative properties are independent of temperature, using molality makes calculations straightforward and consistent.
- Molality helps determine how much the freezing point of a solvent will be lowered when you add a solute.
Solution Chemistry
Solution chemistry examines how substances dissolved in liquids behave, which includes physical changes like freezing and boiling point alterations. It's vital in both academic and practical applications, from laboratory experiments to industrial processes.
When studying freezing point depression, we delve into how solutes interact with solvents to change the solution's properties.
When studying freezing point depression, we delve into how solutes interact with solvents to change the solution's properties.
- The study of solutions covers various phenomena, including how solutes affect the flow of electricity or the solution's color.
- Understanding these interactions allows chemists to manipulate conditions for specific outcomes, like creating antifreeze solutions or formulating new products.
Other exercises in this chapter
Problem 41
Describe how intermolecular forces affect solvation.
View solution Problem 43
Sumarize If a seed crystal was added to a supersaturated solution, how would you characterize the resulting solution?
View solution Problem 48
Explain the nature of colligative properties.
View solution Problem 49
Describe four colligative properties of solutions.
View solution