Chapter 13: Problem 107
What is the freezing point of vinegar, which is an aqueous solution of \(5.00 \%\) acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), by mass \(\left(d=1.006 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\)
Short Answer
Expert verified
Answer: The freezing point of vinegar is approximately -1.631°C.
Step by step solution
01
Calculate the mass of acetic acid and water in the solution
Given that the solution contains \(5.00\%\) acetic acid (\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)) by mass, we can calculate the mass of acetic acid and water in the solution. Assuming the total mass of the solution is \(100\,\mathrm{g}\), the mass of acetic acid in the solution is \(\left(\dfrac{5.00\,\mathrm{g}}{100\,\mathrm{g}}\right) \times 100\,\mathrm{g}=5.00\,\mathrm{g}\). The mass of water in the solution is \(100\,\mathrm{g} - 5.00\,\mathrm{g} = 95.00\,\mathrm{g}\).
02
Calculate the number of moles of acetic acid
Convert the mass of acetic acid to moles, using the molecular weight of acetic acid, which is \(12.01\,\mathrm{g/mol} + 2 \cdot 1.008\,\mathrm{g/mol} + 3 \cdot 1.008\,\mathrm{g/mol} + 2 \cdot 16.00\,\mathrm{g/mol} = 60.05\,\mathrm{g/mol}\). Therefore, the number of moles of acetic acid is \(\displaystyle \dfrac{5.00\,\mathrm{g}}{60.05\,\mathrm{g/mol}} \approx 0.08327\,\mathrm{mol}\).
03
Calculate the molality of the solution
Molality is defined as the number of moles of solute per kilogram of solvent. Since the solvent in this case is water, we can calculate the molality of the solution as \(\displaystyle \dfrac{0.08327\,\mathrm{mol}}{95.00\,\mathrm{g}\cdot 0.001\,\mathrm{kg/g}} \approx 0.8757\,\mathrm{mol/kg}\).
04
Use the freezing-point depression formula
The freezing-point depression formula is \(\delta T_{f} = K_{f}\cdot m\), where \(m\) is the molality of the solution, and \(K_{f}\) is the molal freezing-point depression constant for water, which has a value of approximately \(1.86\,\mathrm{°C\cdot kg/mol}\). Therefore, we get the freezing-point depression as \(\delta T_{f}=1.86\,\mathrm{°C\cdot kg/mol} \times 0.8757\,\mathrm{mol/kg} \approx 1.631\,\mathrm{°C}\).
05
Calculate the freezing point of the solution
The freezing point of pure water is \(0\,\mathrm{°C}\). Subtract the freezing-point depression from the freezing point of pure water to get the freezing point of the solution, i.e., \(0\,\mathrm{°C} - 1.631\,\mathrm{°C} \approx -1.631\,\mathrm{°C}\).
The freezing point of vinegar, which is an aqueous solution of \(5.00\%\) acetic acid, is approximately \(-1.631\,\mathrm{°C}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molality Calculation
Understanding molality is essential for delving into solutions and their behaviors, particularly freezing point depression. Molality, denoted as 'm', is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by temperature changes because it's based on mass rather than volume, which can expand or contract with temperature.
Here's how to calculate molality: First, determine the number of moles of solute present in the solution. This is done by dividing the mass of the solute (in grams) by its molecular weight (in grams per mole). Next, find the mass of the solvent (not the solution) in kilograms. The molality is then calculated by dividing the moles of solute by the kilograms of solvent.
Here's how to calculate molality: First, determine the number of moles of solute present in the solution. This is done by dividing the mass of the solute (in grams) by its molecular weight (in grams per mole). Next, find the mass of the solvent (not the solution) in kilograms. The molality is then calculated by dividing the moles of solute by the kilograms of solvent.
Example Application
In the given problem, to find the acetic acid's molality in vinegar, you divide the moles of acetic acid by the mass of water (the solvent) in kilograms. This results in the molality, which is the basis for calculating the aqueous solution's freezing point depression.Colligative Properties
Colligative properties stem from the number of solute particles present in a solution, irrespective of their identity. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. They are crucial for understanding how solutes can alter the physical properties of the solvent.
Freezing point depression is a particularly interesting colligative property because it explains why adding substances like salt to ice lowers its melting point. The principle can be applied to other scenarios, such as determining the freezing point of an aqueous solution, like the vinegar example.
When a solute is dissolved in a solvent, it disrupts the solvent's ability to form a solid structure, in this case, ice. This disruption requires a lower temperature to achieve the transition from liquid to solid, thus depressing the freezing point. Calculating the freezing point depression involves knowing the molality of the solution and the freezing-point depression constant of the solvent.
Freezing point depression is a particularly interesting colligative property because it explains why adding substances like salt to ice lowers its melting point. The principle can be applied to other scenarios, such as determining the freezing point of an aqueous solution, like the vinegar example.
When a solute is dissolved in a solvent, it disrupts the solvent's ability to form a solid structure, in this case, ice. This disruption requires a lower temperature to achieve the transition from liquid to solid, thus depressing the freezing point. Calculating the freezing point depression involves knowing the molality of the solution and the freezing-point depression constant of the solvent.
Aqueous Solution Concentration
The concentration of a substance in a solution is a quantitative measure of the amount of solute dissolved in a solvent. For aqueous solutions, where water acts as the solvent, concentration can be expressed in various ways including molarity, molality, mass percent, and others.
Mass percent, as used in the provided vinegar problem, indicates the mass of solute as a percentage of the total mass of the solution. It's calculated by dividing the mass of the solute by the total mass of the solution and then multiplying by 100 to get a percentage. This can be particularly useful for preparing solutions of a certain concentration or analyzing the composition of existing solutions, such as commercial vinegar.
Mass percent, as used in the provided vinegar problem, indicates the mass of solute as a percentage of the total mass of the solution. It's calculated by dividing the mass of the solute by the total mass of the solution and then multiplying by 100 to get a percentage. This can be particularly useful for preparing solutions of a certain concentration or analyzing the composition of existing solutions, such as commercial vinegar.