Question

A student determines the molar mass of acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), by the method used in this experiment. She found that the equilibrium temperature of a mixture of ice and water was \(1.0^{\circ} \mathrm{C}\) on her thermometer. When she added \(11.1 \mathrm{g}\) of her sample to the mixture, the temperature, after thorough stirring, fell to \(-3.0^{\circ} \mathrm{C} .\) She then poured off the solution through a screen into a beaker. The mass of the solution was 90.4 g. a. What was the freezing point depression? \(^{\circ} \mathrm{C}\)___________. b. What was the molality of the acetone? m____________. c. How much acetone was in the decanted solution? g_____. d. How much water was in the decanted solution? g_____. e. How much acetone would there be in a solution containing \(1 \mathrm{kg}\) of water and acetone at the same concentration as she had in her experiment? \(g\) acetone____________. f. What did she find to be the molar mass of acetone, assuming she made the calculation properly? g_____.

Short answer

a. The freezing point depression is \(4.0^{\circ}C\). b. The molality of acetone is \(2.15\, \mathrm{mol\, Kg^{-1}}\). c. The amount of acetone in the decanted solution is \(10.7 \, \mathrm{g}\). d. The amount of water in the decanted solution is \(79.7\, \mathrm{g}\). e. The amount of acetone in a solution containing 1 kg of water and acetone is \(124.87\, \mathrm{g}\). f. The molar mass of acetone is \(58.08\, \mathrm{g/mol}\).

Step-by-step solution

1. Calculate the freezing point depression

The freezing point depression can be calculated by subtracting the final temperature from the initial temperature. Initial temperature = \(1.0^{\circ}C\) Final temperature = \(-3.0^{\circ}C\) Freezing point depression = Initial Temperature - Final Temperature Freezing point depression = \(1.0^{\circ}C - (-3.0^{\circ}C) = 4.0^{\circ}C\). #a# Freezing point depression = \(4.0^{\circ}C\)

2. Calculate the molality of acetone

To calculate the molality of acetone, we can use the formula for freezing point depression: Freezing point depression = \(K_{f} \times m\), where \(K_{f}\) is the cryoscopic constant for water (\(1.86\, \mathrm{kg \, K\, mol^{-1}}\)) and \(m\) is the molality of the solution. Rearranging for the molality, \(m\): \(m = \frac{\text{Freezing point depression}}{K_{f}}\) \(m = \frac{4.0^{\circ}C}{1.86\, \mathrm{ Kg\, K\, mol^{-1}}} = 2.15\, \mathrm{mol\, Kg^{-1}}\) #b# Molality of acetone = \(2.15\, \mathrm{mol\, Kg^{-1}}\)

3. Calculate the amount of acetone in the decanted solution

Let \(x\) be the mass of acetone in the decanted solution. Since the total mass of the solution is 90.4 grams and the mass of the sample added to the mixture was 11.1 grams: \(x + (90.4 - x) = 11.1\) \(x = 11.1 - 90.4 + x\) \(x= 10.7\, \mathrm{g}\) #c# Amount of acetone in the decanted solution = \(10.7 \, \mathrm{g}\)

4. Calculate the amount of water in the decanted solution

Using the total mass of the decanted solution (90.4 grams) and the mass of acetone that we found in the previous step (10.7 grams): Mass of water = Total mass of solution - Mass of acetone Mass of water = \(90.4\, \mathrm{g} - 10.7\, \mathrm{g} = 79.7\, \mathrm{g}\) #d# Amount of water in the decanted solution = \(79.7\, \mathrm{g}\)

5. Calculate the amount of acetone in a solution containing 1 kg of water and acetone

From the molality, we know that there are \(2.15 \, \mathrm{mol}\) of acetone per kg of water. To find the mass of acetone in a solution containing 1 kg of water and acetone, we can use the equation: Mass of acetone = Molality \(\times\) Mass of solution Mass of acetone = \(2.15\, \mathrm{mol\, Kg^{-1}} \times 1\, \mathrm{Kg} = 2.15\, \mathrm{mol}\) Next, we need to find the mass of acetone in grams: Molar mass of acetone = \((3 \times 12.01\, \mathrm{g/mol}) + (6 \times 1.01\, \mathrm{g/mol}) + (1 \times 16.00\, \mathrm{g/mol}) = 58.08\, \mathrm{g/mol}\) Mass of acetone = \(2.15\, \mathrm{mol} \times 58.08\, \mathrm{g/mol} = 124.87\, \mathrm{g}\) #e# Amount of acetone in a solution containing 1 kg of water and acetone = \(124.87\, \mathrm{g}\)

6. Calculate the molar mass of acetone

As we found in the previous step, the molar mass of acetone is given as: Molar mass = \(\frac{\text{Mass of acetone}}{\text{Moles of acetone}}\) Molar mass = \(\frac{124.87\, \mathrm{g}}{2.15\, \mathrm{mol}} = 58.08\, \mathrm{g/mol}\) #f# Molar mass of acetone = \(58.08\, \mathrm{g/mol}\)

Key concepts

Freezing Point Depression

Freezing point depression is an important concept in solution chemistry. It refers to the lowering of a solvent's freezing point when a solute is dissolved in it. In simpler terms, adding something like acetone to water decreases the temperature at which water turns to ice. This happens because the solute particles disrupt the formation of the solid structure, which is why the freezing point goes down.
To calculate freezing point depression, you subtract the final, lower temperature of the solution from the initial, higher temperature. In the student's experiment, this meant:

• Initial temperature: \(1.0^{\circ}C\)
• Final temperature: \(-3.0^{\circ}C\)
• Freezing point depression: \(1.0^{\circ}C - (-3.0^{\circ}C) = 4.0^{\circ}C\)

Understanding freezing point depression is crucial for comprehending how solutes affect the physical properties of solutions.

Molality Calculation

Molality is another key aspect when discussing solution chemistry. It measures 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, not volume.
To find molality, you can use the formula related to freezing point depression:

• Freezing point depression = \(K_{f} \times m\),
• Where \(K_{f}\) is the cryoscopic constant (for water, \(1.86\, \mathrm{kg\, K\, mol^{-1}}\)).
• Rearranging it gives: \(m = \frac{\text{Freezing point depression}}{K_{f}}\)
• Using the student's results: \(m = \frac{4.0^{\circ}C}{1.86\, \mathrm{kg\, K\, mol^{-1}}} = 2.15\, \mathrm{mol\, kg^{-1}}\)

This method provides a reliable way to quantify how concentrated a solution is, based on its freezing behavior.

Acetone Properties

Acetone, known chemically as \(\mathrm{CH}_{3}\mathrm{COCH}_{3}\), is a common solvent and an important substance in chemistry labs. It’s known for its distinct smell and rapid evaporation. Acetone is used extensively in manufacturing and cleaning due to its ability to dissolve many substances.
Some key properties include:

• Boiling point: Approximately \(56 \degree C\)
• Molecular weight: \(58.08\, \mathrm{g/mol}\)
• Highly flammable and should be handled with care.

In the context of freezing point depression, acetone disrupts the water's ability to solidify, effectively lowering the freezing point. This interaction makes acetone an excellent substance for studying colligative properties, as its effects can be easily observed.

Solution Chemistry

Solution chemistry involves understanding how substances interact when mixed together to form a solution. It includes studying properties like solubility, concentration, and the effects of solutes on solvents.
In the exercise, key solution chemistry principles are illustrated by:

• Measuring the freezing point depression to understand solute-solvent interactions.
• Determining molality to quantify concentration and derive molar mass.
• Recognizing how properties of solutes like acetone affect experiment outcomes.

Understanding these concepts can help predict how a solution will behave in different conditions, aiding in practical real-world applications such as chemical synthesis and product formulation. Solution chemistry is integral for solving everyday problems in scientific and industrial settings.