Question

The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) Usually \(t\) -butanol absorbs water on exposure to air. If the freezing point of a 10.0 -g sample of \(t\) -butanol is \(24.59^{\circ} \mathrm{C},\) how many grams of water are present in the sample?

Short answer

The presence of absorbed water in the 10.0-g sample of t-butanol causes a freezing point depression of \(0.91^{\circ} \mathrm{C}\). By using the freezing point depression formula and given Kf value, we calculate the molality (0.1 mol/kg) and moles of absorbed water (0.001 mol) in the sample. Finally, we find that there are approximately 0.018 g of water present in the sample.

Step-by-step solution

1. Identify the given information

The three given pieces of information are: - The initial freezing point of t-butanol: 25.50°C. - The freezing point depression constant (Kf) for t-butanol: 9.1°C·kg/mol. - The freezing point of the 10.0 g t-butanol sample with absorbed water: 24.59°C.

2. Determine the change in freezing point (∆Tf)

The change in freezing point of t-butanol due to the presence of water can be calculated as: ∆Tf = (Freezing point of pure t-butanol) – (Freezing point of t-butanol with absorbed water) = \(25.50^{\circ} \mathrm{C} - 24.59^{\circ} \mathrm{C} = 0.91^{\circ} \mathrm{C}\)

3. Calculate the molality of the absorbed water

Using the formula ∆Tf = Kf × molality and solving for molality, we get: Molality = ∆Tf / Kf = \(0.91^{\circ} \mathrm{C} /(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol})\) = \(0.1~\mathrm{mol/kg}\)

4. Convert the mass of t-butanol to kg

To calculate the moles of absorbed water, we need to convert the sample mass of t-butanol from grams to kg: Mass of t-butanol = \(10.0~\mathrm{g}~(1 \mathrm{kg} / 1000 \mathrm{g}) = 0.01~\mathrm{kg}\)

5. Calculate the moles of absorbed water

The moles of absorbed water can be calculated using the formula: Moles of absorbed water = molality × mass of t-butanol (in kg) = (0.1 \(\mathrm{mol}/\mathrm{kg}\)) × (0.01 \(\mathrm{kg}\)) = \(0.001~\mathrm{mol}\)

6. Calculate the mass of absorbed water

To find the mass of absorbed water, we'll use the molar mass of water (18.015 g/mol) along with the previously calculated moles of absorbed water: Mass of absorbed water = moles of absorbed water × molar mass of water = \(0.001~\mathrm{mol} \times 18.015~\mathrm{g/mol} = 0.018~\mathrm{g}\) So, there are approximately 0.018 g of water present in the 10.0 g sample of t-butanol.

Key concepts

Molality Calculation

Molality is a useful concentration unit in chemistry, especially when dealing with colligative properties like freezing point depression. Molality refers to the amount of solute present in a solution and is expressed in moles of solute per kilogram of solvent. Using molality instead of other concentration units like molarity (which is based on volume) is advantageous in scenarios involving temperature changes because volume can fluctuate with temperature, while mass does not.

To calculate molality in our exercise, we first needed to determine the change in freezing point that occurs when water is absorbed by t-butanol. We used the formula:

• ∆T_f = (Freezing point of pure t-butanol) - (Freezing point with absorbed water)
• Plugging in the values, ∆T_f = 25.50\(^\circ\)C − 24.59\(^\circ\)C = 0.91\(^\circ\)C

Using the freezing point depression constant (\(K_f = 9.1\, ^\circ\mathrm{C} \cdot \mathrm{kg}/\mathrm{mol}\)), we set up the equation:

• Molality = \( \frac{\Delta T_f}{K_f} \)
• Molality = \( \frac{0.91\, ^\circ\mathrm{C}}{9.1\, ^\circ\mathrm{C} \cdot \mathrm{kg}/\mathrm{mol}} \ = 0.1\, \mathrm{mol/kg} \)

This calculation enables us to find how much solute (water in this case) has dissolved.

Molar Mass

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance, measured in grams per mole (g/mol). It combines atomic masses of all atoms in a chemical formula. Knowing the molar mass allows us to connect the amount of substance in moles to its physical mass.

In our problem, we needed to calculate the mass of water absorbed by t-butanol. Once we had determined the molality and thus the moles of water, we could use the molar mass of water (18.015 g/mol) to convert moles into grams. We used the formula:

• Mass of absorbed water = (Moles of absorbed water) × (Molar mass of water)
• Calculated as: 0.001 mol × 18.015 g/mol = 0.018 g

This calculation tells us how much water is present in the sample by weight.

T-butanol

T-butanol, also known as tert-butanol, is an alcohol with the molecular formula C\(_4\)H\(_{9}\)OH. It is a colorless solid that melts near room temperature and is known for being easily obtained and used in laboratories due to its properties.

Its unique structure includes a tertiary butyl group, which affects its physical properties, such as its relatively high melting point compared to other alcohols. T-butanol is hygroscopic, meaning it readily absorbs moisture from the air. This characteristic makes t-butanol a great candidate for studying colligative properties such as freezing point depression.

• The initial freezing point of pure t-butanol is 25.50\(^\circ\)C.
• With absorbed water, the freezing point can lower, showing its adaptability to studying changes in solution properties.

Understanding these properties helps in using t-butanol in various chemical applications.

Phase Change

The concept of phase change is crucial in understanding how substances transition between different states of matter, primarily solid, liquid, and gas. During these transitions, certain properties of the substance, like temperature and enthalpy, change significantly. Freezing point depression is a prime example of a phase change related phenomena, where a liquid's freezing point is lowered by adding a solute. This is a colligative property, meaning it depends on the number of particles in solution, not their identity.

In our exercise, when water is absorbed by t-butanol, it becomes effectively a mixture. The presence of water as a solute lowers the freezing point of t-butanol, demonstrating phase change influenced by solute addition. Understanding these changes helps in measuring and predicting solubility and mixture behavior in real-world conditions, such as antifreeze in car engines. Anticipating how the substance transitions at different temperatures is key to applying this knowledge in practical, everyday applications.