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Chapter 11: Problem 74

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

Expert verified
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

01

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.
02

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) = \(25.50^{\circ} \mathrm{C} - 24.59^{\circ} \mathrm{C} = 0.91^{\circ} \mathrm{C}\)
03

3. Calculate the molality of the absorbed water

Using the formula ∆Tf = Kf × molality and solving for molality, we get: Molality = ∆Tf / Kf = \(0.91^{\circ} \mathrm{C} /(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol})\) = \(0.1~\mathrm{mol/kg}\)
04

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 = \(10.0~\mathrm{g}~(1 \mathrm{kg} / 1000 \mathrm{g}) = 0.01~\mathrm{kg}\)
05

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 \(\mathrm{kg}\)) = \(0.001~\mathrm{mol}\)
06

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.

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