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Chapter 11: Problem 68

The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} /\) 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 short answer to the question is: \(0.018\) g of water is present in the \(10.0\) g sample of t-butanol.

Step by step solution

01

Calculate the Freezing Point Depression

First, we'll calculate the change in freezing point (ΔTf) by subtracting the freezing point of the t-butanol sample with unknown water content from the pure t-butanol freezing point. ΔTf = Tf(pure) - Tf(sample) ΔTf = 25.50 - 24.59 ΔTf = 0.91 °C
02

Calculate the Molality

Next, we will use the freezing point depression equation to find the molality. Rearranging the equation to solve for molality (m) gives: molality = ΔTf / Kf Now substitute the given values and calculate the molality: molality = 0.91 °C / 9.1 °C⋅kg/mol molality = 0.1 mol/kg
03

Determine the Moles of Water Present in the Sample

As we have the molality, we know there are 0.1 moles of water per kg of t-butanol. Now, we calculate the moles of water present in the 10.0 g sample: moles of water = molality × (mass of t-butanol / 1000) moles of water = 0.1 mol/kg × (10.0 g / 1000) moles of water = 0.001 mol
04

Calculate the Grams of Water Present in the Sample

To calculate the grams of water present in the sample, we multiply the moles of water by the molar mass of water (18.015 g/mol): grams of water = moles of water × molar mass of water grams of water = 0.001 mol × 18.015 g/mol grams of water = 0.018 g So, there are 0.018 g of water present in the 10.0 g sample of t-butanol.

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