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What is the vapor pressure (kPa) of water at 25 °C above an aqueous solution that is 10.6% by mass of glucose (molar mass = 180.2 g/mol)? The vapor pressure of pure water at 25 °C is 3.17 kPa. а. 3.168 b. 0.037 с. 3.133 d. 0.081 e. 3.064

What is the vapor pressure (kPa) of water at 25 °C above an aqueous solution that is 10.6% by mass of glucose (molar mass = 180.2 g/mol)? The vapor pressure of pure water at 25 °C is 3.17 kPa. а. 3.168 b. 0.037 с. 3.133 d. 0.081 e. 3.064

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Understanding Vapor Pressure of an Aqueous Solution:** **Problem Statement:** What is the vapor pressure (kPa) of water at 25°C above an aqueous solution that is 10.6% by mass of glucose (molar mass = 180.2 g/mol)? The vapor pressure of pure water at 25°C is 3.17 kPa. **Options:** - **a.** 3.168 - **b.** 0.037 - **c.** 3.133 - **d.** 0.081 - **e.** 3.064 **Explanation:** To solve this problem, one must use the concept of the colligative properties of solutions, specifically Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the mole fraction of the solvent in the solution. **Steps to Calculate:** 1. Calculate the mass of glucose (solute) and water (solvent) in the solution: - Assume we have 100 g of the solution. - Mass of glucose = 10.6 g - Mass of water = 100 g - 10.6 g = 89.4 g 2. Convert these masses to moles: - Moles of glucose = \(\frac{10.6 \, \text{g}}{180.2 \, \text{g/mol}} = 0.0588 \, \text{moles}\) - Moles of water = \(\frac{89.4 \, \text{g}}{18 \, \text{g/mol}} \approx 4.9667 \, \text{moles}\) 3. Calculate the mole fraction of water (solvent): - Mole fraction of water \( (\chi_{water}) = \frac{\text{moles of water}}{\text{total moles}} \) - Total moles = 0.0588 mol (glucose) + 4.9667 mol (water) = 5.0255 mol - Mole fraction of water \( (\chi_{water}) = \frac{4.9667}{5.0255} \approx 0.9883 \) 4. Use Raoult's Law to find the vapor pressure of the solution: - \(P_{solution
Transcribed Image Text:**Understanding Vapor Pressure of an Aqueous Solution:** **Problem Statement:** What is the vapor pressure (kPa) of water at 25°C above an aqueous solution that is 10.6% by mass of glucose (molar mass = 180.2 g/mol)? The vapor pressure of pure water at 25°C is 3.17 kPa. **Options:** - **a.** 3.168 - **b.** 0.037 - **c.** 3.133 - **d.** 0.081 - **e.** 3.064 **Explanation:** To solve this problem, one must use the concept of the colligative properties of solutions, specifically Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the mole fraction of the solvent in the solution. **Steps to Calculate:** 1. Calculate the mass of glucose (solute) and water (solvent) in the solution: - Assume we have 100 g of the solution. - Mass of glucose = 10.6 g - Mass of water = 100 g - 10.6 g = 89.4 g 2. Convert these masses to moles: - Moles of glucose = \(\frac{10.6 \, \text{g}}{180.2 \, \text{g/mol}} = 0.0588 \, \text{moles}\) - Moles of water = \(\frac{89.4 \, \text{g}}{18 \, \text{g/mol}} \approx 4.9667 \, \text{moles}\) 3. Calculate the mole fraction of water (solvent): - Mole fraction of water \( (\chi_{water}) = \frac{\text{moles of water}}{\text{total moles}} \) - Total moles = 0.0588 mol (glucose) + 4.9667 mol (water) = 5.0255 mol - Mole fraction of water \( (\chi_{water}) = \frac{4.9667}{5.0255} \approx 0.9883 \) 4. Use Raoult's Law to find the vapor pressure of the solution: - \(P_{solution
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