Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

14. Solutions

Henry's Law Calculations

General Chemistry

14. Solutions

Henry's Law Calculations

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Multiple Choice

What would be the vapor pressure at 25.0 °C of a solution of 5.00 g of glucose (C₆H₁₂O₆) in 100.0 g of ethanol (C₂H₅OH)? Pure ethanol has a vapor pressure of 54.68 mmHg at 25 °C.

0.776 mmHg

52.1 mmHg

54.0 mmHg

54.7 mmHg

4280 mmHg

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Verified step by step guidance

First, calculate the number of moles of glucose (C6H12O6) using its molar mass. The molar mass of glucose is approximately 180.18 g/mol. Use the formula: \( \text{moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} \).

Next, calculate the number of moles of ethanol (C2H5OH) using its molar mass. The molar mass of ethanol is approximately 46.07 g/mol. Use the formula: \( \text{moles of ethanol} = \frac{\text{mass of ethanol}}{\text{molar mass of ethanol}} \).

Determine the mole fraction of ethanol in the solution. The mole fraction of ethanol (\( X_{\text{ethanol}} \)) is given by: \( X_{\text{ethanol}} = \frac{\text{moles of ethanol}}{\text{moles of ethanol} + \text{moles of glucose}} \).

Apply Raoult's Law to find the vapor pressure of the solution. Raoult's Law states that the vapor pressure of the solution (\( P_{\text{solution}} \)) is equal to the mole fraction of the solvent (ethanol) times the vapor pressure of the pure solvent: \( P_{\text{solution}} = X_{\text{ethanol}} \times P^0_{\text{ethanol}} \), where \( P^0_{\text{ethanol}} \) is the vapor pressure of pure ethanol (54.68 mmHg).

Finally, calculate the vapor pressure of the solution using the values obtained from the previous steps. This will give you the vapor pressure of the ethanol-glucose solution at 25.0 °C.