Multiple Choice
Henry's Law Constant for nitrogen in water is 1.67 × 10-4 M • atm–1. If a closed canister contains 0.103 M nitrogen, what would be its pressure in atm?
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14. Solutions
Henry's Law Calculations
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.
A
0.776 mmHg
B
52.1 mmHg
C
54.0 mmHg
D
54.7 mmHg
E
4280 mmHg
Verified step by step guidance1
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.
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