Open Question
What is the boiling point of an aqueous solution with a vapor pressure of 20.5 torr at 25 °C? (Assume a nonvolatile solute.)
14. Solutions
Vapor Pressure Lowering (Raoult's Law)
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A KCl solution is prepared by dissolving 40.0 g KCl in 250.0 g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of pure water at 25°C is 23.76 mm Hg?
A
24.50 mm Hg
B
21.50 mm Hg
C
22.98 mm Hg
D
23.76 mm Hg
Verified step by step guidance1
First, calculate the number of moles of KCl. Use the formula: \( \text{moles of KCl} = \frac{\text{mass of KCl}}{\text{molar mass of KCl}} \). The molar mass of KCl is approximately 74.55 g/mol.
Next, calculate the number of moles of water. Use the formula: \( \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} \). The molar mass of water is approximately 18.02 g/mol.
Determine the mole fraction of water in the solution. The mole fraction of water is given by: \( X_{\text{water}} = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of KCl}} \).
Apply Raoult's Law to find the vapor pressure of the solution. Raoult's Law states: \( P_{\text{solution}} = X_{\text{water}} \times P^0_{\text{water}} \), where \( P^0_{\text{water}} \) is the vapor pressure of pure water at 25°C, which is 23.76 mm Hg.
Finally, calculate the vapor pressure of the solution using the mole fraction of water and the vapor pressure of pure water. This will give you the vapor pressure of the KCl solution.
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