At 25°C, the vapor pressure of chloroform is 0.259 atm and the vapor pressure of methanol is 0.185 atm. Assuming ideal behavior, what is the vapor pressure of chloroform above the solution containing 1.00 mol chloroform and 3.00 mol methanol?
The correct answer and explanation is :
To find the vapor pressure of chloroform above a solution containing 1.00 mol chloroform and 3.00 mol methanol, we can use Raoult’s Law. Raoult’s Law states that the partial vapor pressure of each volatile component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. The formula for Raoult’s Law is:
[
P_{\text{A}} = X_{\text{A}} \times P_{\text{A}}^{\text{pure}}
]
Where:
- ( P_{\text{A}} ) is the partial vapor pressure of component A in the solution,
- ( X_{\text{A}} ) is the mole fraction of component A in the solution,
- ( P_{\text{A}}^{\text{pure}} ) is the vapor pressure of the pure component A.
Step-by-Step Solution:
- Calculate the mole fraction of chloroform (CHCl₃): The mole fraction ( X_{\text{CHCl₃}} ) of chloroform in the solution is calculated as: [
X_{\text{CHCl₃}} = \frac{\text{moles of CHCl₃}}{\text{total moles of solution}} = \frac{1.00}{1.00 + 3.00} = \frac{1.00}{4.00} = 0.25
] - Use Raoult’s Law to find the partial vapor pressure of chloroform: The vapor pressure of pure chloroform at 25°C is given as 0.259 atm. Therefore, using Raoult’s Law: [
P_{\text{CHCl₃}} = X_{\text{CHCl₃}} \times P_{\text{CHCl₃}}^{\text{pure}} = 0.25 \times 0.259 \, \text{atm} = 0.06475 \, \text{atm}
] - Conclusion: The vapor pressure of chloroform above the solution containing 1.00 mol chloroform and 3.00 mol methanol is approximately 0.065 atm.
Explanation:
Raoult’s Law is based on the assumption that the solution behaves ideally, meaning that the intermolecular forces between the components (chloroform and methanol) are similar, and they mix uniformly. The mole fraction represents the proportion of each component in the solution, and the vapor pressure depends on this proportion. Since the mole fraction of chloroform is 0.25, this means that 25% of the vapor above the solution will be chloroform molecules, leading to a partial vapor pressure of 0.065 atm.