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Answer :
Final answer:
The estimated vapor pressure of ethyl acetate at 50.0∘C is approximately 22.6 Torr.
Explanation:
To estimate the vapor pressure of ethyl acetate at 50.0∘C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Given:
- P1 = 100 Torr (vapor pressure at 27.0∘C)
- T1 = 27.0∘C + 273.15 = 300.15 K
- T2 = 50.0∘C + 273.15 = 323.15 K
- ΔHvap = 37.6 kJ/mol
- R = 8.314 J/(mol·K)
Substituting the values into the equation:
ln(P2/100) = -(37.6 * 10^3)/(8.314) * (1/323.15 - 1/300.15)
Simplifying the equation:
ln(P2/100) = -4.527
Using the natural logarithm property, we can rewrite the equation as:
P2/100 = e^(-4.527)
Solving for P2:
P2 = 100 * e^(-4.527)
Calculating the value:
P2 ≈ 22.6 Torr
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