Multiple Choice
Using Raoult's Law, calculate the vapor pressure of water at 50.0°C if the pure vapor pressure of water at this temperature is 12.3 kPa and a non-volatile solute is added to form a solution with a mole fraction of water of 0.8.
14. Solutions
Vapor Pressure Lowering (Raoult's Law)
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Using Raoult's Law, calculate the vapor pressure of SiCl4 at 33.5 °C given that its vapor pressure is 100 mm Hg at 5.4 °C and ΔHvap = 30.2 kJ/mol.
A
200 mm Hg
B
250 mm Hg
C
150 mm Hg
D
300 mm Hg
Verified step by step guidance1
Identify the given values: the vapor pressure of SiCl4 at 5.4 °C is 100 mm Hg, the enthalpy of vaporization (ΔHvap) is 30.2 kJ/mol, and the temperature change from 5.4 °C to 33.5 °C.
Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. This gives T1 = 278.55 K and T2 = 306.65 K.
Use the Clausius-Clapeyron equation to relate the vapor pressures at two different temperatures: \( \ln \left( \frac{P2}{P1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right) \), where R is the ideal gas constant (8.314 J/mol·K).
Rearrange the equation to solve for P2, the vapor pressure at the new temperature: \( P2 = P1 \times \exp \left( -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right) \right) \).
Substitute the known values into the equation: P1 = 100 mm Hg, ΔHvap = 30.2 kJ/mol (convert to J/mol by multiplying by 1000), T1 = 278.55 K, and T2 = 306.65 K. Calculate P2 to find the vapor pressure at 33.5 °C.
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