Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1 °C. At what temperature does benzene boil when the external pressure is 410 torr?

• A. 70.0 °C
• B. 85.0 °C
• C. 75.5 °C
• D. 78.0 °C

Answer: (B)

Verified step by step guidance

1. First, understand that the problem involves calculating the boiling point of benzene at a different pressure using the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature.
2. The Clausius-Clapeyron equation is given by: $\frac{dP}{dT}=\frac{\mathrm{\Delta H}v}{R{T}^2}$, where $\mathrm{\Delta H}v$ is the heat of vaporization, $R$ is the gas constant, and $T$ is the temperature in Kelvin.
3. Convert the normal boiling point of benzene from Celsius to Kelvin: $T$₁ = 80.1 + 273.15.
4. Use the Clausius-Clapeyron equation in its integrated form: , where $P_1$ is the normal pressure (760 torr) and $P_2$ is the given pressure (410 torr).
5. Solve for $T$₂ using the equation above, which will give you the boiling temperature of benzene at 410 torr. Convert the result back to Celsius by subtracting 273.15.