For this problem, we can use the ** Clausius-Clapeyron Equation**:

$\overline{){\mathbf{ln}}\frac{{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{P}}_{\mathbf{1}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{-}\mathbf{\u2206}{\mathbf{H}}_{\mathbf{vap}}}{\mathbf{R}}\mathbf{[}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{]}}$

where:

**P _{1}** = vapor pressure at T

**P _{2}** = vapor pressure at T

**ΔH _{vap}** = heat of vaporization (in J/mol)

**R** = gas constant (8.314 J/mol•K)

**T _{1} and T_{2}** = temperature (in K).

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 405 torr ?

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 405 torr ?

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