The heat of vaporization of benzene (C₆H₆) is 30.7 kJ/mol at its boiling point of 80.1 °C. How much energy in the form of heat is required to vaporize 110 g benzene at its boiling point?

Question

Quizplus · Accepted Answer

To find the energy required, first convert the mass of benzene to moles using its molar mass (78.11 g/mol). Then multiply the moles by the heat of vaporization (30.7 kJ/mol). Calculation: $ \frac{110 \text{ g}}{78.11 \text{ g/mol}} \approx 1.409 \text{ mol} $. Energy = $ 1.409 \text{ mol} \times 30.7 \text{ kJ/mol} \approx 43.2 \text{ kJ} $.

The Heat of Vaporization of Benzene (C6H6) Is 30 3.38×1033.38 \times 10 ^ { 3 }3.38×103

The Heat of Vaporization of Benzene (C₆H₆) Is 30 $3.38 \times 10 ^ { 3 }$