Answer :
Final answer:
The enthalpy of vaporization of benzene at its boiling point is found from the heat supplied and the mass of benzene vaporized, which leads to a value of 30.8 kJ/mol.
Explanation:
The enthalpy of vaporization of benzene can be found by using the heat supplied and the mass of benzene vaporized. The heat supplied is given as 15.4 kJ and the mass vaporized is given as 39.1g. To find the enthalpy of vaporization in kJ/mol, first convert the mass of benzene vaporized to moles (39.1g / 78.1g/mol = approximately 0.5 moles). The enthalpy of vaporization can be calculated by dividing the heat supplied by the number of moles vaporized.
Enthalpy of Vaporization
= 15.4 kJ / 0.5 moles = 30.8 kJ/mol. Therefore, the enthalpy of vaporization of benzene at its boiling point is 30.8 kJ/mol.
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