Final answer:
The molar entropy of vaporization of acetone at its boiling point can be calculated using the enthalpy of vaporization divided by the boiling temperature in Kelvin.
Step-by-step explanation:
Calculating the Molar Entropy of Vaporization
The question asks for the molar entropy of vaporization of acetone under 1 atm of pressure. To calculate this, we use the thermodynamic relation:
\(\Delta S_{vap} = \frac{\Delta H_{vap}}{T_{boil}}\)
Where \(\Delta S_{vap}\) is the molar entropy of vaporization, \(\Delta H_{vap}\) is the molar enthalpy of vaporization, and \(T_{boil}\) is the boiling temperature in Kelvin. For acetone, we have \(\Delta H_{vap} = 29.1\ kJ/mol\) and the boiling temperature \(T_{boil} = 56 + 273.15 = 329.15\ K\). Thus, the calculation will be:
\(\Delta S_{vap} = \frac{29.1\times 10^3}{329.15}\)
This will give us the molar entropy of vaporization of acetone in \(J/K\cdot mol\).