The question is incomplete, here is a complete question.
Calculate the standard entropy of vaporization of ethanol at its boiling point 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol.
Answer : The standard entropy of vaporization of ethanol is, 115 J/mol.K
Explanation :
Formula used :
[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
[tex]\Delta H_{vap}[/tex] = change in enthalpy of vaporization = 40.5 kJ/mol
[tex]T_b[/tex] = boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]
[tex]\Delta S=\frac{40.5kJ/mol}{352K}[/tex]
[tex]\Delta S=\frac{40.5\times 10^3J/mol}{352K}[/tex]
[tex]\Delta S=115J/mol.K[/tex]
Therefore, the standard entropy of vaporization of ethanol is, 115 J/mol.K