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Answer :
The boiling point of ethanol will be 78.85 °C.
This can be calculated using Clausius-Clapeyron equation, which is:
[tex]ΔHvap / Tb = ΔSvap / R[/tex] ,
where [tex]ΔHvap[/tex] is the heat of vaporization, [tex]ΔSvap[/tex] is the entropy of vaporization, [tex]Tb[/tex] is the boiling point in Kelvin, and [tex]R[/tex] is the gas constant.
Plugging in the values given, we can solve for [tex]Tb[/tex]:
[tex]Tb = ΔHvap / (ΔSvap / R) = (38.6 kJ/mol) / (110 J/mol K) / (8.314 J/mol K)[/tex]
[tex]Tb = 351 K[/tex]
Converting to Celsius:
Tb (°C) = Tb (K) - 273.15 = 351 K - 273.15 = 78.85 °C
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