Answer:
209.8 kilo Joules heat is required to vaporize 250 g of ethanol at 78.5 °C.
Explanation:
Mass of an ethanol = 250 g
Molar mass of ethanol = 46 g/mol
Moles of an ethanol = [tex]n=\frac{250 g}{46 g/mol}=5.435 mol[/tex]
Enthalpy of vaporization of ethanol = [tex]\Delta H_{vap} =38.6 kJ/mol[/tex]
Heat required to vaporize 250 g of ethanol at 78.5 °C : Q
[tex]Q=n\times \Delta H_{vap} [/tex]
[tex]=5.435 mol\times 38.6 kJ/mol=209.7826 kJ\approx 209.8kJ[/tex]
Q = 209.8 kilo Joules
