How much heat is required to vaporize 335 g of liquid ethanol at its boiling point? δhvap = 38.6 kj/mol?

Answer is: 281.1 kJ is required to vaporize of liquid ethanol.

1) calculate amount of substance of ethanol:
m(ethanol - C₂H₅OH) = 335 g.
n(C₂H₅OH) = m(C₂H₅OH) ÷ M(C₂H₅OH).
M(C₂H₅OH) = 2Ar(C) + 6Ar(H) + Ar(O) · g/mol.
M(C₂H₅OH) = 2·12 + 6·1 + 16 · g/mol.
M(C₂H₅OH) = 46 g/mol; molar mass.
n(C₂H₅OH) = 335 g ÷ 46 g/mol.
n(C₂H₅OH) = 7.28 mol.
2) calculate heat:
Q = ΔHvap · n(C₂H₅OH).
Q = 38.6 kJ/mol · 7.28 mol.
Q = 281.1 kJ.

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