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Answer :
Final answer:
To vaporize 335 g of ethanol at its boiling point, one must first convert the mass to moles using ethanol's molar mass and then multiply the moles by the enthalpy of vaporization to obtain a total of 280.83 kJ of heat.
Explanation:
The question asks about the amount of heat required to vaporize a given mass of liquid ethanol at its boiling point, given the enthalpy of vaporization (δhvap). Initially, it is important to convert the mass of ethanol to moles by using ethanol's molar mass (46.07 g/mol). After determining the number of moles, we multiply it by the given δhvap value to find the total heat required.
First, calculate the number of moles:
335 g ethanol × (1 mol/46.07 g) = 7.27 mol ethanol
Then, calculate the total heat (q) required:
q = moles × δhvap
q = 7.27 mol × 38.6 kJ/mol
q = 280.83 kJ
Therefore, 280.83 kJ of heat is necessary to vaporize 335 g of ethanol at its boiling point.
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Rewritten by : Barada
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.
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.
