Answer :
Final answer:
Approximately 41.88 kJ of heat is required to vaporize 50.0 g of ethanol based on the given molar heat of vaporization and the molar mass of ethanol.
Explanation:
To calculate how many kilojoules of heat are required to vaporize 50.0 g of ethanol (C₂H₆O), we can use the molar heat of vaporization and the molar mass of ethanol. The molar heat of vaporization for ethanol is given as 38.6 kJ/mol, and the molar mass of ethanol is approximately 46.07 g/mol.
First, we need to determine the number of moles in 50.0 g of ethanol:
Number of moles = mass (g) / molar mass (g/mol) = 50.0 g / 46.07 g/mol
≈ 1.085 mol
Next, we use the molar heat of vaporization to find the total heat required:
Total heat (Q) = number of moles × molar heat of vaporization (kJ/mol) = 1.085 mol × 38.6 kJ/mol
≈ 41.88 kJ
Therefore, approximately 41.88 kJ of heat is required to vaporize 50.0 g of ethanol.
