Calculate the enthalpy change (ΔH) for the vaporization of 50.0 g of water (H2O) at its boiling point, given the heat of vaporization (ΔHvap) of water is 40.7 kJ/mol.
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To calculate the enthalpy change (ΔH) for the vaporization of 50.0 g of water, we need to use the molar mass of water and the given heat of vaporization.
The molar mass of water (H2O) is calculated as follows:
1 mol H2O = 2(1.01 g/mol H) + 16.00 g/mol O = 18.02 g/mol
Next, we need to calculate the number of moles of water (H2O) in 50.0 g:
moles of H2O = mass / molar mass = 50.0 g / 18.02 g/mol = 2.7752 mol
Finally, we can calculate the enthalpy change (ΔH) using the equation:
ΔH = ΔHvap x moles of H2O
substituting the values:
ΔH = 40.7 kJ/mol x 2.7752 mol
ΔH = 112.8 kJ
Therefore, the enthalpy change (ΔH) for the vaporization of 50.0 g of water at its boiling point is 112.8 kJ.
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