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How much heat is required to evaporate 224.6 g of water at 100.0°C? The molar heat...

How much heat is required to evaporate 224.6 g of water at 100.0°C? The molar heat of vaporization for water is 4.07 × 104 J/mol. × 10 kJ

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Answer #1


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass(H2O)= 224.6 g

use:
number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(2.246*10^2 g)/(18.02 g/mol)
= 12.47 mol

Given:
ΔH = 4.07*10^4 J/mol

use:
Q = ΔH * number of mol
= 40700 J/mol * 12.4667 mol
= 5.07*10^5 J
Answer: 5.07*10^5 J

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