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Sagot :
We are given Mass of water = 42.3 g
ΔHvap(H2O)=40.7kJ/mol) indicates that 1 mol of H2O(l) requires 40.7kJ/mol of heat in order to heat up.
Similary, we can find mol of H2O since we are given mass of 42.3g :
n = mass H2O (given) /Molecular mass H2O
= 42.3g /18g/mol
= 2.35 moles
If 1 mol H2O requires 40.7kJ/mol of heat ;
then 2.35 mol H2O will require x kJ/mol to vaporise
Therefore x = (40.7kJ/mol * 2.35 mol)/1
= 95.64kJ
• This means that the heat required to vaporize 42.3 g of water at 100 ∘C = 95.63kJ.,
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