1.

18 g of water is taken to prepare, the tea. Find out the internal energy of vaporisation at 100^(@)C. (Delta_(vap)H^(ө) for water at 373K=40.66kJmol^(-1))

Answer»

37.56 kJ `mol^(-1)`
`-37.56" kJ "mol^(-1)`
`43.76" kJ "mol^(-1)`
`-43.76" kJ "mol^(-1)`

Solution :`underset((18g))(H_(2)O(l)) to underset((18g))(H_(2)O(G))`
Number of moles in 18 g of `H_(2)O(l)=(18g)/(18g" "mol^(-1))=1` mol
`Delta_(vap)U=Delta_(vap)H^(ө)-Deltan_(g)RT`
`=40.66-(1)(8.314)(373)J" "mol^(-1)`
`=40.66-3.10=37.56` kJ `mol^(-1)`


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