Standard enthalpy of vapourisation `DeltaH^(@)` forwater is `40.66 KJ mol^(-1)`.The internal energy of vapourisation of water for its 2 mol will be :-

Options

A. `+43.76`
B. `+40.66`
C. `+37.56`
D. None of these

Answer

D

Explanation

`H_(2)O(l)toH_(2)O(g)`
`DeltaH^(@)=DeltaE^(@)+Deltan_(g)RT`
`40.66=66DeltaE^(@)+(1xx8.314xx373)/(1000)`
`DeltaE^(@)=+37.56KJ//mol`
for 2 mole `DeltaE^(@)=2xx37.26=75.12KJ`