18.0 g of water completely vaporises at `100^(@)C` and 1 bar pressure and the enthalpy change in the process is `40.79 kJ mol^(-1)`. What will be the enthalpy change for vaporising two moles of water under the same conditions ? What is the standard enthalpy of vaporisation for water ?

Explanation

Moles of water = `(18.0)/(18) = 1.0 mol`
Enthalpy change for vapourisation of 1 mol of water = `40.79 kJ " mol^(-1)`
(i) Enthalpy change for vapourisation of 2 mol of water = `40.79 xx 2 = 81.58 kJ " mol^(-1)`
(ii) Standard enthalpy of vapourisation = Enthalpy change for vapourisation of 1mol of water = `40.79 kJ mol^(-1)`.