18.0 g water completely vaporises at 100 °C and 1 bar pressure

18.0 g water completely vaporises at 100 °C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ/ mol. 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 ?

18.0 g $H _{ 2 }O$ = 1 mol $H _{ 2 }O$
Enthalpy change for vaporising 1 mole of $H _{ 2 }O$ = 40.79 kJ
Enthalpy change for vaporising 2 moles of . $H { 2 }O$ = 2 x 40.79 kJ = 81.58 kJ
Standard enthalpy of vaporisation at 100°C and 1 bar pressure,
${ ∆ }
{ vap }$H° = + 40.79 kJ/ mol