18.0 g of water completely vapourises at 100^(@)C and 1 bar pressure and the enthalpy change for the process is 40.79 kJ mol^(-1). (i) What will be the enthalpy change for vapourising 2 moles of water under the same conditions? (ii) What is the standard enthalpy of vapourisation for water?

18.0 g of water completely vapourises at 100^(@)C and 1 bar pressure a

1.	18.0 g of water completely vapourises at 100^(@)C and 1 bar pressure and the enthalpy change for the process is 40.79 kJ mol^(-1). (i) What will be the enthalpy change for vapourising 2 moles of water under the same conditions? (ii) What is the standard enthalpy of vapourisation for water?
Answer» SOLUTION :18.0 of water = 1 MOL Standard enthalpy of VAPORISATION at `100^(@)C` and 1 bar PRESSURE = `40.79 kJ"mol"^(-1)`. Enthalpy change in vaporisation of 2 mol of water `= 2 XX 40.79 = 81.58` kJ.