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

Given that, quantity of water `= 18.0 g`, pressure = 1 bar
As we know that, `18.0 g H_(2)O = 1 " mole " H_(2)O`
Enthalpy change for vaporising 1 mole of `H_(2)O = 40.79 kJ mil^(-1)`
`:.` Enthalpy change for vaporising 2 moles of `H_(2)O = 2 xx 40.79 kJ = 81.358 kJ`
Standard enthalpy of vaporisation at `100^(@)C` and 1 bar pressure, `Delta_("vap") H^(@) = + 40.79 kJ mol^(-1)`