ΔfH  of  H2Ol  is –286 kJ / mole and ΔfH  of  H2Og  is –246 kJ / mole, then the enthalpy change when 9 grams of water vapour condenses liquid water is

H2(g) +  12O2(g) → H2Ol ;ΔH=−286  kJ        ……. (1)H2(g) +  12O2(g)→ H2Og ;ΔH=−246  kJ        ……..(2)Eq(1) – Eq(2)H2Og → H2Ol ; ΔH=−40  kJ/moleFor 18g , enthalpy change is  -40kJFor 9 grams, enthalpy change will be – 20 kJ.