${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{H}\mathrm{of}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$  is –286 kJ / mole and ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{H}\mathrm{of}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$  is –246 kJ / mole, then the enthalpy change when 

9 grams of water vapour condenses to liquid water is

${\mathrm{H}}_2\left(\mathrm{g}\right)$ +  $\frac{1}{2}$${\mathrm{O}}_2\left(\mathrm{g}\right)$ $\stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);\mathrm{\Delta H}=-286\mathrm{kJ}$        ……. (1)

${\mathrm{H}}_2\left(\mathrm{g}\right)$ +  $\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)$$\stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right);\mathrm{\Delta H}=-246\mathrm{kJ}$        ……..(2)

Eq(1) – Eq(2)

$\underset{ }{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\stackrel{ }{\to }\underset{ }{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);}\mathrm{\Delta H}=-40\mathrm{kJ}/\mathrm{mole}$

For 18g , enthalpy change is  -40kJ

For 9 grams, enthalpy change will be – 20 kJ.