Given :
A 75.0 mL sample of water is heated to its boiling point.
To Find :
The heat (in kJ ) is required to vaporize it .
Solution :
Heat of vaporization of water , [tex]\Delta H_{vap}=40.65\ kJ/mol[/tex].
Molar mass of [tex]H_2O[/tex] , [tex]M=18\ g/mol[/tex] .
Density of water , [tex]\rho=1\ g/ml[/tex] .
Mass of 75 ml of water sample , [tex]m=75\ g[/tex] .
Moles of 75 g water = [tex]\dfrac{75}{18}=4.17\ mol[/tex] .
Now , heat required to vaporize 4.17 mol of water :
[tex]H=4.17\times 40.65\ kJ\\\\H=169.51\ kJ[/tex]
Therefore , heat required to boil 75 ml sample of water is 169.51 kJ.
Hence , this is the required solution .
