Question

The standard molar heat of vaporization
for water is 40.79 kJ/mol. How
much energy would be required to vaporize
8.51 ×1010 molecules of water?
Answer in units of kJ.

Thanks for the help!!!

Answer 1

How many mols in 8.51E10 molecules?
That's 8.51E10 moleceles x (1 mol molecules/6.02E23 molecules) = ?
Then q in kJ = mols H2O x Hvap(kJ/mol)

Answer 2

6.02*10^23 molecules/mol
so
40.79kJ /6.02*10^23 molecules

times I suspect you mean 8.51*10^10 molecules

40.79 (8.51/6.02)10^(10-23)

57.66 * 10^-13
or
5.766 * 10^-12 kJ