Understanding Calculation of N (Number of Atoms)

[KiNGGeexD]

I was looking through my notes and came across this (above) I understand the calculation etc but I don't see where the number of atoms N was calculated?

I don't see where it has come fromAny help would be great thanks in advanced

 

[BvU]

Neat handwriting you are using to produce your notes...not nice you (they?) break the line between the minus sign and the ²³, though...

28 gram is one mole
density is 0.81 g/cm3
so 28 gram is 28/0.81 = 35 cm3, right ? i.e. 35 cm3/mole

one mole is N_avogadro molecules and 35/(6 e23) = 5.8 e-23 cm3/molecule.

Take the $\root 3 \of \$ and you get something like 3.9 e-8 cm or 0.39 nm

 

[KiNGGeexD]

It's the top calculation for latent heat I'm not sure about I get the surface tension one, but the one above makes no mention of N but must use it:)

 

[BvU]

Why don't you type the part of "your" notes that you have a question about ? I can hardly read what is to the left of ${1\over 2} nN\epsilon$. Anyway, since we are talking J/mol, N=N_A.

 

[KiNGGeexD]

Would N in the top equation not be just Avergadros number?

 

[BvU]

Yes. Since we are talking J/mol, N=N_A.

 

[KiNGGeexD]

Ahh ok this calculation must just be rounded up because using avergadros number just now I obtained

1.95E-21 J

Thanks