1. Mar 14, 2009

### fluidistic

Hi PF,
I'm getting an headache on this problem.
1. The problem statement, all variables and given/known data
Rayleigh put a milligram of oil which density is $$0.9g/cm^3$$ on a water surface and found out that the oil covered an area of $$0.9 m^2$$.
1)What is the height of the oil film?
2)The height of the oil film is of the same order as a molecule of oil, which lead us to suppose that the oil film is one molecule thin.
Suppose that the three dimensions of the oil molecule are the same, find the molecular volume of oil.
3)If the molar mass of oil is $$282.5g/mol$$, estimate Avogadro's number through the comparison of molecular volume and molar volume.

2. Relevant equations Avogadro's number is about $$6.02 \times 10^{23}$$.

3. The attempt at a solution
1)Density $$\rho$$ is worth $$\frac{m}{V}=0.9$$ so $$V=\frac{0.001}{0.9}=0.00111111...cm^3$$ where $$V$$ is the volume of the oil drop.
Now $$V=base \cdot h$$ hence $$h=0.00001234567889cm$$.
2)Calculating the volume of a molecule lead me nowhere. Instead in order to find Avogadro's number I think it's more effective to follow this : as the dimensions of the molecule are the same and that the diameter or height or length of it is worth $$0.00001234567889cm$$, each molecule cover a surface of $$(0.00001234567889cm)^2$$. Now I can find how many molecules are in the drop. Because it has a surface of $$0.9m^2$$ or $$90cm^2$$. $$\frac{90}{(0.00001234567889cm)^2}=5.904900127\times 10^{11}$$.
Now for part 3), we know that the oil drop was $$0.001g$$ and that a mole of oil is $$282.5g$$. So that I must multiply by $$282500$$ the number of molecules in the drop to find the number of molecule in a mole, that is Avogadro's number. But doing so I find $$1.661320857 \times 10^{17}$$...

P.S.: I also tried via the way they suggested, but found something like $$3.5\times 10 ^{17}$$ and I had to suppose that the form of the molecules was spherical.
Of course I made errors and I'm missing something. If you could help me that would be kind! Thank you.

2. Mar 14, 2009

### Redbelly98

Staff Emeritus
Hi,

You're not converting 0.9 m2 into cm2 correctly.

3. Mar 14, 2009

### fluidistic

Thank you SO much. I'll do the algebra tomorrow when I wake up. (too late for now).

4. Mar 15, 2009

### fluidistic

$$0.9m^2=8100cm^2$$.
My final result becomes $$1.50\times 10^{19}$$. I'm getting closer but I'm still very far from what I should get.

5. Mar 15, 2009

### Redbelly98

Staff Emeritus
I'm getting much closer to Avogadro's number.

What do you get now for h and also for the number of molecules in the oil drop?

6. Mar 15, 2009

### fluidistic

I forgot to change the volume! Now I got the right result : $$1.216069868 \times 10^{23}$$. If you still mind I got a number of $$4.3046721 \times 10^{17}$$ for the number of molecules in the drop and h is worth $$\frac{0.00111...}{8100}cm$$.
Thanks for your help and time.

7. Mar 15, 2009

### Redbelly98

Staff Emeritus