Clausius-Mossotti Equation and Dielectric Constants of Air

[Blastrix91]

Homework Statement

http://img809.imageshack.us/img809/5676/unavngivetk.png

Additional table info:

The molarmass $M_{air}$ is $29 * 10^{-3} g/mol = 29*\frac{10^{-3}}{1000} kg/mol$
The mass density $\rho_{air}$ is $1.3 kg/m^{3}$
The dielectric constant $K_{air}$ is 1.00059

My teacher told me to consider the air as 1 molecule.

Homework Equations

http://img94.imageshack.us/img94/1667/unavngivet2w.png

The Attempt at a Solution

Here is my approach:

Basically it is only N we need which is molecules per volume:

N = \frac{Molecules}{V} = \frac{N_a n}{V} = \frac{m}{M_{air}} \frac{N_a}{V} = \frac{\rho_{air}V}{M_{air}} \frac{N_a}{V} = \frac{\rho_{air}N_a}{M_{air}}

\alpha = \frac{3 \varepsilon_0 M_{air}}{\rho_{air}N_a} \frac{K_{air}-1}{K_{air}+1}

Inserting the values:\frac{3(8.854*10^{-12})(29*\frac{10^{-3}}{1000})}{1.3(6.022*10^{23})} * \frac{1.00059 - 1}{1.00059 + 1} = 2.902 * 10^{-43}

The answer in the back of my book says it is supposed to be: $\alpha = 9.7 * 10^{-41}$

Can any of you guys spot where I did wrong? (The table values were some I got from my teacher and if my approach is right, then he might have wrote them wrong)

 

[TSny]

The molarmass $M_{air}$ is $29 * 10^{-3} g/mol = 29*\frac{10^{-3}}{1000} kg/mol$

A quick Google check on the molar mass of air will show an error in this value.

You used
\alpha = \frac{3 \varepsilon_0 M_{air}}{\rho_{air}N_a} \frac{K_{air}-1}{K_{air}+1}

but the denominator of $K_{air}+1$ does not agree with the form of the Claussius-Mossotti equation that you quoted.Also, does your answer for $\alpha$ correspond to a molecule or an atom?

 

[Blastrix91]

All three things you pointed out gave me the right answer. Thank you ^^