Calculating collision diameter with viscosity, but not having the mass given I've got a homework question i'm stumped on.


  The viscosity of the ethylene gas was measured at $99.8\times10^{-7}\ \mathrm{N\ s\ m^{-2}}$ at $288\ \mathrm K$. Calculate the collision diameter in $\overset{\circ}{\mathrm{A}}$ for this gas molecule.


I started with

$$\eta=\frac{m\overline v}{3\sqrt2\pi d^2} $$

then i said

$$\overline v=\sqrt{\frac{8RT}{\pi M}}$$

Molar mass $M=28.05\ \mathrm{g\ mol^{-1}}$

So my equation looks like 

$$\eta=\frac{m}{3\sqrt2\pi d^2}\sqrt{\frac{8RT}{\pi M}} $$

but $m$ is still a variable. What should I do from here? Is there another substitution I can do?



So apparently my confusion about solving this problem, came from a lack of understanding what the $m$ variable was supposed to represent. I had thought that it was the total mass of the system, but it is actually the mass of one molecule.
Answering my own question here (just received the answer to the problem.)

the value for $(\mathrm{m})$ was supposed to be in $\mathrm{kg}$, and could be found by taking the Molar mass and converting:

$\mathrm{M} \cdot 1.661 x10^{-27}\mathrm{kg\cdot amu^{-1}}$

The solution came out looking like

$\overset{-}{v} = \sqrt{\frac{8x8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} x 288\mathrm{K}}{3.14x28.05x10^{-3}kg \cdot mol^{-1}}}$

$d=\sqrt{\frac{m \overset{-}{v}}{3 \sqrt{2} \pi \eta}}$

$d=\sqrt{\frac{(28.05 \mathrm{amu} \space x \space 1.661x10^{-27} \mathrm{kg} \cdot \mathrm{amu}^{-1}) \space x \space 466.23 \mathrm{m} \cdot \mathrm{s}^{-1}}{3 \sqrt{2} \pi \space x \space 99.8x10^{-7} \mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}}}$

$=4.04x10^{-10} \mathrm{m} = 4.04 \overset {\circ}{A}$