Using osmotic pressure to find the molar mass of a compound Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms $(\pu{12.5E-6 g})$ of what you hope is pure saxitoxin (a non-electrolyte) from a poisonous (and quite cross) puffer fish. You dissolve this sample in $\pu{3.10 mL}$ of water and determine that the osmotic pressure of
  the resulting solution is $\pu{0.236 Torr}$ at $\pu{19 °C}$ $(\pu{760 Torr} = \pu{1.00 atm}).$ What is the molar mass of the compound?


I am having trouble reaching the correct answer of $\pu{310.8 g/mol}$ for this question. I first found the pressure to be $\pu{0.0003105 atm}$ and plugged numbers in for the formula $\Pi = MRT.$ I got $\pu{2.04E-5 mol L-1}.$ I used this to find the amount of substance and got $\pu{6.3E-8 mol}.$

Lastly, I divided $\pu{12.5E-6 g}$ by $\pu{6.3E-8 mol}$ and got $\pu{1.98E-14 g/mol}$ which is not the right answer.

I feel like I have the right process so if you can help show me where I messed up, I would appreciate it.
I couldn't resist. The osmotic pressure equation is

$$Π = cRT,$$

where $Π$ is osmotic pressure in atmospheres, $c$ is molar concentration, $R = \pu{0.082057366080960 L atm mol-1 K-1}$ (exactly) and $T$ is temperature in Kelvin. 

In this problem, $Π = \pu{0.236 torr}$ times exactly $\pu{1 atm}/\pu{760 torr}$ and $T = \pu{(273.15 + 19.0) K}.$ So $c = \pu{1.2953E-5 M}$. But $\pu{12.5E-6 g}$ of saxitoxin was dissolved in $\pu{3.10 mL}$ of water, which is $\pu{4.032E-3 g/L}$. Hence $\pu{1 mol}$ of saxitoxin has a mass of $\pu{311 g}.$