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.

• Note that calling the physical quantities by the units is incorrect, e.g. saying "I found atm" or "I found the number of moles" is illiterate and should be avoided. Instead, use the exact terms such as "pressure" and "amount of substance" (see my edit). – andselisk Feb 5 '20 at 5:30
• Also, it is generally advised not to perform intermediate calculations (solve algebraically first, plug in the numbers at the end) unless you are explicitly asked to do so, and if you decided to go against the recommended approach, make sure you aren't using textual description for the calculations; post proper math formulas and notations instead. – andselisk Feb 5 '20 at 5:30

$$Π = cRT,$$
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 $$\pu{311 g}.$$