A small protein molecule, code-named "sloth", has a MM of $\mathrm{1.50 x 10^4}$ g/mol. What is the osmotic pressure exerted at $\mathrm{24.0^oC}$ by 25.0 mL of an aqueous solution that contains $\mathrm{3.75 x 10^{10}}$ nanograms of "sloth"? R = 0.08206 (atm L)/(mol K).

The equation my professor gave me is: $\mathrm{O.P. = T*R*M}$. So I did:

$\mathrm{O.P. = (24.0 + 273) K * 0.08206 (atm*L)/(mol*K) * ([(3.75 x 10^{10})x 10^{-9} g] / 1.50 x 10^4 g/mol) / 0.025 L}$

I keep on trying and I get 2.44 atm, but my professor marked the correct answer as $\mathrm{2.44 x 10^{-3}}$ atm. I think he may have divided the moles of sloth by 25 instead of .025 L, which would make his answer $10^3$ smaller than mine. Did I do something wrong?

  • $\begingroup$ Hello and welcome to Chemistry.SE! If you have any questions about how this site works, a good starting point is taking the short tour. If you have any questions about homework-type of questions in particular, you should read through this discussion. Good luck! $\endgroup$
    – airhuff
    Commented Feb 21, 2017 at 1:29

1 Answer 1


Ok let's start with $\text{OP} = \text{T}\times\text{R}\times\text{M}$

$\text{T} = (24.0 + 273) \text{K} = 297 \text{K}$

$\text{R} = 0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}$

$\text{M} = \dfrac{\text{moles}}{\text{L}} = \dfrac{\frac{37.5 \text{ g}}{1.50 \times 10^4 \text{ g/mol}}}{0.025 \text {L}} = \dfrac{2.50\times10^{-3}\text{ mol}}{0.025\text{ L}} = 0.100 \dfrac{\text{mol}}{\text{L}} $

$\text{OP} = 297\text{ K}\times0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}\times0.100 \dfrac{\text{mol}}{\text{L}} = 2.44 \text{ atm} $

So I get your answer too...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.