Insulin (#("C"_2"H"_10"O"_5)_n#) is dissolved in a suitable solvent at #20^@ "C"#. The slope of a plot of osmotic pressure (#Pi#) against #c# is found to be #4.65xx10^(-3)#. The molecular weight of insulin is?

Well, I got #5.17 xx 10^6# #"g/mol"#, but you have got some explaining to do...

Well, osmotic pressure is given by:

#Pi = icRT#

where:

• The van't Hoff factor #i# indicates the effective number of particles per solute particle. It is not necessarily #1#, but would be #1# for perfect nonelectrolytes.
• #c# is the concentration in the appropriate units.
• #R = "0.082057 L"cdot"atm/mol"cdot"K"# is the universal gas constant.
• #T# is the temperature in #"K"#.

So if you plot #Pi# vs. #c#, the slope SHOULD simply be #iRT#... if the graph was made using #"mol/L"# concentrations... But what you neglected to tell us is that the graph was made using concentrations of #"g/cm"^3#...

Thus,

#"slope" = 4.65 xx 10^(-3) "cm"^3cdot"atm/g"#

#= iRT# #(cancel"L"cdot"atm")/cancel"mol" cdot (1 cancel("dm"^3))/cancel"L" xx ("10 cm"/(1 cancel"dm"))^3 cdot 1/M cancel"mol"/cancel"g"#

#= (1000iRT)/M#

where #M# is the molar mass in #"g/mol"# of the insulin.

We assume that #i ~~ 1# for insulin, because we don't know any better value. We have no idea what the solvent is.

As a result,

#color(blue)(M) = (1000iRT)/"slope"#

#~~ (1000 cdot 1 cdot 0.082057 cancel("cm"^3)cdotcancel"atm""/""mol"cdotcancel"K" cdot 293.15 cancel"K")/(4.65 xx 10^(-3) cancel("cm"^3)cdotcancel"atm""/""g"#

#~~ color(blue)(5.17 xx 10^6 "g/mol")#

And after searching around and seeing sloppy, incorrect, and not proofread questions, I found only one that was remotely similar.

The other linked questions either #(i)# had the incorrect answer choices or #(ii)# listed the wrong units for the concentration.