What is the osmotic pressure in #"mm Hg"# of a solution of #"45.0 g"# ribose dissolved into #"800.0 g"# of water at #40^@"C"#? Assume the density is #"1.00 g/mL"#. #MW = "150.13 g/mol"#

2 Answers
Aug 8, 2017

Answer:

We assume that the ribose is involatile..........

Explanation:

And we ALSO MUST KNOW that the vapour pressure of water at #40# #""^@C# is #55.4*mm*Hg# (these data really should have been quoted with the question........)

Now the vapour pressure exerted by a solution is proportional to the mole fraction of the volatile component........

#"Moles of ribose"=(45.0*g)/(150.13*g*mol^-1)=0.300*mol#

#"Moles of water"=(800.0*g)/(18.01*g*mol^-1)=44.42*mol#

Now #chi_"component"="Moles of component"/"Total moles in solution"#

And thus #chi_"ribose"=(0.300*mol)/(0.300*mol+44.42*mol)=6.71xx10^-3#

#chi_"water"=(44.42*mol)/(0.300*mol+44.42*mol)=0.993#

And so the vapour pressure of the solution is #0.993xx55.4*mm*Hg=55.0*mm*Hg#. The diminution in vapour pressure is VERY SLIGHT. I would challenge any scientist to detect such a pressure difference. Use a more volatile solvent......

Aug 8, 2017

#Pi = "9.63 atm" = ??? "mm Hg"#


The osmotic pressure is given by:

#Pi = icRT#,

where:

  • #i# is the van't Hoff factor, presumably one for ribose, a nonelectrolyte...
  • #c# is the molarity in #"mol/L"#.
  • #R# and #T# are known from the ideal gas law.

and it is the pressure required to stop the flow of solvent through a semi-permeable membrane from low to high concentration.

Thus, the osmotic pressure is (assuming the solution volume doesn't change, which is completely unreasonable!!):

#color(blue)(Pi) = (1) cdot (45.0 cancel"g" xx cancel"1 mol ribose"/(150.13 cancel"g"))/(0.800 cancel"L") cdot 0.082057 cancel"L"cdot"atm/"cancel"mol"cdotcancel"K" cdot (40 + 273.15 cancel"K")#

#=# #color(blue)("9.63 atm")#

I now dare you to multiply this by the appropriate conversion factor to get your answer in #"mm Hg"#. That is, I dare you to read the back cover of your textbook, or google "#"mm Hg in an atm"#".