Question

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"`

Answer 1

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......

Answer 2

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

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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"`".