# 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

#### Answer:

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

#### Explanation:

And we ALSO MUST KNOW that the vapour pressure of water at

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

Now

And thus

And so the vapour pressure of the solution is

#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