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

The osmotic pressure is given by:

#Pi = icRT#,

where:

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"#".
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Answer 2

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

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......
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Answer 3

The osmotic pressure of the solution is approximately "5.32 mm Hg".

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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