# 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"

Aug 8, 2017

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

#### Explanation:

And we ALSO MUST KNOW that the vapour pressure of water at $40$ ""^@C is $55.4 \cdot m m \cdot H g$ (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........

$\text{Moles of ribose} = \frac{45.0 \cdot g}{150.13 \cdot g \cdot m o {l}^{-} 1} = 0.300 \cdot m o l$

$\text{Moles of water} = \frac{800.0 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 44.42 \cdot m o l$

Now ${\chi}_{\text{component"="Moles of component"/"Total moles in solution}}$

And thus ${\chi}_{\text{ribose}} = \frac{0.300 \cdot m o l}{0.300 \cdot m o l + 44.42 \cdot m o l} = 6.71 \times {10}^{-} 3$

${\chi}_{\text{water}} = \frac{44.42 \cdot m o l}{0.300 \cdot m o l + 44.42 \cdot m o l} = 0.993$

And so the vapour pressure of the solution is $0.993 \times 55.4 \cdot m m \cdot H g = 55.0 \cdot m m \cdot H g$. 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

$\Pi = \text{9.63 atm" = ??? "mm Hg}$

The osmotic pressure is given by:

$\Pi = i c R T$,

where:

• $i$ is the van't Hoff factor, presumably one for ribose, a nonelectrolyte...
• $c$ is the molarity in $\text{mol/L}$.
• $R$ 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!!):

$\textcolor{b l u e}{\Pi} = \left(1\right) \cdot \left(45.0 \cancel{\text{g" xx cancel"1 mol ribose"/(150.13 cancel"g"))/(0.800 cancel"L") cdot 0.082057 cancel"L"cdot"atm/"cancel"mol"cdotcancel"K" cdot (40 + 273.15 cancel"K}}\right)$

$=$ $\textcolor{b l u e}{\text{9.63 atm}}$

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