How to calculate the osmotic pressure of a sucrose solution?


> 
> The boiling point of an aqueous solution containing sucrose, $\ce{C12H22O11}$, is $101.45~^\circ\mathrm{C}$. Calculate the osmotic pressure of this solution at $35~^\circ\mathrm{C}$, at which the solution density is $\pu{1.036 g/mL}$. The boiling point elevation constant is $k\_b = 0.512~^\circ\pu{C\cdot m^{-1}}$.
>  The expected answer is
>  $\Pi = 37.6~\mathrm{atm}$.
> 
> 
> 


I tried it this way:


1. I used $\Delta T\_b = k\_b \cdot m$ and substituted $T\_b = 101.45$ and $k\_b = 0.512$
and got $m = 198.144~\mathrm{mol/kg}$.
2. Density equal to $\pu{1.036 g/ml}$ equals $\pu{1.036 kg/L}$, which means $\pu{1.036 kg}$ for $\pu{1 L}$ and $\pu{1 kg} = \pu{0.965 L}$.
3. $M = \pu{198.144 mol}/\pu{0.965 L} = \pu{205.33 mol/L}$
4. The osmotic pressure equation $\Pi = MRT = 205.33 \cdot 0.08206 \cdot (35+273)
 = \pu{640 atm}$.


Where have I gone wrong?





> 
> I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg
> 
> 
> 


This should be the first clue that something went wrong. 198 mol of sucrose is about 67 ***kilograms*** of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose. That can't be right. (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)


I think you forgot to use the *change* in boiling point temperature in your $\Delta T\_b = k\_b m$ equation. The *change* is only 1.45 °C, not 101.45 °C.


What happens if you plug in that value and redo your calculations?


***Update***: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.



> 
> Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.
> 
> 
> 


This is true, but keep in mind the definition of *molality*: it is moles of solute per kg of *solvent*. However the density is obviously referring to the kg of *total solution*. So, it is inappropriate to find the molarity by simply dividing the molality by the density. As I described [in a past answer on chem.se](https://chemistry.stackexchange.com/questions/26483/whose-concentration-is-more-1-molar-or-1-molal), if the *molarity* is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density. So you will have to invert this relationship to find the molarity from the molality.


 2

After more than five years, it seems safe to assume that an explicit answer will no longer serve to furnish the OP with a homework solution, so I will post numerical values for all the steps already given in Curt F.'s answer:


**Molality of the solution**


The change in boiling temperature is $\pu{1.45 K}$. Together with the boiling point elevation constant, we can calculate the molality of the solution as $\pu{2.83 mol/kg}$.


**Concentration of the solution**


To figure out the concentration, let's assume we have a solution containing $\pu{1 kg}$ of solvent. So we would have $\pu{2.83 mol}$ or $\pu{0.9694 kg}$ of sucrose. We can get the mass of the solution as the sum of the mass of solvent and solute, and the volume of the solution because we also know the density. The concentration comes out as $\pu{1.49 M}$. If you write down the calculation, you will see that the mass of solvent cancels out, giving you the following relationship between the molality $b$ and the concentration $c$:


$$c = \frac{b \cdot \rho}{1 + M\_\mathrm{sucrose} \cdot b}$$


with $M\_\mathrm{sucrose}$ the molar mass of sucrose and $\rho$ the density of the solution at the relevant temperature. Density and concentration are temperature-dependent, while molality is not.


**Osmotic pressure**


The osmotic pressure comes out as $\pu{37.6 atm}$ when you plug in the correct concentration, temperature and the universal gas constant into $\Pi = c R T$.


**Step-by-step math**


$ΔT = 1.45\ \mathrm{K}$


$k\_{\mathrm{b}} = 0.512\ \frac{\mathrm{K}\ \mathrm{kg}}{\mathrm{mol}}$


$b = \dfrac{ΔT}{k\_{\mathrm{b}}}$


$\ \ \ =\dfrac{1.45\ \mathrm{K}}{0.512\ \frac{\mathrm{K}\ \mathrm{kg}}{\mathrm{mol}}}$


$\ \ \ =2.83\ \frac{\mathrm{mol}}{\mathrm{kg}}$


$ρ = 1.036\ \frac{\mathrm{kg}}{\mathrm{L}}$


$m\_{\mathrm{solvent}}$ $= 1\ \mathrm{kg}$


$c = \dfrac{ρ}{\dfrac{1}{b} + M\_{\mathrm{\ce{C12H22O11}}}}$


$\ \ \ =\dfrac{1.036\ \frac{\mathrm{kg}}{\mathrm{L}}}{\dfrac{1}{2.83\ \frac{\mathrm{mol}}{\mathrm{kg}}} + 342.30\ \frac{\mathrm{g}}{\mathrm{mol}}}$


$\ \ \ =\dfrac{1.036\ \frac{\mathrm{kg}}{\mathrm{L}}}{0.3531\ \frac{\mathrm{kg}}{\mathrm{mol}} + 342.30\ \frac{\mathrm{g}}{\mathrm{mol}}}$


$\ \ \ =\dfrac{1.036\ \frac{\mathrm{kg}}{\mathrm{L}}}{0.6954\ \frac{\mathrm{kg}}{\mathrm{mol}}}$


$\ \ \ =1.490\ \frac{\mathrm{mol}}{\mathrm{L}}$


$R = 0.08205\ \frac{\mathrm{atm}\ \mathrm{L}}{\mathrm{K}\ \mathrm{mol}}$


$T = 35.\ \mathrm{°aC}$


$Π = c \cdot R \cdot T$


$\ \ \ =1.490\ \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.08205\ \frac{\mathrm{atm}\ \mathrm{L}}{\mathrm{K}\ \mathrm{mol}} \cdot 35.\ \mathrm{°aC}$


$\ \ \ =0.12224\ \frac{\mathrm{atm}}{\mathrm{K}} \cdot 308.\ \mathrm{K}$


$\ \ \ =37.7\ \mathrm{atm}$


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