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A water solution is prepared by mixing $\pu{200 g}$ of a ethylene glycol (ethan-1,2-diol; $\ce{C2H6O2}$) solution (20% by weight) with $\pu{500 g}$ of glycerol (propan-1,2,3-triol; $\ce{C3H8O3}$) solution (15% by weight). Calculate the freezing temperature and the osmotic pressure at $\pu{25 ^\circ C}$ knowing that the density of the prepared solution is $\pu{1.25 g/ml}$ (For water, $K_c = \pu{1.86 K kg mol-1}$).

I know all the formulas but how do I proceed to start with these types of exercises?

I know, for example, that $\Pi=MRTi$, with $M$ being molarity. But how do I find the volume of the solution?
And what about $i$, how do I calculate it here?

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  • $\begingroup$ Hint: You know that masses are always additive. You also know the density of the total mixture. Determine the volume (L). $\endgroup$ – M. Farooq Jun 11 at 4:30
  • $\begingroup$ The volume you need here is to calculate the molarity, that is for finding osmotic pressure. Use the weight % to calculate molality, which needs to calculate freezing point. $\endgroup$ – Mathew Mahindaratne Jun 11 at 6:18
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First, the $i$ in the equation you cited is the van't Hoff factor of the solute(s) (ethylene glycol and glycerol in this case). The van't Hoff factor accounts for how many species are generated by the solute. Specifically, it is the important for electrolytes such as $ \ce{NaCl} $. Because $ \ce{NaCl} $ dissociates into $ \ce{Na^{+}} $ and $ \ce{Cl^{-}} $ in the solution, its van't Hoff factor should be 2. Similarly, you can see that $ \ce{CaCl2} $'s factor should be 3 (1 $ \ce{Ca^{2+}} $ and 2 $ \ce{Cl^{-}} $ )

However, since it is obvious that neither ethylene glycol nor glycerol dissociate in the solution, you can safely ignore it because their van't Hoff factors are all 1.

About the volume, it is actually fairly easy. They gave the density of the prepared solution to be $ \pu{1.25 g/ml} $, which is data you need to calculate the volume. Since density equals mass divided by volume, you will need mass and density to calculate the volume. The mass of the resulting solution is equal to the sum of the masses of the original solutions, i.e

$$ m_{\text{result}} = \pu{(200 + 500) g} = \pu{700 g } $$

From this we can calculate the solution's volume:

$$ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{\pu{700 g}}{\pu{1.25 g/ml}} = \pu{560 ml = 0.56 L} $$

As you said you know all further formulas I will leave the rest for you.

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  • $\begingroup$ Thank you! I guess i was just confused from the %wt. DO they not play any role in here at all? $\endgroup$ – Ani Lici Jun 11 at 4:39
  • $\begingroup$ The %wt is the concentration of the original solutions. From this you can calculate the masses and moles of the two solutes. For example the mass of ethylene glycol should be 200 * 20% = 40 g. $\endgroup$ – ĐỨc Lê Hồng Jun 11 at 4:42
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In the answer elsewhere, ĐỨc Lê Hồng has given how to find the volume of the solution and why van't Hoff factor is not important the case of ethylene glycol and glycerol are involved as solutes in the solution. However, you must know how to find the molality to find the freeding point depression and molarity to find the osmotic pressure.

To do so, you need to find the amounts of these two solutes in the solutions. First look at the amount of ethylene glycol (ethan-1,2-diol; $\ce{C2H6O2}$) solution. It is said that $\pu{200 g}$ of a $20\%$ $(w/w)$ ethylene glycol solution is mixed. That means $200 \times \frac{20}{100} = \pu{40 g}$ of ethylene glycol in there. The rest, $\pu{160 g}$ is water. The molar mass of ethylene glycol is $\pu{62.07 g mol-1}$. Thus amount of ethylene glycol in the final solution is $\frac{40}{62.07} = \pu{0.644 mol}$.

You can do the same calculations to find the amount of glycerol (propan-1,2,3-triol; $\ce{C3H8O3}$) in the final solution. It is said that $\pu{500 g}$ of a $15\%$ $(w/w)$ glycerol solution is mixed with ethylene glycol solution to obtain the final solution. That means $500 \times \frac{15}{100} = \pu{75 g}$ of glycerol in the mixture. The rest, $500 - 75 = \pu{425 g}$ is water. The molar mass of glycerol is $\pu{92.09 g mol-1}$. Thus amount of glycerol in the final solution is $\frac{75}{92.09} = \pu{0.814 mol}$.

Thus, the amount of solute in the final solution $= \pu{(0.644 + 0.814) mol} = \pu{1.458 mol} $.
And, the amount of pure water in the final solution $= \pu{(160 + 425) g} = \pu{585 g} = \pu{0.585 kg}$.
Therefore, molality $(m)$ of the final solution $= \frac{\pu{1.458 mol}}{\pu{(0.585 kg}} = \pu{2.492 mol kg-1} $.

Since you know the equations you can calculate the $\Delta T$ of melting point, and hence freezing point of the solution.

Using same data, you can calculate the molarity of the solution as well. The final volume of the solution was found to be $\pu{0.560 L}$ (see ĐỨc Lê Hồng's answer).
The amount of mixed solutes in the final solution $= \pu{(0.644 + 0.814) mol} = \pu{1.458 mol} $.
Therefore, the molarity $(M)$ of the final solution $= \frac{\pu{1.458 mol}}{\pu{(0.560 L}} = \pu{2.604 mol L-1} $.

Since you know the equations you can calculate the $\Pi$ (osmotic pressure) of the solution as well.

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  • $\begingroup$ thanks a lot! Unfortunately i cant vote yet cuz i dont have 15 points yet sorry! $\endgroup$ – Ani Lici Jun 11 at 23:55

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