50 mL of 0.2 molal urea solution (density = 1.012 g mL^-1 at 300 K) is mixed with 250 mL of a soluti

Q: Why did we need the density of the first solution?

Because molality is based on the mass of the solvent, not the volume of the solution. To relate the given $50 \text{ mL}$ volume to molality, we first had to convert the volume into the total mass of the solution using density.

Q: How did you get $1012 \text{ g}$ for the mass of the $0.2 \text{ molal}$ solution?

A $0.2 \text{ molal}$ solution contains $0.2 \text{ moles}$ of solute per $1000 \text{ g}$ of solvent. The molar mass of urea is $60 \text{ g/mol}$, so $0.2 \text{ moles}$ is $12 \text{ g}$. Adding the mass of the solute ($12 \text{ g}$) to the mass of the solvent ($1000 \text{ g}$) gives a total solution mass of $1012 \text{ g}$.

Q: Why can we just add the volumes of the two solutions?

The problem explicitly states $\Delta_{\text{mix}}V = 0$, which means the volumes are perfectly additive. In real, non-ideal solutions, mixing $50 \text{ mL}$ and $250 \text{ mL}$ might not give exactly $300 \text{ mL}$ due to intermolecular interactions.

Q: Why didn't we convert the temperature to Celsius?

The ideal gas constant $R$ is given with units involving Kelvin ($\text{K}^{-1}$). Therefore, the temperature must be in Kelvin to ensure the units cancel out correctly.

Q: What if the solute was NaCl instead of urea?

Urea is a non-electrolyte, so it doesn't dissociate (van 't Hoff factor $i = 1$). If it were NaCl, it would dissociate into $\text{Na}^+$ and $\text{Cl}^-$, so we would have to multiply the total moles of particles by $i \approx 2$ when calculating the osmotic pressure.

Question

50 mL of 0.2 molal urea solution (density = $1.012 	ext{ g mL}^{-1}$ at 300 K) is mixed with 250 mL of a solution containing 0.06 g of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in Torr) of the resulting solution at 300 K is ________. [JEE(Advanced) 2023]
[Use : Molar mass of urea = $60 	ext{ g mol}^{-1}$; gas constant, $R = 62 	ext{ L Torr K}^{-1} 	ext{ mol}^{-1}$ ; Assume, $\Delta_{	ext{mix}}H = 0$, $\Delta_{	ext{mix}}V = 0$]

Sigma Percentile · Accepted Answer

Mixing It Up: Calculating Osmotic Pressure of a Blended Urea Solution

### The Setup: Two Beakers, One Goal

Imagine you are in a chemistry lab, standing in front of two beakers. The first beaker contains $50 	ext{ mL}$ of a $0.2 	ext{ molal}$ urea solution. The second beaker holds $250 	ext{ mL}$ of a solution containing just $0.06 	ext{ g}$ of urea. Your task is to mix them together and find the exact osmotic pressure of the resulting concoction at $300 	ext{ K}$.

To find the osmotic pressure, we rely on the master equation:
$$ \pi = CRT $$

Here, $C$ represents the molarity of the final solution, which is the total moles of urea divided by the total volume. So, our mission is clear: we need to find the total moles of urea and the total volume of the mixed solution.

### Decoding the First Solution

Let's focus on the first solution. We know its volume is $50 	ext{ mL}$ and its density is $1.012 	ext{ g/mL}$. By multiplying the volume and density, we can find the total mass of this solution:
$$ m_{	ext{sol1}} = 50 	ext{ mL} 	imes 1.012 	ext{ g/mL} = 50.6 	ext{ g} $$

Now, what exactly does $0.2 	ext{ molal}$ mean? It tells us that there are $0.2 	ext{ moles}$ of urea dissolved in $1000 	ext{ g}$ of solvent. Since the molar mass of urea is $60 	ext{ g/mol}$, $0.2 	ext{ moles}$ weigh exactly $12 	ext{ g}$. Therefore, the total mass of such a reference solution would be $1000 	ext{ g} + 12 	ext{ g} = 1012 	ext{ g}$.

This means that in every $1012 	ext{ g}$ of this solution, there are $12 	ext{ g}$ of urea. Using a simple ratio, we can find the mass of urea in our $50.6 	ext{ g}$ sample:
$$ m_{	ext{urea}} = \frac{12}{1012} 	imes 50.6 = 0.6 	ext{ g} $$

### Bringing It All Together

Now we can calculate the total amount of urea. We have $0.6 	ext{ g}$ from the first solution and $0.06 	ext{ g}$ from the second solution. Adding them up gives us a total mass of $0.66 	ext{ g}$. To find the total moles, we divide this by the molar mass of urea:
$$ n_{	ext{total}} = \frac{0.66}{60} = 0.011 	ext{ mol} $$

What about the total volume? The problem explicitly states that the change in volume upon mixing is zero ($\Delta_{	ext{mix}}V = 0$). This is a crucial piece of information because it allows us to simply add the volumes together:
$$ V_{	ext{total}} = 50 	ext{ mL} + 250 	ext{ mL} = 300 	ext{ mL} = 0.3 	ext{ L} $$

### The Final Calculation

Finally, we have everything we need to plug into our osmotic pressure formula. We substitute the total moles, the total volume, the given gas constant $R = 62 	ext{ L Torr K}^{-1} 	ext{ mol}^{-1}$, and the temperature $T = 300 	ext{ K}$:
$$ \pi = \frac{0.011}{0.3} 	imes 62 	imes 300 $$

Simplifying the expression:
$$ \pi = 0.011 	imes 62 	imes 1000 = 11 	imes 62 = 682 	ext{ Torr} $$

And there we have it! The osmotic pressure of the resulting solution is exactly **$682 	ext{ Torr}$**. Always remember to keep a close eye on the units of the gas constant $R$, as they dictate the units of your final answer.

Enter Numerical Value:

Visualized Solution

Visualizing the Mixture

The Master Equation

Mass of Solution 1

Urea in Solution 1

Total Moles of Urea

Total Volume

Calculating Osmotic Pressure

The Way Forward

The Sigma Insight: Colligative Properties

The Setup

Decoding the First Solution

Bringing It All Together

The Final Calculation

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