Animated Solution for Chemistry - Solutions: A solution is prepared by dissolving $0.6\text{ g}$ of urea (molar mass $=60{\text{ g mol}}^{-1}$) and $1.8\text{ g}$ of glucose (molar mass $=180{\text{ g mol}}^{-1}$) in $100\text{ mL}$ of water at $27^{\circ }\text{C}$. The osmotic pressure of the solution is ($R=0.08206{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$)

Imagine you are standing in a laboratory, holding a beaker filled with exactly $100\text{ mL}$ of pure water. On its own, this water has specific, predictable physical properties. But the moment you dissolve a solute into it—whether it's a pinch of salt, a spoonful of sugar, or in our case, urea and glucose—you fundamentally alter the physical reality of that liquid.

This phenomenon is governed by what chemists call colligative properties. These are properties of a solution that depend strictly on the ratio of the number of solute particles to the number of solvent molecules, and completely ignore the chemical identity of the solute. It doesn't matter if the particle is a massive glucose molecule or a tiny sodium ion; to the solvent, a particle is a particle. One of the most fascinating colligative properties is osmotic pressure.

Osmotic Pressure

Osmotic pressure, denoted by the Greek letter $\pi$, is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.

Interestingly, the mathematical behavior of dilute solutions closely mirrors the behavior of ideal gases. Just as the ideal gas law is $PV=nRT$, the equation for osmotic pressure is:

Here, $C$ represents the total molar concentration of all solute particles in the solution ($C=\frac{n_{\text{total}}}{V}$), $R$ is the universal gas constant, and $T$ is the absolute temperature in Kelvin.

Moles of Urea and Glucose

In our specific problem, we are dissolving two different non-electrolytes into the water: $0.6\text{ g}$ of urea and $1.8\text{ g}$ of glucose. Because osmotic pressure cares only about the total number of particles, our first mission is to calculate the moles of each substance and add them together.

Let's calculate the moles of urea ($n_{\text{urea}}$):

Now, let's calculate the moles of glucose ($n_{\text{glucose}}$):

Since both urea and glucose are non-electrolytes, they do not dissociate into ions in water. This means their van't Hoff factor ($i$) is exactly $1$. Therefore, the total number of moles of solute particles in our beaker is simply the sum of the two:

Now that we have our total moles, we can plug our values into the master equation. But beware—this is where many students make a fatal silly mistake! The universal gas constant $R$ is given as $0.08206{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$. Because $R$ uses liters and Kelvin, we must convert our volume and temperature to match.

Our volume is $100\text{ mL}$, which converts to $0.1\text{ L}$. Our temperature is $27^{\circ }\text{C}$, which converts to $300\text{ K}$ ($27+273$).

Let's substitute these pristine values into our expanded osmotic pressure formula:

And there we have it! The osmotic pressure of this mixed solution is exactly $4.92\text{ atm}$.

As you reflect on this problem, ask yourself: what would have happened if we had dissolved $0.01\text{ mol}$ of Sodium Chloride (NaCl) instead of urea? Because NaCl is a strong electrolyte, it dissociates completely into ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions. This means one mole of NaCl produces two moles of particles, giving it a van't Hoff factor of $i=2$. The osmotic pressure would have been significantly higher! Always keep an eye out for the chemical nature of your solute when dealing with colligative properties.