Problem 37 - Chemistry: Principles and Reactions

Question

Calculate the osmotic pressure of the following solutions of urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), at \(22^{\circ} \mathrm{C} .\) (a) \(0.217 \mathrm{M}\) urea (b) \(25.0 \mathrm{~g}\) urea dissolved in enough water to make \(685 \mathrm{~mL}\) of solution. (c) \(15.0 \%\) urea by mass (density of the solution \(=1.12 \mathrm{~g} / \mathrm{mL}\) )

Step-by-Step Solution

Answer

Question: Calculate the osmotic pressures of the following urea solutions at 22°C: (a) 0.217 M urea, (b) 25.0 g urea in 685 mL of solution, and (c) 15.0% urea by mass (density of the solution = 1.12 g/mL). Answer: The osmotic pressures of the urea solutions are (a) 4.996 atm, (b) 14.60 atm, and (c) 67.71 atm.

Step 1: Conversion of temperature to Kelvin

First, we will convert the temperature from Celsius to Kelvin using the following formula: \(T(K) = T(°C) + 273.15\). Hence, \(T = 22 + 273.15 = 295.15 K\)

Step 2a: Calculating molar concentration for (a) 0.217 M urea

For the first part, the concentration is already given in molar (M) units, so we don't need to perform any calculations. The molar concentration, \(c = 0.217 M\).

Step 3a: Calculate osmotic pressure for (a) 0.217 M urea

Now, we can calculate the osmotic pressure using the formula \(Π = cRT\). Plugging in the values, we get \(Π = (0.217\,M)(0.0821\, \frac{L\cdot atm}{mol\cdot K})(295.15\,K) = 4.996\, atm\).

Step 2b: Calculate molar concentration for (b) 25.0 g urea in 685 mL of solution

First, let's find the moles of urea. The molar mass of urea \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\) is \(2(14.01) + 2(1.01) + 12.01 + 16.00 = 60.06\, g/mol\). So, moles of urea \(= \frac{25.0\,g}{60.06\,g/mol} = 0.4161\,mol\). Next, convert total volume of solution from mL to L: \(685\,mL = 0.685\,L\). Now, we can find the molar concentration: \(c = \frac{0.4161\,mol}{0.685\,L} = 0.6072\,M\).

Step 3b: Calculate osmotic pressure for (b) 25.0 g urea in 685 mL of solution

Using the osmotic pressure formula \(Π = cRT\), we compute the osmotic pressure for this case as: \(Π = (0.6072\,M)(0.0821\, \frac{L\cdot atm}{mol\cdot K})(295.15\,K) = 14.60\, atm\).

Step 2c: Calculate molar concentration for (c) 15.0 % urea by mass (density of the solution = 1.12 g/mL)

Given a 15.0% urea solution by mass, this means that 15.0 g urea is present in every 100 g solution. The density of the solution is \(1.12\, g/mL\). To find the volume of 100 g solution, we use the relationship: \(density = \frac{mass}{volume}\). Rearranging and solving for volume: \(volume = \frac{mass}{density} = \frac{100\,g}{1.12\,g/mL} = 89.29\, mL\). Convert the volume to liters: \(0.08929\, L\). Now, we can calculate the moles of urea in the solution: \(moles\, urea = \frac{15.0\,g}{60.06\,g/mol} = 0.2498\, mol\). Finally, calculate the molar concentration: \(c = \frac{0.2498\, mol}{0.08929\, L} = 2.798\,M\)

Step 3c: Calculate osmotic pressure for (c) 15.0 % urea by mass (density of the solution = 1.12 g/mL)

Using the osmotic pressure formula \(Π = cRT\), we compute the osmotic pressure for this case as: \(Π = (2.798\,M)(0.0821\, \frac{L\cdot atm}{mol\cdot K})(295.15\,K) = 67.71\, atm\). So, the osmotic pressures of the solutions are: (a) \(4.996\, atm\) (b) \(14.60\, atm\) (c) \(67.71\, atm\)

Key Concepts

Colligative Properties

Colligative properties are characteristics of solutions that depend on the number of dissolved particles in the solution, rather than the identity of those particles. This means that colligative properties are the same for all solvents regardless of what solute is dissolved. The four main colligative properties are vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
Osmotic pressure is the pressure required to stop the flow of water through a semipermeable membrane from the pure solvent into the solution. It's a pivotal concept when calculating the behavior of solutions in biological, chemical, and physical contexts. We use osmotic pressure to predict how a solution will react when it's separated from a pure solvent by a membrane that only allows the solvent to pass through. In educational material, it is crucial to highlight the importance of osmotic pressure in everyday phenomena, such as plant nutrient absorption and kidney filtration.
In the exercise provided, the osmotic pressure calculation uses the molar concentration of urea in the solution and the temperature in Kelvin. It is essential to note that the higher the concentration of solute particles, the higher the osmotic pressure - a direct reflection of the colligative nature of osmotic pressure.

Molar Concentration

Molar concentration, often represented by the symbol 'c' and expressed in moles per liter (M), is a measure of the concentration of a solute in a solution. It's a core part of understanding chemistry and solution dynamics. Calculating the molar concentration involves dividing the amount of substance (in moles) by the volume of the solution (in liters).
For students learning about molar concentration for the first time, using real-world examples, such as the sweetness of a sugar solution for cooking or the strength of a saltwater solution for gargling, can be helpful. In the textbook exercise, the molar concentration for scenario (b), for instance, is found by taking the given mass of urea, converting it to moles using the molarity of urea, and then dividing by the total volume of solution created.
Understanding molar concentration allows students to predict how a given amount of a solute will affect the overall solution, which is critical in many scientific and industrial processes. In biological systems, molar concentration plays a significant role in processes such as enzyme-substrate interactions and gradient formation across cell membranes.

Raoult's Law

Raoult's Law is a principle that relates to the vapor pressure of a solution to the concentration of the solute and the pure solvent. It states that the presence of a non-volatile solute lowers the vapor pressure of a solvent. The more solute particles present, the more the solvent's vapor pressure is lowered. This lowering is proportional to the mole fraction of the solute in an ideal solution.
What is important for students to grasp is that Raoult's Law underlines the predictability of a solution's behavior when a solvent is mixed with a non-volatile solute. For instance, adding salt to water decreases the water's vapor pressure, which in turn raises the boiling point and lowers the freezing point of the water. While Raoult's Law does not directly apply to the osmotic pressure, it creates a foundational understanding that helps explain why colligative properties, including osmotic pressure, are influenced by the molar concentration of the solute.
In teaching environments, it can be beneficial to explain Raoult's Law using practical examples, such as adding antifreeze to a car radiator or explaining why we salt roads in winter. These applications of Raoult's Law demonstrate its relevance to everyday life and help solidify the student's comprehension of the law's implications for various colligative properties.