Question

Equal volumes of \(M / 20\) glucose solution at \(300 \mathrm{~K}\) and \(M / 20\) sucrose solution at \(300 \mathrm{~K}\) are mixed without change in temperature. If the osmotic pressure of glucose solution, sucrose solution and the mixture of two solutions are \(\pi_{1}, \pi_{2}\) and \(\pi_{3}\) respectively, then (a) \(\pi_{1}=\pi_{2}=\pi_{3}\) (b) \(\pi_{1}>\pi_{2}>\pi_{3}\) (c) \(\pi_{1}<\pi_{2}<\pi_{3}\) (d) \(\pi_{1}=\pi_{2}<\pi_{3}\)

Short answer

\(\pi_{1}=\pi_{2}=\pi_{3}\)

Step-by-step solution

Understand the properties of osmotic pressure

Osmotic pressure (\( \text{symbol: } \) is directly proportional to the molarity (\( M \) of the solute in the solution, according to the formula \( \pi = MRT \) where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. Because the temperature and gas constant are the same for both the glucose and sucrose solutions, their osmotic pressures will be directly related to their molarity.

Comparing osmotic pressures of individual solutions

Since both the glucose and sucrose solutions have the same molarity (\( M/20 \) and are at the same temperature, their osmotic pressures will be equal. This can be represented as \( \pi_{1} = \pi_{2} \) where \( \pi_{1} \) is the osmotic pressure of the glucose solution and \( \pi_{2} \) is the osmotic pressure of the sucrose solution.

Determine the osmotic pressure of the mixture

When equal volumes of the two solutions are mixed, the total volume doubles, but the amount of solute remains the same for both glucose and sucrose. This means that the molarity of each solute in the mixture is halved to \( M/40 \). Because osmotic pressure is a colligative property (it depends on the number of particles), the osmotic pressure of the mixture will be the sum of the osmotic pressures of the individual solutions in the mixed volume. But since each solute now contributes half to the osmotic pressure it did before, we have \( \pi_{3} = \pi_{1}/2 + \pi_{2}/2 \) which simplifies to \( \pi_{3} = \pi_{1} \) using the relation \( \pi_{1} = \pi_{2} \) from Step 2.

Key concepts

Colligative Properties

Colligative properties are those properties of solutions that depend on the number of solute particles in a given amount of solvent, regardless of the identity of the solute particles. Key colligative properties include vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure. These properties are fundamentally important because they relate to changes in the physical properties of the solvent caused by the presence of dissolved particles.

For example, when a non-volatile solute is added to a solvent, it causes the solvent to boil at a higher temperature and freeze at a lower temperature than pure solvent. This is because the solute particles interfere with the solvent molecules as they change state. Osmotic pressure, in particular, reflects the tendency of solvent molecules to flow through a semipermeable membrane from a less concentrated solution into a more concentrated one. Understanding colligative properties is essential when predicting how a solution will behave under different conditions, and they have practical applications in industries such as food, pharmacology, and environmental science.

In the context of the textbook exercise, osmotic pressure is the colligative property under examination. We utilize the relationship between osmotic pressure and the number of solute particles to solve problems related to the mixing of solutions, as demonstrated in the given problem.

Molarity and Osmotic Pressure Relationship

The relationship between molarity and osmotic pressure is direct and described by the equation \[ \pi = MRT \], where:\

\
• \( \pi \) is the osmotic pressure,\<\/li>\
• \( M \) is the molarity of the solution,\<\/li>\
• \( R \) is the ideal gas constant,\<\/li>\
• \( T \) is the absolute temperature in Kelvin.\<\/li>\<\/ul>\The ideal gas constant (\( R \)) and temperature (\( T \)) are often fixed for a given problem, indicating that the osmotic pressure of a solution is dependent on the concentration of solute particles in that solution. This relationship is particularly important in the realm of chemistry and biochemistry where osmotic pressure must be taken into account when preparing solutions that will be physiologically compatible, such as intravenous fluids.
  
  In the exercise provided, both glucose and sucrose solutions have the same molarity and are mixed at the same temperature, implying that the initial osmotic pressures (\( \pi_1 \) for glucose and \( \pi_2 \) for sucrose) should be equal due to their molarity being the factor that determines osmotic pressure.

Osmotic Pressure Calculations

Osmotic pressure calculations often involve using the formula \[ \pi = MRT \.\] However, solving problems also requires an understanding of how mixing solutions affects the concentration of the solute and hence, the osmotic pressure. When two solutions of equal volumes and molar concentrations are mixed, the final molarity of each solute becomes half of its initial molarity since the volume of the solution has doubled. Consequently, the osmotic pressure produced by each solute halves as well.

In the instance of a mixture containing more than one solute, as in our subject exercise, the total osmotic pressure is the sum of the pressures due to each solute. Therefore, for the mixture of glucose and sucrose solutions, the osmotic pressure will be \( \pi_3 = \frac{\pi_1}{2} + \frac{\pi_2}{2} \), which simplifies to \( \pi_3 = \pi_1 \) assuming that \( \pi_1 = \pi_2 \) due to the same initial molarity of the solutions. Students need to remember these key points when performing osmotic pressure calculations to better understand and successfully solve complex problems involving the mixing of solutions and their impacts on osmotic pressure.