Question

A solution having \(54 \mathrm{~g}\) of glucose per litre has an osmotic pressure of \(4.56\) bar. If the osmotic pressure of a urea solution is \(1.52\) bar at the same temperature, what would be its concentration? (a) \(1.0 \mathrm{M}\) (b) \(0.5 \mathrm{M}\) (c) \(0.3 \mathrm{M}\) (d) \(0.1 \mathrm{M}\)

Short answer

The concentration of the urea solution is 0.1 M, corresponding to choice (d).

Step-by-step solution

Identify the relevant formula for osmotic pressure

The osmotic pressure (Pi) of a solution can be calculated using the van't Hoff equation which is \(Pi = iCRT\), where \((i)\) is the van't Hoff factor (which is 1 for glucose and urea since they do not dissociate in solution), \((C)\) is the molar concentration of the solute, \((R)\) is the gas constant, and \((T)\) is the temperature in Kelvin. The gas constant (R) and temperature (T) are assumed to remain constant in both solutions.

Calculate molar concentration of the glucose solution

Calculate molar concentration of Glucose using \((C_1)\) from the given osmotic pressure of the glucose solution (Pi1) using the formula \(Pi_1 = CRT_1\). Since \((i)\) is 1 and assuming \((R)\) and \((T)\) are the same for both solutions, \((C_1)\) will be \((C_1 = \frac{Pi_1}{RT}\).

Express the glucose concentration in moles per litre

Convert the mass concentration of glucose to molar concentration. Since the molar mass of glucose (\(C_6H_{12}O_6\) is approximately 180 g/mol, \((C_1 = \frac{54\mathrm{~g}}{180\mathrm{~g/mol}} = 0.3 \mathrm{M}\).

Set up a proportion between the osmotic pressures and molar concentrations

Using the previously mentioned formula and assuming temperature and gas constant are the same, the osmotic pressures are directly proportional to the molar concentrations of the solutions. Set up the proportion: \((Pi_1/C_1 = Pi_2/C_2\)). We know \((Pi_2\)) (the osmotic pressure of urea solution), and we need to solve for \((C_2\)) (its concentration).

Solve for the concentration of the urea solution

Solve the proportion \((4.56\mathrm{~bar}/0.3\mathrm{~M} = 1.52\mathrm{~bar}/C_2\)) to find the concentration of urea solution \((C_2\)) which gives us \((C_2 = (1.52\mathrm{~bar} \cdot 0.3\mathrm{~M}) / 4.56\mathrm{~bar}\)).

Calculate the final answer

Divide the numbers to find the molar concentration of the urea solution: \((C_2 = \frac{1.52\mathrm{~bar} \cdot 0.3\mathrm{~M}}{4.56\mathrm{~bar}} = 0.1\mathrm{~M}\)). Thus, the concentration of the urea solution corresponds to answer choice (d) 0.1 M.

Key concepts

van't Hoff equation

Understanding the van't Hoff equation is vital for solving osmotic pressure problems in physical chemistry. This equation expresses the osmotic pressure (Pi) as a direct consequence of the solute concentration in a solution. The general form of the van't Hoff equation is Pi = iCRT, where (i) represents the van't Hoff factor indicating the number of particles the solute forms in solution, (C) is the concentration in moles per liter (molarity), (R) is the gas constant, and (T) is the temperature in Kelvin.

Since both glucose and urea do not dissociate into ions, their van't Hoff factor is 1. This simplifies the equation, allowing us to focus on the direct relationship between molar concentration and osmotic pressure, holding (R) and (T) constant. When solving problems of this nature, remember that any change in osmotic pressure is reflective of the change in solute concentration, which is precisely how we can compare the concentration of different solutions given their osmotic pressures.

colligative properties

Colligative properties are physical properties of solutions that depend on the number of dissolved particles in a given amount of solvent, rather than the nature of the particles. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. When approaching osmotic pressure, an essential colligative property, remember it's not about the kind of solute particles but their concentration.

Osmotic pressure can be thought of as the 'pulling' pressure that a solution exerts to draw in solvent through a semipermeable membrane, to equalize solute concentrations on both sides of the membrane. In the given exercise, by using colligative properties, one can determine the molar concentration of a urea solution by comparing its osmotic pressure to that of a known glucose solution.

molarity calculations

Molarity, the measure of concentration used in chemistry, is defined as moles of solute per liter of solution (M = moles/L). It allows for direct comparison between solutions' concentrations. When solving osmotic pressure problems, determining the molarity is a key step.

In the provided exercise, converting the mass of glucose per liter to its molar equivalent required knowledge of the molar mass of glucose. After finding the molarity of glucose, the proportionality between osmotic pressures of glucose and urea solutions could be set up, allowing the calculation of urea's molarity. Always ensure to convert given quantities to moles and account for the volume of solution when dealing with molarity calculations.

physical chemistry

Physical chemistry involves the study of how matter behaves on a molecular and atomic level and how chemical reactions occur. Osmosis, a process essential to understanding osmotic pressure problems, is grounded in the principles of physical chemistry.

Understanding osmosis requires comprehension of how solutes and solvents interact, how pressure can influence movement through semipermeable membranes, and how all these factors relate to the underlying equations, such as the van't Hoff equation. Grasping these principles provides a solid foundation not only for solving complex problems like the one in the exercise but also for appreciating the dynamic interactions in real-world chemical systems.