Problem 86 - Chemistry The Science in Context

Question

Determine the molarity of each of the following solutions from its osmotic pressure at \(25^{\circ} \mathrm{C} .\) Include the van 't Hoff factor for the solution when the factor is given. a. \(\Pi=0.0259\) atm for a solution of urea \(\left[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\right]\) b. \(\Pi=1.56\) atm for a solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) c. \(\Pi=0.697\) atm for a solution of \(\mathrm{KI}, i=1.90\)

Step-by-Step Solution

Answer

Question: Determine the molarity of the following solutions at 25°C: (a) a solution of urea, CO(NH2)2, with osmotic pressure of 0.0259 atm, (b) a solution of sucrose, C12H22O11, with osmotic pressure of 1.56 atm, and (c) a solution of potassium iodide, KI, with osmotic pressure of 0.697 atm and a van 't Hoff factor of 1.90. Answer: (a) The molarity of the urea solution is 0.00105 mol/L. (b) The molarity of the sucrose solution is 0.0643 mol/L. (c) The molarity of the potassium iodide solution is 0.0126 mol/L.

Step 1: Solution for urea (a)

Given: \(\Pi = 0.0259 \ \mathrm{atm}\) The van 't Hoff factor for urea is not given, but since it is a non-electrolyte, \(i = 1\). Using the osmotic pressure equation: \(0.0259 = M·(1)·(0.0821·298.15)\) Solving for M, we get: \(M = \frac{0.0259}{0.0821·298.15} = 0.00105 \ \mathrm{mol/L}\)

Step 2: Solution for sucrose (b)

Given: \(\Pi = 1.56 \ \mathrm{atm}\) The van 't Hoff factor for sucrose is not given, but since it is a non-electrolyte, \(i = 1\). Using the osmotic pressure equation: \(1.56 = M·(1)·(0.0821·298.15)\) Solving for M, we get: \(M = \frac{1.56}{0.0821·298.15} = 0.0643 \ \mathrm{mol/L}\)

Step 3: Solution for KI (c)

Given: \(\Pi = 0.697 \ \mathrm{atm}\) The van 't Hoff factor for potassium iodide is given: \(i = 1.90\) Using the osmotic pressure equation: \(0.697 = M·(1.90)·(0.0821·298.15)\) Solving for M, we get: \(M = \frac{0.697}{1.90·0.0821·298.15} = 0.0126 \ \mathrm{mol/L}\)

Key Concepts

Osmotic Pressure

Osmotic pressure is a key concept in chemistry that helps us understand the behavior of solutions. It is the pressure required to prevent the movement of solvent molecules through a semipermeable membrane into a solution. This movement, known as osmosis, occurs naturally to balance the concentration of solutes on both sides of the membrane.

The formula for calculating osmotic pressure \(\Pi\) is:

• \(\Pi = iMRT\)

where:

• \(i\) is the van 't Hoff factor.
• \(M\) is the molarity of the solution (moles per liter).
• \(R\) is the ideal gas constant \((0.0821 \ L \cdot atm / mol \cdot K)\).
• \(T\) is the temperature in Kelvin.

Osmotic pressure is crucial in biological and chemical processes, influencing how cells interact with their environments.

Van 't Hoff Factor

The van 't Hoff factor \(i\) is vital for understanding how solute particles influence colligative properties like osmotic pressure. It indicates how many particles a compound forms when dissolved.

• For non-electrolytes, such as urea and sucrose, \(i = 1\) because they do not dissociate into ions.
• For electrolytes, like \(\text{KI}\), \(i\) is greater than 1 because they dissociate into ions. For example, \(\text{KI}\) can have a van 't Hoff factor of \(1.90\).

This factor is essential in calculating the correct molarity of a solution when dealing with electrolytes that dissociate in solution.

Non-Electrolyte Solution

Non-electrolyte solutions are simpler to work with since they contain solutes that do not dissociate into ions when dissolved. This behavior affects the van 't Hoff factor, which is \(i = 1\) for non-electrolyte solutions.

Examples of non-electrolytes include urea and sucrose, which maintain their molecular structure in solution. When calculating osmotic pressure for these solutions, you can use the formula:

• \(\Pi = iMRT\)

The calculations are straightforward because you can directly use the molarity of the molecular solute, without adjusting for dissociation.

Non-electrolyte solutions are important in many applications, such as food and pharmaceuticals, where the stability of the compound in solution is critical.