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Chapter 13: Problem 83

The osmotic pressure of a \(0.010 \mathrm{M}\) aqueous solution of \(\mathrm{CaCl}_{2}\) is found to be 0.674 atm at \(25^{\circ}\) C. (a) Calculate the van't Hoff factor, \(i\), for the solution. (b) How would you expect the value of \(i\) to change as the solution becomes more concentrated? Explain.

Short Answer

Expert verified
The van't Hoff factor, i, for a 0.010 M CaCl2 solution is 0.92, calculated using the formula \( i = \frac{\Pi_{experimental}}{\Pi_{ideal}} =\frac{0.674 \, \text{atm}}{0.733 \, \text{atm}} \). As the concentration of the solution increases, the van't Hoff factor is expected to deviate further from the ideal value of 3, due to increased ionic interactions.

Step by step solution

01

Calculate the theoretical osmotic pressure for a 0.010 M CaCl2 solution

Assuming that CaCl2 dissociates ideally into one Ca2+ ion and two Cl- ions (i = 3), we can calculate the theoretical osmotic pressure using the formula: \( \Pi_{ideal} = i_{ideal} \times c \times R \times T \) Where \(i_{ideal}\) is the van't Hoff factor for ideal dissociation, which is 3 in our case. We will plug in the values to obtain the theoretical osmotic pressure: \( \Pi_{ideal} = 3 \times 0.010 \, \text{M} \times 0.0821 \, \frac{\text{L atm}}{\text{mol K}} \times 298.15 \, \text{K} \)
02

Calculate the van't Hoff factor, i

Before calculating the van't Hoff factor, first compute the value of \( \Pi_{ideal} \): \( \Pi_{ideal} = 0.733 \, \text{atm} \) Now we can calculate the van't Hoff factor by comparing the experimental osmotic pressure to the theoretical osmotic pressure: \( i = \frac{\Pi_{experimental}}{\Pi_{ideal}} \) \( i = \frac{0.674 \, \text{atm}}{0.733 \, \text{atm}} \)
03

Compute the van't Hoff factor

Divide the experimental osmotic pressure by the theoretical osmotic pressure: \( i = 0.92 \) Thus, the van't Hoff factor for this 0.010 M CaCl2 solution is 0.92.
04

Discuss the change in the van't Hoff factor as the concentration increases

As the concentration of the CaCl2 solution increases, the ions will be more crowded in the solution, leading to a higher chance of the ions interacting with one another. These interactions may decrease the extent to which the ions are separated from one another and, consequently, the van't Hoff factor could deviate further from the ideal value of 3. Thus, as the concentration increases, we would expect the van't Hoff factor to decrease from the ideal value.

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