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Chapter 13: Problem 79

The osmotic pressure of a \(0.010 \mathrm{M}\) aqueous solution of \(\mathrm{CaCl}_{2}\) is found to be \(0.674 \mathrm{~atm}\) at \(25^{\circ} \mathrm{C}\). (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 aqueous solution of CaCl鈧� with an osmotic pressure of 0.674 atm at 25掳C is approximately 2.73. As the solution becomes more concentrated, the value of i is expected to decrease due to the increase in electrostatic interactions between the ions, suggesting that the extent of dissociation of the solute decreases as the concentration increases.

Step by step solution

01

Understand and note the given values

We are given the osmotic pressure (蟺) as 0.674 atm, the temperature (T) as 25掳C (converted to Kelvin: 273.15 + 25 = 298.15 K), and the concentration of CaCl鈧� (C) as 0.010 M.
02

Recall the osmotic pressure formula and the formula for the van't Hoff factor

The osmotic pressure (蟺) is related to the molar concentration, temperature, and the van't Hoff factor, i, through the formula: 蟺 = iCRT where: 蟺 = Osmotic pressure i = van't Hoff factor C = Molar concentration R = Gas constant (0.0821 L atm K鈦宦� mol鈦宦�) T = Temperature (in Kelvin) The van't Hoff factor, i, can be calculated by rearranging this equation as: i = 蟺 / (CRT)
03

Calculate the van't Hoff factor

Plug in the values of osmotic pressure, molar concentration, gas constant, and temperature into the equation: i = (0.674 atm) / (0.010 M * 0.0821 L atm K鈦宦� mol鈦宦� * 298.15 K) i 鈮� 2.73
04

Discuss how the van't Hoff factor would change as the concentration increases

As the solution concentration increases, more CaCl鈧� particles are present in the solution. Since the solute dissociates into ions, it is expected that the degree of dissociation could decrease as the concentration increases, primarily due to the increase in electrostatic interactions between the ions. Consequently, the van't Hoff factor (i) could decrease as the solution becomes more concentrated, implying that the extent of dissociation of the solute decreases as the concentration increases.

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