Question

The osmotic pressure of a 0.01m sucrose solution at 298K is about 0.2 atm. What would you estimate the osmotic pressure of a 0.01m CaCl2 solution assuming the dissociation degree is 1?

a) 0.4 atm
b) 0.2 atm
c) 0.1 atm
d) 0.8 atm

Answer 1

The osmotic pressure of a 0.01m CaCl2 solution at 298K would be approximately 0.6 atm because CaCl2 dissociates into three particles per formula unit, and osmotic pressure depends on the total concentration of dissolved particles.

Step-by-step explanation:

The osmotic pressure (Π) of a solution can be calculated using the formula Π = MRT, where M is the molarity of the solution, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin.

In the case of a 0.01m CaCl2 solution, the dissociation degree is assumed to be 1, meaning that each mole of CaCl2 completely dissociates into three moles of ions (one Ca2+ and two Cl−). Therefore, the total concentration of particles in solution would be three times the concentration of CaCl2, resulting in a concentration of (3)(0.01m) = 0.03m.

Using the osmotic pressure equation and considering the given temperature of 298K, we would calculate the osmotic pressure as follows: Π = MRT = (0.03 mol/L) [0.0821 (L· atm)/(K · mol)] (298 K) = 0.6 atm. Therefore, the correct answer is option b) 0.6 atm, which was not listed among the provided options.