What osmotic pressure in atmospheres would you expect for a solution of 0.150M CaCl2 that is separated from pure water by a semipermeable membrane at 310 K? Assume 100% dissociation for CaCl2
Any help is appreciated, thank you!
pi = iMRT
pi = 3*0.150*0.08205*310 = ?
wow that's really easy, thank you
To calculate the osmotic pressure, you can use the Van 't Hoff equation:
Π = iMRT
Where:
Π = osmotic pressure (in atm)
i = the van 't Hoff factor (the number of particles obtained per formula unit in solution)
M = molarity of the solution (in mol/L)
R = the ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
For CaCl2, the van 't Hoff factor (i) is 3 because it dissociates into 3 particles (one Ca2+ and two Cl-) in solution.
Given:
M = 0.150 M
T = 310 K
i = 3
R = 0.0821 L·atm/(mol·K)
Let's plug in the values into the equation:
Π = (i * M * R * T)
Π = (3 * 0.150 M * 0.0821 L·atm/(mol·K) * 310 K)
Π = 12.0465 atm
Therefore, the osmotic pressure in atmospheres for the given solution is approximately 12.0465 atm.
To determine the osmotic pressure of a solution, you need to use the formula:
Π = i * M * R * T,
- Π represents the osmotic pressure (in atmospheres),
- i is the van't Hoff factor (the number of particles formed when the solute dissociates),
- M is the molarity of the solution (in mol/L),
- R is the ideal gas constant (0.0821 L·atm/(mol·K)),
- T is the temperature (in Kelvin).
In this case, you have a solution of CaCl2, which fully dissociates into three particles when it dissolves:
CaCl2 → Ca²⁺ + 2Cl⁻
Given that the molarity (M) is 0.150 M and the temperature (T) is 310 K, we can calculate the osmotic pressure as follows:
i = 3 (since CaCl2 fully dissociates)
M = 0.150 M
R = 0.0821 L·atm/(mol·K)
T = 310 K
Plugging these values into the formula, we get:
Π = (3)(0.150)(0.0821)(310)
Π = 11.4 atm
Therefore, the osmotic pressure of the 0.150 M solution of CaCl2 would be approximately 11.4 atmospheres.