Osmotic Pressure in atm. = Molarity ( R ) (Kelvin Temp.)
Do this for all three solutions, then add the pressures.
Do this for all three solutions, then add the pressures.
Î = i * M * R * T
Where:
- Î is the osmotic pressure (in atm)
- i is the van't Hoff factor (the number of particles into which each solute molecule dissociates)
- M is the molarity of the solution (in mol/L)
- R is the ideal gas constant (0.0821 L * atm / K * mol)
- T is the temperature in Kelvin (25°C = 298 K)
For each compound, you need to determine the van't Hoff factor (i) and calculate the osmotic pressure, and then sum up the contributions from all solutes.
Let's calculate the osmotic pressure for each solute:
1. Sodium chloride (NaCl):
- The chemistry of NaCl tells us that it dissociates completely into Naâș and Clâ» ions in water.
- Therefore, the van't Hoff factor (i) for NaCl is 2 (2 particles: Naâș and Clâ»).
- Molarity (M) = 0.4 mol/L
- R = 0.0821 L * atm / K * mol
- T = 298 K
- Î (NaCl) = i * M * R * T = 2 * 0.4 * 0.0821 * 298 = 19.355 atm
2. Glucose (CâHââOâ):
- Glucose does not dissociate into ions and remains as a single molecule in water.
- Therefore, the van't Hoff factor (i) for glucose is 1.
- Molarity (M) = 0.3 mol/L
- R = 0.0821 L * atm / K * mol
- T = 298 K
- Î (glucose) = i * M * R * T = 1 * 0.3 * 0.0821 * 298 = 7.319 atm
3. Calcium chloride (CaClâ):
- The chemistry of CaClâ tells us that it dissociates into CaÂČâș and 2 Clâ» ions in water.
- Therefore, the van't Hoff factor (i) for CaClâ is 3 (3 particles: CaÂČâș and 2 Clâ»).
- Molarity (M) = 0.2 mol/L
- R = 0.0821 L * atm / K * mol
- T = 298 K
- Î (CaClâ) = i * M * R * T = 3 * 0.2 * 0.0821 * 298 = 14.846 atm
Finally, add up the osmotic pressures for all solutes:
Total osmotic pressure (Î ) = Î (NaCl) + Î (glucose) + Î (CaClâ)
Total osmotic pressure = 19.355 + 7.319 + 14.846 = 41.52 atm
So, the osmotic pressure for the given solution is 41.52 atm.