Final answer:
The osmolarity of a 0.25 M solution of trisodium phosphate is calculated by multiplying the molarity of the solution by the number of particles per formula unit that dissociate in solution. Since trisodium phosphate dissociates into four ions, the osmolarity is four times the molarity, resulting in an osmolarity of 1.00 Osmol/L.
Step-by-step explanation:
To calculate the osmolarity of a solution that is 0.25 molar in trisodium phosphate (Na3PO4), one must consider the dissociation of trisodium phosphate in solution. Trisodium phosphate will dissociate into four ions: three sodium ions (Na+) and one phosphate ion (PO43-). Therefore, the osmolarity, which is the total concentration of solute particles in solution, will be four times the molarity of the trisodium phosphate.
The calculation is as follows:
- Molarity (M) of trisodium phosphate = 0.25 M
- Number of particles per formula unit = 4 (3 Na+ + 1 PO43-)
- Osmolarity = Molarity × Number of particles per formula unit
- Osmolarity = 0.25 M × 4
- Osmolarity = 1.00 Osmol/L
Hence, the osmolarity of a 0.25 M solution of trisodium phosphate is 1 Osmol/L.