Final answer:
The osmotic pressure of a 20 g/L glucose solution at 298.15 K is 2.74 bar. Adding NaCl to the solution increases the osmotic pressure. For a 1 mM glucose solution, the osmotic pressure is 247.57 Pa.
Step-by-step explanation:
To calculate the osmotic pressure (Π) of a 20 g/L glucose solution at 298.15 K, we use the formula Π = iMRT, where i is the van't Hoff factor which is 1 for glucose, M is the molarity, R is the gas constant (0.0831 L·bar/K·mol), and T is the temperature in Kelvin. The molarity (M) is calculated by dividing the mass of glucose (20 g) by its molecular weight (180.156 g/mol) and then by the volume of solution in liters (1 L), resulting in a molarity of 0.111 mol/L. The osmotic pressure is then Π = (1)(0.111 mol/L)(0.0831 L·bar/K·mol)(298.15 K) = 2.74 bar.
Adding sodium chloride (NaCl) to the glucose solution will change the osmotic pressure because NaCl dissociates into Na+ and Cl- ions, increasing the number of particles in solution and thus the osmotic pressure.
For a 1 mM glucose solution, we repeat the osmotic pressure calculation with the new molarity of 1 mM which is 0.001 mol/L. The osmotic pressure, in this case, converts to Pascals using the conversion 1 bar = 100,000 Pa. So the osmotic pressure is Π = (1)(0.001 mol/L)(0.0831 L·bar/K·mol)(298.15 K) × 100,000 Pa/bar = 247.57 Pa.