What is the osmotic pressure of a solution made from dissolving 60.0g of glucose, C6H12O6, in 450 mL of water at 10.0∘C?
pi = MRT
pi in mm Hg.
You know R (0.08206) and T (remember in kelvin).
M = mols/L solution.
where mols = grams glucose/molar mass glucose and L solution is 0.450
Solve for pi
To find the osmotic pressure of a solution, we need to use the formula:
π = i × M × R × T
Where:
- π represents the osmotic pressure of the solution.
- i is the van't Hoff factor, which represents the number of particles into which the solute dissociates (e.g., glucose does not dissociate, so its van't Hoff factor is 1).
- M is the molarity of the solution, calculated by dividing the number of moles of solute by the volume of the solution in liters.
- R is the ideal gas constant, which has a value of 0.0821 L·atm/(mol·K).
- T is the temperature in Kelvin.
To determine the molarity of the solution, we need to calculate the number of moles of glucose (C6H12O6) first. The molar mass of glucose is:
(6 × 12.01 g/mol) + (12 × 1.01 g/mol) + (6 × 16.00 g/mol) = 180.18 g/mol
Now, we can calculate the number of moles:
moles = mass / molar mass
moles = 60.0 g / 180.18 g/mol
Next, we need to convert 450 mL of water to liters:
volume = 450 mL / 1000 mL/L
Now, we can calculate the molarity (M):
M = moles / volume
Finally, with the values of i (1), M, R (0.0821), and T (10.0 + 273 = 283 K), we can substitute them into the osmotic pressure formula to find the answer: π.