Which of the solutions has the greatest osmotic pressure?
Choose one answer.
a. 10.0 g glucose in 1.00 L at 298 K
b. 10.0 g glucose in 1.00 L at 303 K
c. 10.0 g glucose in 0.500 L at 298 K
d. 5.0 g glucose in 1.00 L at 298 K
pi=MRT
R is a constant. T is given. M = moles/L and moles = g/molar mass.
Calculate pi (osmotic pressure) for each and you will have your answer.
To determine which of the solutions has the greatest osmotic pressure, we need to calculate the osmotic pressure for each solution using the formula:
Osmotic Pressure (π) = (n/V)RT
Where:
π = osmotic pressure
n = number of moles of solute
V = volume of solution
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
First, let's find the number of moles of glucose in each solution:
a. 10.0 g glucose in 1.00 L at 298 K:
Molar mass of glucose (C6H12O6) = 180.16 g/mol
10.0 g / 180.16 g/mol = 0.0555 mol
b. 10.0 g glucose in 1.00 L at 303 K:
Same as solution a:
0.0555 mol
c. 10.0 g glucose in 0.500 L at 298 K:
Same as solution a:
0.0555 mol
d. 5.0 g glucose in 1.00 L at 298 K:
Molar mass of glucose: 180.16 g/mol
5.0 g / 180.16 g/mol = 0.0277 mol
Now, let's calculate the osmotic pressures for each solution:
a. Osmotic pressure (π) = (0.0555 mol / 1.00 L) * (0.0821 L.atm/mol.K) * 298 K = 1.614 atm
b. Osmotic pressure (π) = (0.0555 mol / 1.00 L) * (0.0821 L.atm/mol.K) * 303 K = 1.637 atm
c. Osmotic pressure (π) = (0.0555 mol / 0.500 L) * (0.0821 L.atm/mol.K) * 298 K = 3.229 atm
d. Osmotic pressure (π) = (0.0277 mol / 1.00 L) * (0.0821 L.atm/mol.K) * 298 K = 0.406 atm
Comparing the calculated osmotic pressures, we can see that option c (10.0 g glucose in 0.500 L at 298 K) has the greatest osmotic pressure with a value of 3.229 atm.