An aqueous solution containing 10.50 g of a compound in 100.0 ml of solution has
an osmotic pressure of 0.0187 atm. at 25°C. What is the molecular weight of the compound?
pi = MRT
You know pi, R, and T. Solve for M = mols/L.
Then you know M and L, solve for mols.
Finally, mols = grams/molar mass. You know mols and grams, solve for molar mass.
To find the molecular weight of the compound, we need to use the equation for osmotic pressure:
π = (n/V)RT
Where:
π = osmotic pressure
n = number of moles of solute
V = volume of the solution in liters
R = ideal gas constant (0.0821 L·atm·mol^(-1)·K^(-1))
T = temperature in Kelvin
First, we need to convert the volume of the solution from milliliters to liters:
V = 100.0 ml = 0.1 L
Next, we rearrange the equation to solve for the number of moles of solute:
n = (πV)/(RT)
Now, we can substitute the given values into the equation:
π = 0.0187 atm
V = 0.1 L
R = 0.0821 L·atm·mol^(-1)·K^(-1)
T = 25°C = 25 + 273 = 298 K
n = (0.0187 atm * 0.1 L) / (0.0821 L·atm·mol^(-1)·K^(-1) * 298 K)
Performing the calculation, we find:
n ≈ 0.00074 mol
Next, we need to find the molar mass of the compound by dividing the mass of the compound by the number of moles:
Molar mass = mass / moles
Given:
Mass = 10.50 g
Moles = 0.00074 mol
Molar mass = 10.50 g / 0.00074 mol
Performing the calculation, we find:
Molar mass ≈ 14,189 g/mol
Therefore, the molecular weight of the compound is approximately 14,189 g/mol.