Calculate the osmotic pressure of this solution.
18..6 grams of solute with molecular mass of 8940 grams are dissolved in enough water to make 1.00 dm3 of solution at 25 celsius. What is the osmotic pressure of the solution?
pi = MRT
moles = grams/molar mass.
Solve for moles. Then use moles in
M = moles/L to solve for M, then pi.
To calculate the osmotic pressure of a solution, we can use the formula:
Osmotic Pressure (π) = (n/V)RT
Where:
- π is the osmotic pressure
- n is the number of moles of solute
- V is the volume of the solution in liters
- R is the ideal gas constant (0.0821 L.atm/mol.K)
- T is the temperature in Kelvin
First, we need to determine the number of moles of solute (n). We can do this by dividing the mass of the solute by its molar mass.
Given:
Mass of solute (m) = 18.6 grams
Molar mass (M) = 8940 grams
n = m/M
n = 18.6g / 8940g/mol
Now, let's convert the volume of the solution to liters:
1.00 dm^3 = 1.00 L
Next, we need to convert the temperature to Kelvin:
25 Celsius = 25 + 273.15 = 298.15 K
Now, we can substitute the values into the formula:
π = (n/V)RT
π = (18.6g/8940g/mol) / 1.00 L * 0.0821 L.atm/mol.K * 298.15 K
By solving this equation, you'll obtain the osmotic pressure (π) of the solution.