A protein has a molecular weight of 98,300 grams/ mole. What will be the osmotic pressure if 0.128 grams of this protein I dissolved in 15.0 ml of water at 20 degrees calcias?
Note the correct spelling of celsius.
pi = osmotic pressure in atm.
R is 0.08206 L*atm/mol*K
pi = MRT
M = mols/L. mols = grams/molar mass. Solve for mols, then M, then pi. Technically this won't be quite right since M = mols/L soln and the problem quotes adding the solute to 15.0 mL.
To calculate the osmotic pressure, we can use the formula:
Osmotic pressure = (n/V)RT
Where:
n = number of moles of solute
V = volume of the solution (in liters)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
To find the number of moles of solute (protein), we first need to find the molar mass of the protein:
Molar mass = mass of protein / moles of protein
Molar mass = 98,300 g/mol
Now we can find the number of moles:
Moles of protein = mass of protein / molar mass of protein
Moles of protein = 0.128 g / 98,300 g/mol
Next, we need to convert the volume of the solution to liters:
Volume of solution = 15.0 ml / 1000 ml/L
Now we have all the values we need to calculate the osmotic pressure:
Osmotic pressure = (n/V)RT
Osmotic pressure = (moles of protein / volume of solution) * R * T
However, we need to convert the temperature from Celsius to Kelvin:
Temperature (Kelvin) = Temperature (Celsius) + 273.15
So, in this case:
Temperature (Kelvin) = 20°C + 273.15
Now we can substitute the values into the formula and calculate the osmotic pressure. Please note that the unit of osmotic pressure is typically measured in atmospheres (atm).
I hope this helps you understand how to calculate the osmotic pressure!