A solution containing 29.00 mg of an unknown protein per 22.0 mL solution was found to have an osmotic pressure of 3.10 torr at 18 ∘C. What is the molar mass of the protein?
(3.10 torr* 22mL)=.682
I do not know what to do next
First, I wouldn't start that way.
Convert torr to atm; i.e., 310/760 = pi = ?
pi = MRT. You know pi, R (0.0820), and T (in kelvin please). Solve for M = molarity.
Then M = mols/L.
You know M and L (or can convert to L), solve for mols.
Then mols = grams/molar mass. You know mols from above, the problem gives you mass in mg (convert to g), solve for molar mass.
To find the molar mass of the protein, you need to use the formula for osmotic pressure:
π = nRT/V
Where:
π is the osmotic pressure
n is the number of moles of solute
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
V is the volume of the solution in liters
To convert the osmotic pressure from torr to atm, you divide by 760 (since 1 atm = 760 torr):
3.10 torr / 760 torr/atm = 0.00408 atm
Now, rearrange the formula to solve for the number of moles of solute (n):
n = πV / RT
n = (0.00408 atm) * (0.0220 L) / (0.0821 L·atm/(mol·K) * (18 + 273.15) K)
n = 3.314 x 10^-6 moles
Now, use the number of moles of solute (n) and the mass of the solute (29.00 mg) to find the molar mass:
molar mass = mass / moles
molar mass = 29.00 mg / 3.314 x 10^-6 moles
molar mass = 8.756 x 10^6 g/mol
Therefore, the molar mass of the unknown protein is approximately 8.756 x 10^6 g/mol.