An aqueous solution is prepared by dissolving 1.08 g of human serum albumin, a protein obtained from blood plasma, in 50.0 mL of water. The solution has an osmotic pressure of 0.00770 atm at 298 K. What is the molar mass of the albumin?
Answer: 6.86 x 10^4 g/mol
W2=1.08 g
π=5.85 mm of Hg
1 atm=760 mm
So, 1 mm =0.00131 atm.
Therefore,5.85 mm =7.6635*10^-3 atm
T=298 k
Volume of solution=50 cm^3= 50/1000=0.05 L
Then apply the above values in the formula,
π=CRT then calculate the answer
Answer will be 68655 g/mol
pi = MRT
Substitute and solve for M
Then M = mols/L and solve for mols.
Then mols = grams/molar mass. You have mols and grams, solve for molar mass.
To find the molar mass of the human serum albumin, we can use the osmotic pressure of the solution. The osmotic pressure of an ideal dilute solution is given by the equation:
π = (n/V)RT
Where π is the osmotic pressure, n is the number of moles of solute, V is the volume of solvent in liters, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
We are given:
π = 0.00770 atm
V = 50.0 mL = 0.0500 L
T = 298 K
First, let's find the number of moles of solute (human serum albumin):
π = (n/V)RT
Rearranging the equation, we get:
n = (πV) / (RT)
Plugging in the given values:
n = (0.00770 atm * 0.0500 L) / (0.0821 L·atm/mol·K * 298 K)
Calculating this:
n ≈ 0.000160 mol
Now, we can find the molar mass (M) of human serum albumin using the equation:
M = mass / moles
Given:
mass = 1.08 g
moles = 0.000160 mol
Plugging in these values:
M = 1.08 g / 0.000160 mol
Calculating this:
M ≈ 6.75 x 10^3 g/mol
Therefore, the molar mass of human serum albumin is approximately 6.75 x 10^3 g/mol.