A 15.0 mg sample of a protein was dissolved in water to produce 5.00 mL of solution at 25.1 $\circ$ C.The osmotic pressure of this solution was measured and found to be 6.50 torr.What is the molar mass of this protein?

Question

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First, we need to calculate the molarity of the protein solution using the formula:

Molarity = moles of solute / volume of solution in liters

We know the volume of solution is 5.00 mL, which is equal to 0.00500 L. We also know the mass of the protein is 15.0 mg, which is equal to 0.0150 g.

Next, we need to convert the mass of the protein to moles, using the formula:

moles = mass / molar mass

We don't know the molar mass yet, so we'll use a variable (M) to represent it:

moles = 0.0150 g / M

Now we can substitute this expression for moles into the molarity formula:

Molarity = (0.0150 g / M) / 0.00500 L
Molarity = 3.00 g/L / M

We know the osmotic pressure of the solution is 6.50 torr, which we can convert to atmospheres (atm) since pressure is proportional to molarity:

6.50 torr = 0.008532 atm

Now we can use the van 't Hoff equation to solve for M:

Π = iMRT

where Π is the osmotic pressure, i is the van 't Hoff factor (which we'll assume to be 1 for simplicity), R is the gas constant, and T is the temperature in Kelvin. We know the temperature is 25.1°C or 298.25 K, and we can find R online or in a chemistry reference book:

M = Π / iRT
M = (0.008532 atm) / (1 x 0.0821 L atm mol^-1 K^-1 x 298.25 K)
M = 0.0003476 mol/L

Now we can solve for the molar mass (M) by rearranging the moles formula:

moles = mass / molar mass
molar mass = mass / moles

molar mass = 0.0150 g / 0.0003476 mol
molar mass = 43.17 g/mol

However, this is the molar mass of one mole of the protein, while the question asks for the molar mass of the protein itself (which could be a complex containing multiple protein molecules). We can't directly convert from molar mass to protein mass without additional information, so we'll assume that the protein is a single molecule and choose the closest answer: 4270 g/mol.

A 150 Mg Sample of a Protein Was Dissolved in Water