1. A .50 - g sample of immunoglobulin G, a nonvolatile nonelectrolyte, is dissol

Question

1. A .50 - g sample of immunoglobulin G, a nonvolatile nonelectrolyte, is dissolved in enough water to make .100 L of solution, and the osmotic pressure of the solution at 25 degrees C is found to be .619 torr. Calculate the molecular mass of immunoglobulin G.

2. When 2.74 g of phosphorus is dissolved in 100.00 mL of carbon disulfide, the boiling point is 319.71 K. Given that the normal boiling point of pure varbon disulfide is 319.30 K, its density is 1.261 g/ml, and its boiling-point elevation contant is Kb = 2.34 K/m, determine the molar mass of phosphorus.

3. A solution of biphenyl (C12H10), a nonvolatile nonelectrolyte, in denzene has a freezing point of 5.4 degrees C. Determine the osmotic p ressure of the solution at 10 degrees C if its density is .88 g/cm3. Note: normal freezing pt of benzene = 5.5 degrees C: Kf= 5.12 degrees C

4. Consider these two solutions: Solution A is prepared by dissolving 5.00 g of MgCl2 in enough water to make .250 L of solutions, and Solution B is prepared by dissolving 5.00 g of KCl in enough water to make .250 L of solution. Which direction will solvent initally flow if these two solutions are seperated by a semipermeable membrane?

5. Assuming that the volumes of the solutions described in question 4 are additive and igoring any effects that gravity may have on the osmotic pressure of the solutions, what will be the final volume of solution A when the net solvent flow through the semipermeable membrane stops?

Explanation / Answer

Question:

A .50 - g sample of immunoglobulin G, a nonvolatile nonelectrolyte, is dissolved in enough water to make .100 L of solution, and the osmotic pressure of the solution at 25 degrees C is found to be .619 torr. Calculate the molecular mass of immunoglobulin G.

Answer :

The osmotic pressure is related as

Osmotic pressure = molarity of solution X gas constant X temperature

Gas constant = 62.36 L Torr K1 mol1

molarity = mass of solute / molecular weight of solute X volume of solution

So 0.619 = molarity X 62.36 X 298 K

Molarity = 3.17 X 10^-5 molar

3.17 X 10^-5 = 0.5 / mol wt X 0.1

Molecular weight = 1.578 X 10^5 g / mole

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