Question

(i) Explain Henry's law. (ii) 200 ml of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar mass of the protein. (1+2=3)

Hint:

Henry's law: p = K_Hx. For part (ii) use the osmotic pressure relation M = wRT/(πV) with R = 0.083 L bar K^-1 mol^-1 and V in litres.

Answer 1

Part (i): Henry's law (alternate wording). At a fixed temperature, the mass of a gas dissolved in a given volume of liquid is proportional to the pressure of the gas in equilibrium with the liquid. Written as \(p = K_H x\), it tells us that raising the pressure pushes more gas into solution, and that gases with a large Henry constant \(K_H\) are only sparingly soluble. It applies to gas-in-liquid solutions where the gas does not react with the solvent.

Part (ii): Solve by finding moles first.
Step 1: Use the osmotic-pressure equation in the form \(\pi V = nRT\), so the number of moles of protein is \(n = \dfrac{\pi V}{RT}\).
Step 2: Put in the numbers. \(n = \dfrac{(2.57\times10^{-3})(0.200)}{(0.083)(300)} = \dfrac{5.14\times10^{-4}}{24.9}\).
Step 3: \(n = 2.064\times10^{-5}\ \text{mol}\).
Step 4: Molar mass = mass / moles. \(M = \dfrac{w}{n} = \dfrac{1.26}{2.064\times10^{-5}}\).
\[\boxed{M \approx 6.10 \times 10^{4}\ \text{g mol}^{-1}}\]
The large value is expected, since proteins are macromolecules. Note that osmotic pressure is chosen for such measurements because it is measurably large even at very low (dilute) concentrations.