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.
Henry's law describes how much of a gas dissolves in a liquid. It states that, at a constant temperature, the solubility of a gas in a liquid (or the mole fraction of the gas in solution) is directly proportional to the partial pressure of that gas over the liquid.
\[ p = K_H \cdot x \]
where \(p\) = partial pressure of the gas above the solution, \(x\) = mole fraction of the gas in the solution, and \(K_H\) = Henry's law constant (a value that depends on the nature of the gas and the temperature). A higher \(K_H\) means the gas is less soluble. This law explains, for example, why soda bottles are sealed under high pressure and why fizzing occurs when the cap is opened.

Part (ii): Molar mass from osmotic pressure.
Step 1: Formula. Osmotic pressure \(\pi\) is related to concentration by \(\pi = CRT\), where \(C = \dfrac{n}{V} = \dfrac{w}{M V}\). Rearranging for molar mass:
\[ M = \frac{wRT}{\pi V} \]
Step 2: List the data (in consistent units).
\(w = 1.26\ \text{g}\); \(R = 0.083\ \text{L bar K}^{-1}\text{mol}^{-1}\); \(T = 300\ \text{K}\); \(\pi = 2.57\times10^{-3}\ \text{bar}\); \(V = 200\ \text{mL} = 0.200\ \text{L}\).
Step 3: Substitute.
\[ M = \frac{1.26 \times 0.083 \times 300}{2.57\times10^{-3} \times 0.200} \]
Step 4: Arithmetic. Numerator \(= 1.26 \times 0.083 \times 300 = 31.374\). Denominator \(= 2.57\times10^{-3} \times 0.200 = 5.14\times10^{-4}\).
\[ M = \frac{31.374}{5.14\times10^{-4}} \approx 61039\ \text{g mol}^{-1} \]
\[\boxed{M \approx 6.10 \times 10^{4}\ \text{g mol}^{-1}}\]