Molar Mass of Hemoglobin from Osmotic Pressure

To find the molar mass of hemoglobin, we first calculate the osmotic pressure ($\Pi$) exerted by the height difference of the solution column at equilibrium.

The osmotic pressure is given by the hydrostatic pressure equation: $\Pi = \rho g h$

Given:

• Density of solution, $\rho = 1000 \text{ kg m}^{-3}$
• Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$
• Height difference, $h = 80.0 \text{ mm} = 80.0 \times 10^{-3} \text{ m}$

Substituting these values: $\Pi = 1000 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} \times 80.0 \times 10^{-3} \text{ m} = 800 \text{ Pa}$

Using the osmotic pressure formula: $\Pi = C R T = \frac{m}{M \cdot V} R T$

where:

• $\Pi = 800 \text{ Pa}$
• Mass of hemoglobin, $m = 20 \text{ g} = 0.02 \text{ kg}$
• Volume of solution, $V = 1 \text{ L} = 10^{-3} \text{ m}^3$
• Gas constant, $R = 8.3 \text{ kPa dm}^3 \text{ K}^{-1} \text{ mol}^{-1} = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$
• Temperature, $T = 300 \text{ K}$
• $M$ is the molar mass of hemoglobin in $\text{kg mol}^{-1}$

Rearranging the formula to solve for $M$: $M = \frac{m \cdot R \cdot T}{\Pi \cdot V}$

Substitute the values into the equation: $M = \frac{0.02 \text{ kg} \times 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \times 300 \text{ K}}{800 \text{ Pa} \times 10^{-3} \text{ m}^3}$

$M = \frac{49.8}{0.8} = 62.25 \text{ kg mol}^{-1}$

Rounding off to the nearest integer, we get: $M \approx 62 \text{ kg mol}^{-1}$