Molar Mass Determination of Macromolecule via Osmotic Pressure Height

To calculate the molar mass of the macromolecule, we use the relation for osmotic pressure ($\Pi$):

$\Pi = \rho g h$

where:

• $\rho = 1.00 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3}$ (density of the solution)
• $g = 10 \text{ m s}^{-2}$ (acceleration due to gravity)
• $h = 2.00 \text{ cm} = 0.02 \text{ m}$ (height of the solution column)

Substituting the given values into the equation for osmotic pressure:

$\Pi = 1000 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} \times 0.02 \text{ m} = 200 \text{ Pa} \text{ (or N m}^{-2}\text{)}$

The osmotic pressure is also related to the molar mass ($M$) of the solute by the formula:

$\Pi = C R T = \left(\frac{c}{M}\right) R T$

where:

• $c$ is the concentration of the solution $= 2.00 \text{ g dm}^{-3} = 2.00 \text{ g L}^{-1} = 2.00 \text{ kg m}^{-3} = 2000 \text{ g m}^{-3}$
• $R = 8.3 \text{ J K}^{-1}\text{ mol}^{-1}$
• $T = 300 \text{ K}$

Rearranging the formula to solve for the molar mass $M$:

$M = \frac{c R T}{\Pi}$

Substituting the known parameters using standard units:

$M = \frac{2000 \text{ g m}^{-3} \times 8.3 \text{ J K}^{-1}\text{ mol}^{-1} \times 300 \text{ K}}{200 \text{ Pa}}$

$M = \frac{2000 \times 8.3 \times 300}{200} \text{ g mol}^{-1}$

$M = 10 \times 8.3 \times 300 \text{ g mol}^{-1} = 24900 \text{ g mol}^{-1} = 2.49 \times 10^4 \text{ g mol}^{-1}$

Given that the molar mass is expressed as $X \times 10^4 \text{ g mol}^{-1}$, we find:

$X = 2.49$