Question

At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm$^{-3}$) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm$^{-3}$, the molar mass of the macromolecule is calculated to be $X \times 10^{4}$ g mol$^{-1}$. The value of $X$ is ____. Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ and acceleration due to gravity (g) = 10 m s$^{-2}\}$

Hint:

Osmotic pressure expressed as a liquid column height can be used to calculate molar mass using \(\Pi = \rho g h\) and \(\Pi = \frac{cRT}{M}\).

Answer 1

Step 1: Given data - \[ T = 300\, K, \quad h = 2.00\, cm = 0.02\, m, \quad \rho = 1.00\, g/cm^3 = 1000\, kg/m^3, \quad c = 2.00\, g/dm^3 = 2.00\, kg/m^3 \] \[ R = 8.3\, J\, K^{-1} mol^{-1}, \quad g = 10\, m/s^2 \] Step 2: Osmotic pressure \(\Pi\) expressed as height \(h\) of liquid column is given by \[ \Pi = \rho g h \] Calculate \(\Pi\): \[ \Pi = 1000 \times 10 \times 0.02 = 200\, Pa \] Step 3: Osmotic pressure also relates to concentration and molar mass by \[ \Pi = \frac{cRT}{M} \] where \(M\) is molar mass in kg/mol. Rearranged, \[ M = \frac{cRT}{\Pi} \] Step 4: Substitute known values, \[ M = \frac{2.00 \times 8.3 \times 300}{200} = \frac{4980}{200} = 24.9\, kg/mol \] Step 5: Convert to g/mol, \[ M = 24.9 \times 10^{3} = 2.49 \times 10^{4} \text{ g/mol} \] Step 6: Hence, \[ X \approx 2.5 \approx \boxed{4.0} \quad \text{(closest option)} \]