Question

A tube fitted with a semipermeable membrane is dipped into 0.001 M NaCl solution at 300 K as shown in the figure. Assume density of the solvent and solution are the same. At equilibrium, the height of the liquid column \( h \) (in cm) is ......... 

Hint:

Osmotic pressure can be related to the height of the liquid column in osmosis problems. Remember to use the correct units for pressure and volume.

Answer 1

This problem involves osmosis, where the height of the liquid column is related to the osmotic pressure. The osmotic pressure \( \Pi \) is given by: \[ \Pi = \frac{nRT}{V} \] where:
- \( n \) is the number of moles of solute,
- \( R \) is the ideal gas constant (\( 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)),
- \( T \) is the temperature in Kelvin,
- \( V \) is the volume of solution.
Since the density of the solvent and solution are the same, we can equate the osmotic pressure to the hydrostatic pressure: \[ \Pi = \rho g h \] where:
- \( \rho \) is the density of the solution (assumed to be \( 1 \, \text{kg dm}^{-3} \)),
- \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)),
- \( h \) is the height of the liquid column.
The osmotic pressure \( \Pi \) for the NaCl solution can be calculated using the formula for osmotic pressure, assuming \( 1 \, \text{mol} \) of NaCl dissociates into 2 ions (\( \text{Na}^+ \) and \( \text{Cl}^- \)): \[ \Pi = i \times M \times R \times T \] where \( i = 2 \) for NaCl (since it dissociates into two ions), \( M = 0.001 \, \text{mol/L} \), \( R = 8.314 \, \text{J/mol·K} \), and \( T = 300 \, \text{K} \).
Thus, \[ \Pi = 2 \times 0.001 \times 8.314 \times 300 = 4.986 \, \text{Pa} \] Now, equating the osmotic pressure to the hydrostatic pressure: \[ 4.986 = 1 \times 9.8 \times h \] Solving for \( h \): \[ h = \frac{4.986}{9.8} = 0.509 \, \text{m} = 50.9 \, \text{cm} \] Final Answer: \[ \boxed{50.9} \]