Erythrocytes are red blood cells containing hemoglobin. In a saline solution they shrivel when the salt concentration is high and swell when the salt concentration is low. In a \(25^{\circ} \mathrm{C}\) aqueous solution of \(\mathrm{NaCl}\), whose freezing point is \(-0.406^{\circ} \mathrm{C}\), erythrocytes neither swell nor shrink. If we want to calculate the osmotic pressure of the solution inside the erythrocytes under these conditions, what do we need to assume? Why? Estimate how good (or poor) of an assumption this is. Make this assumption and calculate the osmotic pressure of the solution inside the erythrocytes.

We'll assume that the solution inside the erythrocytes behaves as an ideal solution. This means that the solution follows Raoult's Law, and thus we can use colligative properties like freezing depression. Keep in mind that this assumption might not be perfect, but for the purposes of this exercise, it will be sufficient.

The given freezing point depression is -0.406 °C (instead of 0°C for water). We can use the formula for freezing point depression: \(\Delta T_f = K_f \cdot m\) where \(\Delta T_f = (0 - (-0.406))^{\circ} C\), \(K_f\) (the cryoscopic constant for water) is \(1.86\, kg\cdot{} K / mol\), and \(m\) is the molality in \(mol/kg\). \(m = \frac{\Delta T_f}{K_f} = \frac{0.406}{1.86} = 0.218 mol/kg\)

Now we will use the Van't Hoff equation to find the osmotic pressure: \(Π = n \cdot R \cdot T\) First, we need to calculate n, the number of particles per mole of NaCl. As NaCl dissociates into two particles (Na+ and Cl-), n=2. R is the ideal gas constant (0.0821 L∙atm/mol∙K). T is the temperature in Kelvin: \(25^{\circ} C + 273.15 = 298.15 K\) Finally, we need to convert molality to concentration (moles of solute per liter of solvent): \(c = m \cdot 0.0556 \frac{mol}{L}\) (due to the fact that the density of water is approximately \(1\,g/mL\), which means we have almost 1 kg of solvent per liter of solution). Now we find the osmotic pressure: \(Π = n \cdot c \cdot R \cdot T = 2 \cdot 0.218 \cdot 0.0821 \cdot 298.15 = 10.79 atm\)

Estimating how good or poor the assumption of ideal solution behavior is beyond the scope of a high school level problem. However, assuming ideal solution behavior in this case, we can conclude that the osmotic pressure of the solution inside the erythrocytes is approximately 10.79 atm.