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 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 must assume that the solution inside the erythrocytes is isotonic, meaning that the solution inside the erythrocytes has the same salt concentration as the external NaCl solution, where no net movement of water occurs. This assumption is necessary for calculating the osmotic pressure of the solution inside the erythrocytes. However, this might not be a perfect assumption if the freezing point data has some errors or if the intracellular fluid has other solutes that affect its freezing point.

We are given the freezing point depression, \(∆T_f = -0.406^{\circ}\), of the NaCl solution, and the freezing point depression constant for water is \(K_f = 1.86 \frac{K \cdot kg}{mol}\), at \(25^{\circ} \mathrm{C}\) temperature. The molality of the NaCl solution can be found using the following formula: \[∆T_f = K_f \cdot molality\] Rearranging the equation to find the molality: \[molality = \frac{∆T_f}{K_f} = \frac{-0.406^{\circ}}{1.86 \frac{K \cdot kg}{mol}} \approx -0.218 \frac{mol}{kg}\]

Now, we will use the Van't Hoff equation to calculate the osmotic pressure of the NaCl solution inside the erythrocytes: \[\pi = i \cdot M \cdot R \cdot T\] where: - \(\pi\) is the osmotic pressure - \(i\) is the Van't Hoff factor (for NaCl, we assume complete dissociation, so it is 2) - \(M\) is the molar concentration (calculated in step 2) - \(R\) is the gas constant (\(0.0821 \frac{L \cdot atm}{mol \cdot K}\)) - \(T\) is the temperature in kelvin (\(25^{\circ}C + 273.15 = 298.15 K\)) Before calculating osmotic pressure, we need to convert molality (mol/kg) to molarity (mol/L). Assuming the density of the NaCl solution is close to the density of water at \(25^{\circ}C \approx 0.997 \frac{kg}{L}\), we can convert the molality to molarity: \[M = -0.218 \frac{mol}{kg} \cdot 0.997 \frac{kg}{L} \approx -0.217 \frac{mol}{L}\] Now, calculate the osmotic pressure: \[\pi = 2 \cdot (-0.217 \frac{mol}{L}) \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 298.15 K\] \[\pi \approx 10.61 atm \]

The assumption of isotonicity might not be perfect in practice since the intracellular fluid contains other solutes that can affect the freezing point. However, this assumption allows us to simplify the problem and calculate an estimation of the osmotic pressure for the NaCl solution inside the erythrocytes. In conclusion, under the isotonic assumption, the osmotic pressure of the solution inside the erythrocytes is approximately 10.61 atm.