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Chapter 11: Problem 74

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.

Short Answer

Expert verified
We need to assume that the NaCl solution inside the erythrocytes is an ideal solution. This assumption might not be completely accurate at higher concentrations of NaCl, but it simplifies our calculations and provides a reasonable estimation. Using the given freezing point depression and the van't Hoff equation, we can calculate the osmotic pressure of the solution inside the erythrocytes to be approximately 10.62 atm/kg water under the given conditions.

Step by step solution

01

Make an assumption

We will make the assumption that the NaCl solution inside the erythrocytes is an ideal solution. That means the solution obeys Raoult's law and colligative properties such as freezing point depression are directly proportional to the solute concentration.
02

Discuss the validity of the assumption

The assumption that the NaCl solution inside the erythrocytes is an ideal solution might not be completely accurate at higher concentrations of NaCl due to the presence of ions, which can result in deviations from an ideal behaviour. However, we can assume that at the given conditions, this assumption is reasonable, as it simplifies our calculations and provides a reasonable estimation of the osmotic pressure.
03

Calculate the molality of the NaCl solution

We are given the freezing point depression of the NaCl solution as \(\Delta T_f = 0.406^{\circ} \mathrm{C}\). We can use the freezing point depression equation to find the molality of NaCl: \(\Delta T_f = K_f\cdot m\) Where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant for water (1.86 K·kg/mol at \(25^{\circ} \mathrm{C}\)), and \(m\) is the molality of the solution in mol NaCl/kg water. Now we can solve for the molality: \(m = \frac{\Delta T_f}{K_f} = \frac{0.406^{\circ} \mathrm{C}}{1.86 \frac{K \cdot kg}{mol}} = 0.2181 \frac{mol NaCl}{kg water}\)
04

Calculate the osmotic pressure

We can now calculate the osmotic pressure using the van't Hoff equation: \(\Pi = i \cdot m \cdot R \cdot T\) Where \(\Pi\) is the osmotic pressure, \(i\) is the van't Hoff factor (which is 2 for NaCl), \(m\) is the molality of the solution, \(R\) is the ideal gas constant (0.0821 L·atm/mol·K), and \(T\) is the temperature in Kelvin (25°C = 298 K). Plugging in the values, we get: \(\Pi = 2 \cdot 0.2181 \frac{mol NaCl}{kg water} \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 298 K = 10.62 \frac{atm}{kg water}\) Therefore, the osmotic pressure of the solution inside the erythrocytes is approximately 10.62 atm/kg water under the given conditions.

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