Question

Question: The walls of erythrocytes (red blood cells) are permeable to water. In a salt solution, they shrivel (lose water) when the outside salt concentration is high and swell (take up water) when the outside salt concentration is low. In an experiment at $25°C$, an aqueous solution of $NaCl$ that has a freezing point of $-0.046°C$ causes erythrocytes neither to swell nor to shrink, indicating that the osmotic pressure of their contents is equal to that of the $NaCl$ solution. Calculate the osmotic pressure of the solution inside the erythrocytes under these conditions, assuming that its molarity and molality are equal.

Short answer

The molality of the solution is $0.025mol/kg$, morality of the solution is $0.025M$ and osmotic pressure is $0.61atm$.

Step-by-step solution

Given data

In an experiment at $25°C$, an aqueous solution of $NaCl$ that has a freezing point of $-0.046°C$ causes erythrocytes neither to swell nor to shrink.

Concept of Osmotic pressure

Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.

Formula Used:

$\mathrm{\pi }=\mathrm{icRTl}$

Calculation of the molality

The solute temperature is given as:

$$\begin{gathered} ∆T_f=T_{solvent-T_{solution}} \\ ∆T_f=0.0°C-(-0.046°C) \\ ∆T_f=0.046°C \\ ∆T_f=0.046K \end{gathered}$$

Therefore,

$$\begin{gathered} ∆T_f=K_{f}m \\ ∆T_f=K_{f}m \\ 0.046K=1.86Kkgmo{l}^{-1}\times m \end{gathered}$$

$$\begin{gathered} m=\frac{0.046K}{1.86Kkgmo{l}^{-1}} \\ m=0.025mol/kg \end{gathered}$$

Calculation of the volume

The mass of the is given as:

$$\begin{gathered} m\left(NaCl\right)=0.025mol\times 58.44g/mol \\ m\left(NaCl\right)=1.5g \\ m\left(Solution\right)=1.5+1000=1001.5g \end{gathered}$$

Since we know that:

$v\frac{m}{p}$

Substitute the values in the above equation and solve further.

$$\begin{gathered} V=\frac{1,001.5G}{1.0g/mL} \\ V=1,001.5mL \\ V=1.0015L \end{gathered}$$

Calculation of the molarity and osmotic pressure

Molarity can be calculated as:

molarity $=\frac{n}{V}$

molarity $=\frac{0.025mol}{1.0015L}$

molarity $=0.0249mol/L$

molarity $\approx 0.025M$

Osmotic pressure can be calculated as:

$$\begin{gathered} \mathrm{\pi }=\mathrm{cRT} \\ \mathrm{\pi }=0.025\frac{\mathrm{mol}}{\mathrm{L}}\times 0.08206\mathrm{L}\mathrm{atm}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\times 298\mathrm{K} \\ \mathrm{\pi }=0.61\mathrm{atm} \end{gathered}$$

Thus, the molality of the solution is $0.025mol/kg$, morality of the solution is $0.025M$ and osmotic pressure is $0.61atm$.