The average osomotic pressure of human blood is 7.8 bar at `37^(@)C`. What is the concentration of an aqueous `NaCI` solution that could be used in the blood stream?

Options

A. 0.16 molal
B. 0.32 molal
C. 0.60 molal
D. 0.45 molal

Answer

To find the concentration of an aqueous NaCl solution that could be used in the bloodstream, we can use the formula for osmotic pressure:

\[ \Pi = CRTI \]

Where:
- \(\Pi\) = osmotic pressure (in bar)
- \(C\) = concentration of the solution (in mol/L)
- \(R\) = ideal gas constant (0.0821 L·bar/K·mol)
- \(T\) = temperature (in Kelvin)
- \(I\) = van 't Hoff factor (number of particles the solute dissociates into)

### Step 1: Identify the given values
- Osmotic pressure, \(\Pi = 7.8 \, \text{bar}\)
- Temperature, \(T = 37^\circ C\)

### Step 2: Convert temperature to Kelvin
To convert Celsius to Kelvin, use the formula:

\[ T(K) = T(°C) + 273 \]

Calculating:

\[ T = 37 + 273 = 310 \, K \]

### Step 3: Determine the van 't Hoff factor (\(I\)) for NaCl
Sodium chloride (NaCl) dissociates into two ions in solution:

\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]

Thus, the van 't Hoff factor \(I = 2\).

### Step 4: Substitute the values into the osmotic pressure formula
Now we can rearrange the osmotic pressure formula to solve for \(C\):

\[ C = \frac{\Pi}{RTI} \]

Substituting the known values:

\[ C = \frac{7.8}{0.0821 \times 310 \times 2} \]

### Step 5: Calculate the concentration \(C\)
Calculating the denominator:

\[ 0.0821 \times 310 \times 2 = 51.062 \]

Now substituting back into the equation for \(C\):

\[ C = \frac{7.8}{51.062} \approx 0.152 mol/L \]

### Step 6: Round the concentration
Rounding to three significant figures gives:

\[ C \approx 0.153 \, \text{mol/L} \]

### Final Answer
The concentration of an aqueous NaCl solution that could be used in the bloodstream is approximately:

\[ 0.153 \, \text{mol/L} \quad \text{(or nearly } 0.160 \text{ mol/L)} \]