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

Options

A. `0.16mol//L`
B. `0.32mol//L`
C. `0.60mol//L`
D. `0.45mol//L`

Explanation

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

\[ \Pi = iCRT \]

Where:
- \(\Pi\) = osmotic pressure (in bar)
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(C\) = molar concentration of the solution (in mol/L)
- \(R\) = ideal gas constant (0.0831 L·bar/(K·mol))
- \(T\) = temperature in Kelvin

### Step 1: Convert the temperature from Celsius to Kelvin
Given that the temperature is \(37^\circ C\):

\[ T = 37 + 273.15 = 310.15 \, K \]

### Step 2: Identify the van 't Hoff factor for NaCl
Sodium chloride (NaCl) dissociates into two ions in solution: Na\(^+\) and Cl\(^-\). Therefore, the van 't Hoff factor \(i\) for NaCl is:

\[ i = 2 \]

### Step 3: Substitute the known values into the osmotic pressure equation
We know:
- \(\Pi = 7.8 \, \text{bar}\)
- \(R = 0.0831 \, \text{L·bar/(K·mol)}\)
- \(T = 310.15 \, K\)
- \(i = 2\)

Substituting these values into the osmotic pressure equation:

\[ 7.8 = 2 \cdot C \cdot 0.0831 \cdot 310.15 \]

### Step 4: Solve for \(C\)
Rearranging the equation to isolate \(C\):

\[ C = \frac{7.8}{2 \cdot 0.0831 \cdot 310.15} \]

Calculating the denominator:

\[ 2 \cdot 0.0831 \cdot 310.15 \approx 51.62 \]

Now substituting back to find \(C\):

\[ C = \frac{7.8}{51.62} \approx 0.1515 \, \text{mol/L} \]

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

\[ C \approx 0.1515 \, \text{mol/L} \]