A `5.8%` (wt./vol.)NaCl solution will exert an osmotic pressure closest to which one of the following

Options

A. 5.8% (mass/vol) sucrose solution
B. 5.8% (mass/vol) glucose solution
C. 2M sucrose solution
D. 1M glucose solution

Answer

To determine the osmotic pressure exerted by a 5.8% (wt/vol) NaCl solution, we can use the formula for osmotic pressure:

\[ \pi = i \cdot C \cdot R \cdot T \]

Where:
- \(\pi\) = osmotic pressure
- \(i\) = van 't Hoff factor (degree of dissociation)
- \(C\) = concentration of the solute in mol/L
- \(R\) = ideal gas constant (approximately 0.0821 L·atm/(K·mol) or 8.314 J/(K·mol))
- \(T\) = temperature in Kelvin

### Step 1: Calculate the van 't Hoff factor (i)
For NaCl, which dissociates into Na\(^+\) and Cl\(^-\), the van 't Hoff factor \(i\) is:

\[ i = 2 \]

### Step 2: Calculate the concentration (C)
The concentration \(C\) in mol/L can be calculated using the formula:

\[ C = \frac{\text{number of moles of solute}}{\text{volume of solution in liters}} \]

1. **Find the mass of NaCl**: Given that we have a 5.8% (wt/vol) solution, this means there are 5.8 grams of NaCl in 100 mL of solution.
2. **Calculate the number of moles of NaCl**:

\[ \text{Molar mass of NaCl} = 58.5 \, \text{g/mol} \]

\[ \text{Number of moles} = \frac{5.8 \, \text{g}}{58.5 \, \text{g/mol}} \approx 0.0993 \, \text{mol} \]

3. **Convert volume from mL to L**:

\[ 100 \, \text{mL} = 0.1 \, \text{L} \]

4. **Calculate concentration**:

\[ C = \frac{0.0993 \, \text{mol}}{0.1 \, \text{L}} = 0.993 \, \text{mol/L} \]

### Step 3: Substitute values into the osmotic pressure formula
Assuming standard temperature (T = 298 K):

\[ \pi = i \cdot C \cdot R \cdot T \]

Substituting the values:

\[ \pi = 2 \cdot 0.993 \, \text{mol/L} \cdot 0.0821 \, \text{L·atm/(K·mol)} \cdot 298 \, \text{K} \]

Calculating this gives:

\[ \pi \approx 2 \cdot 0.993 \cdot 0.0821 \cdot 298 \approx 48.6 \, \text{atm} \]

### Conclusion
The osmotic pressure exerted by a 5.8% (wt/vol) NaCl solution is approximately 48.6 atm.