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

Options

A. `5.8%(wt.//vol.)`sucrose solution
B. `5.8%(wt.//vol.)`glucose solution
C. `2`molal sucrose solution
D. `1`molal sucrose solution

Explanation

To find the osmotic pressure of a 5.8% (wt/vol) NaCl solution, we can follow these steps:

### Step 1: Understand the concentration
A 5.8% (wt/vol) NaCl solution means there are 5.8 grams of NaCl in 100 mL of solution.

### Step 2: Convert grams to moles
To find the osmotic pressure, we first need to convert the mass of NaCl to moles. The molar mass of NaCl is approximately 58.5 g/mol.

\[ \text{Moles of NaCl} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{5.8 \, \text{g}}{58.5 \, \text{g/mol}} \approx 0.099 \, \text{mol} \]

### Step 3: Calculate the molarity
Since the solution volume is 100 mL, we need to convert this to liters:

\[ \text{Volume in liters} = \frac{100 \, \text{mL}}{1000} = 0.1 \, \text{L} \]

Now, we can calculate the molarity (M):

\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.099 \, \text{mol}}{0.1 \, \text{L}} = 0.99 \, \text{M} \]

### Step 4: Determine the van 't Hoff factor (i)
For NaCl, which dissociates into Na⁺ and Cl⁻ ions in solution, the van 't Hoff factor (i) is 2.

### Step 5: Calculate the osmotic pressure
The osmotic pressure (π) can be calculated using the formula:

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

Where:
- \(i = 2\)
- \(M = 0.99 \, \text{M}\)
- \(R = 0.0821 \, \text{L atm/(K mol)}\) (ideal gas constant)
- \(T\) is the temperature in Kelvin (assume room temperature, approximately 298 K)

Substituting the values:

\[ \pi = 2 \cdot 0.99 \cdot 0.0821 \cdot 298 \]

Calculating this gives:

\[ \pi \approx 2 \cdot 0.99 \cdot 0.0821 \cdot 298 \approx 48.6 \, \text{atm} \]

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

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