If the osmotic pressure of 0.010 M aqueous solution of sucrose at `27^(@)C` is 0.25 atm, then the osmotic pressure of a 0.010 M aqueous solution of NaCl at `27^(@)C` is

Options

A. 0.062 atm
B. 0.12 atm
C. 0.25 atm
D. 0.50 atm

Answer

To solve the problem, we will use the formula for osmotic pressure, which is given by:

\[ \pi = iCRT \]

Where:
- \(\pi\) = osmotic pressure
- \(i\) = Van't Hoff factor (number of particles the solute dissociates into)
- \(C\) = molarity of the solution
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

### Step 1: Identify the given values for sucrose
- Molarity (\(C\)) = 0.010 M
- Temperature (\(T\)) = 27°C = 27 + 273 = 300 K
- Osmotic pressure (\(\pi\)) = 0.25 atm
- Van't Hoff factor for sucrose (\(i\)) = 1 (since sucrose is a non-electrolyte)

### Step 2: Write the osmotic pressure equation for sucrose
Using the formula:

\[ \pi_{\text{sucrose}} = i_{\text{sucrose}} \cdot C \cdot R \cdot T \]

Substituting the known values:

\[ 0.25 = 1 \cdot 0.010 \cdot R \cdot 300 \]

### Step 3: Calculate the value of \(R\)
Rearranging the equation to find \(R\):

\[ R = \frac{0.25}{0.010 \cdot 300} \]

Calculating \(R\):

\[ R = \frac{0.25}{3} = 0.0833 \, \text{L atm K}^{-1} \text{mol}^{-1} \]

### Step 4: Identify the values for NaCl
- Molarity (\(C\)) = 0.010 M
- Van't Hoff factor for NaCl (\(i\)) = 2 (since NaCl dissociates into Na⁺ and Cl⁻)

### Step 5: Write the osmotic pressure equation for NaCl
Using the formula:

\[ \pi_{\text{NaCl}} = i_{\text{NaCl}} \cdot C \cdot R \cdot T \]

Substituting the known values:

\[ \pi_{\text{NaCl}} = 2 \cdot 0.010 \cdot R \cdot 300 \]

### Step 6: Substitute the value of \(R\) and calculate \(\pi_{\text{NaCl}}\)
Substituting \(R = 0.0833\):

\[ \pi_{\text{NaCl}} = 2 \cdot 0.010 \cdot 0.0833 \cdot 300 \]

Calculating:

\[ \pi_{\text{NaCl}} = 2 \cdot 0.010 \cdot 24.99 \approx 0.50 \, \text{atm} \]

### Final Answer
The osmotic pressure of a 0.010 M aqueous solution of NaCl at 27°C is **0.50 atm**.

---