If osmotic pressure of 1M urea is `pi` , what will be the osmotic pressure for 0.1 M NaCl?

Options

A. `pi`
B. `0.02 pi`
C. `2pi`
D. `0.2 pi`

Explanation

To solve the problem of finding the osmotic pressure of a 0.1 M NaCl solution given that the osmotic pressure of a 1 M urea solution is π, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Osmotic Pressure**:
The osmotic pressure (π) of a solution can be calculated using the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
where:
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(C\) = concentration of the solution (in molarity)
- \(R\) = gas constant
- \(T\) = temperature (in Kelvin)

2. **Identify the Given Values for Urea**:
- For urea (a non-electrolyte):
- Concentration \(C_1 = 1 \, M\)
- Osmotic pressure \(π_1 = π\)
- Van 't Hoff factor \(i_1 = 1\) (since urea does not dissociate)
- Therefore, the osmotic pressure for urea can be expressed as:
\[ π = 1 \cdot 1 \cdot R \cdot T \implies π = R \cdot T \]

3. **Identify the Given Values for NaCl**:
- For NaCl (an electrolyte):
- Concentration \(C_2 = 0.1 \, M\)
- Van 't Hoff factor \(i_2 = 2\) (since NaCl dissociates into Na⁺ and Cl⁻)
- The osmotic pressure for NaCl can be expressed as:
\[ π_2 = i_2 \cdot C_2 \cdot R \cdot T = 2 \cdot 0.1 \cdot R \cdot T \]

4. **Calculate the Osmotic Pressure for NaCl**:
- Substitute the values:
\[ π_2 = 2 \cdot 0.1 \cdot R \cdot T = 0.2 \cdot R \cdot T \]

5. **Relate \(R \cdot T\) to π**:
- From the earlier step, we know that \(R \cdot T = π\).
- Therefore, we can substitute \(R \cdot T\) in the equation for \(π_2\):
\[ π_2 = 0.2 \cdot π \]

6. **Final Answer**:
- The osmotic pressure of the 0.1 M NaCl solution is:
\[ π_2 = 0.2π \]

### Conclusion:
The osmotic pressure for the 0.1 M NaCl solution is \(0.2π\).