0.1 M NaCl and `0.1 M CH_(3)COOH` are kept in separate containers. If their osmotic pressures are `P_(1)` and `P_(2)` respectively then what is the correct statement ?

Options

A. `P_(1)gt P_(2)`
B. `P_(1)=P_(2)`
C. `P_(1)lt P_(2)`
D. `P_(1)=P_(2)=0` atm

Answer

To solve the problem of comparing the osmotic pressures \( P_1 \) and \( P_2 \) of 0.1 M NaCl and 0.1 M CH₃COOH respectively, we can follow these steps:

### Step 1: Understand the Formula for Osmotic Pressure
The osmotic pressure (\( P \)) of a solution can be calculated using the formula:
\[ P = i \cdot C \cdot R \cdot T \]
where:
- \( i \) = 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 2: Identify the Components for Each Solution
1. **For NaCl**:
- NaCl dissociates into Na⁺ and Cl⁻ ions.
- Therefore, \( i \) for NaCl = 2 (1 Na⁺ + 1 Cl⁻).
- The concentration \( C \) is 0.1 M.

2. **For CH₃COOH**:
- CH₃COOH (acetic acid) dissociates into CH₃COO⁻ and H⁺ ions.
- However, acetic acid is a weak electrolyte, and it does not dissociate completely.
- Therefore, \( i \) for CH₃COOH will be less than 2 (due to partial dissociation).
- The concentration \( C \) is also 0.1 M.

### Step 3: Compare the van't Hoff Factors
- For NaCl: \( i_{NaCl} = 2 \)
- For CH₃COOH: \( i_{CH₃COOH} < 2 \) (since it is a weak electrolyte)

### Step 4: Determine the Osmotic Pressures
Since both solutions have the same concentration (0.1 M) and are at the same temperature, we can compare their osmotic pressures based on their van't Hoff factors:
\[ P_1 \propto i_{NaCl} \quad \text{and} \quad P_2 \propto i_{CH₃COOH} \]
Thus:
\[ P_1 = i_{NaCl} \cdot C \quad \text{and} \quad P_2 = i_{CH₃COOH} \cdot C \]
Since \( i_{NaCl} > i_{CH₃COOH} \), we conclude that:
\[ P_1 > P_2 \]

### Final Conclusion
The correct statement is:
\[ P_1 > P_2 \]

---