Which is the correct relation between osmotic pressure of 0.1 M NaCl solution and 0.1 M `Na_(2)SO_(4)` solution ?

Options

A. The osmotic pressure of `Na_(2)SO_(4)` is less than that of NaCl solution
B. The osmotic pressure `Na_(2)SO_(4)` is more than that of NaCl solution
C. Both have same osmotic pressure
D. None of these

Answer

To determine the relationship between the osmotic pressures of 0.1 M NaCl solution and 0.1 M Na₂SO₄ solution, we can follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (π) of a solution can be calculated using the formula:
\[ \pi = CRTI \]
where:
- \(C\) = concentration of the solution (in molarity)
- \(R\) = universal gas constant
- \(T\) = temperature (in Kelvin)
- \(I\) = van 't Hoff factor (which accounts for the number of particles the solute dissociates into)

### Step 2: Identify the van 't Hoff factor for NaCl
For NaCl, which dissociates into Na⁺ and Cl⁻ ions:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
This means NaCl produces 2 ions upon dissociation. Therefore, the van 't Hoff factor \(I\) for NaCl is:
\[ I_{\text{NaCl}} = 2 \]

### Step 3: Calculate the osmotic pressure for NaCl
Using the formula for osmotic pressure:
\[ \pi_{\text{NaCl}} = C \cdot R \cdot T \cdot I_{\text{NaCl}} = 0.1 \cdot R \cdot T \cdot 2 = 0.2 \cdot R \cdot T \]

### Step 4: Identify the van 't Hoff factor for Na₂SO₄
For Na₂SO₄, which dissociates into 2 Na⁺ ions and 1 SO₄²⁻ ion:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]
This means Na₂SO₄ produces 3 ions upon dissociation. Therefore, the van 't Hoff factor \(I\) for Na₂SO₄ is:
\[ I_{\text{Na}_2\text{SO}_4} = 3 \]

### Step 5: Calculate the osmotic pressure for Na₂SO₄
Using the formula for osmotic pressure:
\[ \pi_{\text{Na}_2\text{SO}_4} = C \cdot R \cdot T \cdot I_{\text{Na}_2\text{SO}_4} = 0.1 \cdot R \cdot T \cdot 3 = 0.3 \cdot R \cdot T \]

### Step 6: Compare the osmotic pressures
Now we can compare the osmotic pressures:
- For NaCl: \(\pi_{\text{NaCl}} = 0.2 \cdot R \cdot T\)
- For Na₂SO₄: \(\pi_{\text{Na}_2\text{SO}_4} = 0.3 \cdot R \cdot T\)

Since \(0.3 \cdot R \cdot T > 0.2 \cdot R \cdot T\), we conclude that:
\[ \pi_{\text{Na}_2\text{SO}_4} > \pi_{\text{NaCl}} \]

### Final Conclusion
The osmotic pressure of the 0.1 M Na₂SO₄ solution is greater than that of the 0.1 M NaCl solution.