A `0.004M` solution of `Na_(2)SO_(4)` is isotonic with a `0.010 M` solution of glucose at same temperature. The apparent degree of dissociation of `Na_(2)SO_(4)` is

Options

A. `25%`
B. `50%`
C. `75%`
D. `85%`

Explanation

To find the apparent degree of dissociation of \( \text{Na}_2\text{SO}_4 \) in a solution that is isotonic with a glucose solution, we can follow these steps:

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. This means that the osmotic pressure of the \( 0.004 \, M \) solution of \( \text{Na}_2\text{SO}_4 \) is equal to that of the \( 0.010 \, M \) solution of glucose.

### Step 2: Write the formula for osmotic pressure
The osmotic pressure (\( \Pi \)) can be expressed as:
\[ \Pi = i \cdot C \cdot R \cdot T \]
where:
- \( i \) is the van 't Hoff factor (number of particles the solute dissociates into),
- \( C \) is the molarity of the solution,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 3: Set up the equation for both solutions
For glucose (which does not dissociate), the osmotic pressure is:
\[ \Pi_{\text{glucose}} = 1 \cdot 0.010 \cdot R \cdot T \]

For \( \text{Na}_2\text{SO}_4 \), which dissociates into \( 2 \, \text{Na}^+ \) and \( \text{SO}_4^{2-} \), the van 't Hoff factor \( i \) can be expressed as:
\[ i = 2 + \alpha \]
where \( \alpha \) is the degree of dissociation.

Thus, the osmotic pressure for \( \text{Na}_2\text{SO}_4 \) is:
\[ \Pi_{\text{Na}_2\text{SO}_4} = (2 + \alpha) \cdot 0.004 \cdot R \cdot T \]

### Step 4: Set the osmotic pressures equal to each other
Since the solutions are isotonic:
\[ 1 \cdot 0.010 \cdot R \cdot T = (2 + \alpha) \cdot 0.004 \cdot R \cdot T \]

### Step 5: Cancel out \( R \) and \( T \)
Since \( R \) and \( T \) are common on both sides, we can cancel them out:
\[ 0.010 = (2 + \alpha) \cdot 0.004 \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation gives:
\[ 2 + \alpha = \frac{0.010}{0.004} \]
Calculating the right side:
\[ 2 + \alpha = 2.5 \]
Now, isolate \( \alpha \):
\[ \alpha = 2.5 - 2 = 0.5 \]

### Step 7: Convert \( \alpha \) to percentage
To express the degree of dissociation as a percentage, multiply by 100:
\[ \alpha \times 100 = 0.5 \times 100 = 50\% \]

### Conclusion
The apparent degree of dissociation of \( \text{Na}_2\text{SO}_4 \) is \( 50\% \).