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 \( 0.010 \, M \) glucose solution, we can follow these steps:

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. The osmotic pressure (\( \pi \)) can be expressed using the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
where:
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( C \) = molarity of the solution
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

Since both solutions are at the same temperature and using the same gas constant, we can set their osmotic pressures equal to each other.

### Step 2: Set up the equation for osmotic pressure
Let:
- \( C_1 = 0.004 \, M \) for \( \text{Na}_2\text{SO}_4 \)
- \( C_2 = 0.010 \, M \) for glucose
- \( i_1 \) = van 't Hoff factor for \( \text{Na}_2\text{SO}_4 \)
- \( i_2 = 1 \) for glucose (since it does not dissociate)

From the equality of osmotic pressures, we have:
\[ C_1 \cdot i_1 = C_2 \cdot i_2 \]
Substituting the known values:
\[ 0.004 \cdot i_1 = 0.010 \cdot 1 \]

### Step 3: Solve for \( i_1 \)
Rearranging the equation gives:
\[ i_1 = \frac{0.010}{0.004} = 2.5 \]

### Step 4: Determine the degree of dissociation
The van 't Hoff factor \( i \) for \( \text{Na}_2\text{SO}_4 \) can also be expressed in terms of its degree of dissociation \( \alpha \):
\[ i = 1 + (n - 1) \cdot \alpha \]
where \( n \) is the number of particles the solute dissociates into. For \( \text{Na}_2\text{SO}_4 \):
- It dissociates into \( 2 \, \text{Na}^+ \) and \( \text{SO}_4^{2-} \), giving a total of \( n = 3 \).

Substituting into the equation:
\[ 2.5 = 1 + (3 - 1) \cdot \alpha \]
This simplifies to:
\[ 2.5 = 1 + 2\alpha \]
\[ 2.5 - 1 = 2\alpha \]
\[ 1.5 = 2\alpha \]
\[ \alpha = \frac{1.5}{2} = 0.75 \]

### Step 5: Convert to percentage
To express the degree of dissociation as a percentage:
\[ \alpha \times 100 = 0.75 \times 100 = 75\% \]

### Final Answer
The apparent degree of dissociation of \( \text{Na}_2\text{SO}_4 \) is \( 75\% \).