A `0.004M` solution of `Na_(2)SO_(4)` is isotonic with a `0.010M` 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 Na₂SO₄ in a solution that is isotonic with a glucose solution, we will follow these steps:

### Step 1: Understand Isotonic Solutions
Two solutions are isotonic if they have the same osmotic pressure. The osmotic pressure (π) can be expressed using the formula:
\[ \pi = iCRT \]
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: Set Up the Equation
Since the solutions are isotonic, we can equate the osmotic pressures of the two solutions:
\[ \pi_{Na₂SO₄} = \pi_{glucose} \]
This gives us:
\[ i_{Na₂SO₄} \cdot C_{Na₂SO₄} = i_{glucose} \cdot C_{glucose} \]

### Step 3: Identify Values
- For glucose (a non-electrolyte), \( i_{glucose} = 1 \) and \( C_{glucose} = 0.010 \, M \).
- For Na₂SO₄, \( C_{Na₂SO₄} = 0.004 \, M \).

### Step 4: Determine the Van 't Hoff Factor for Na₂SO₄
Na₂SO₄ dissociates into three ions:
\[ Na₂SO₄ \rightarrow 2Na^+ + SO₄^{2-} \]
Thus, the van 't Hoff factor \( i \) for Na₂SO₄ is:
\[ i_{Na₂SO₄} = 3 \]

### Step 5: Plug Values into the Equation
Now we can substitute the values into the equation:
\[ i_{Na₂SO₄} \cdot C_{Na₂SO₄} = i_{glucose} \cdot C_{glucose} \]
\[ 3 \cdot 0.004 = 1 \cdot 0.010 \]

### Step 6: Solve for the Degree of Dissociation (α)
Let \( \alpha \) be the degree of dissociation of Na₂SO₄. The effective concentration of Na₂SO₄ considering dissociation is:
\[ C_{effective} = C_{Na₂SO₄} \cdot (1 + 2\alpha) \]
So we have:
\[ 3 \cdot 0.004 = 0.010 \]
This can be rewritten as:
\[ 0.012 = 0.010 \]
Now, substituting for \( C_{effective} \):
\[ 3 \cdot 0.004 = 0.010 \]
\[ 0.012 = 0.010 \]

### Step 7: Rearranging the Equation
From the equation:
\[ 3 \cdot 0.004 = 0.010 \]
\[ 0.012 = 0.010 \]
We can solve for \( \alpha \):
\[ 1 + 2\alpha = \frac{0.010}{0.004} \]
\[ 1 + 2\alpha = 2.5 \]
\[ 2\alpha = 2.5 - 1 \]
\[ 2\alpha = 1.5 \]
\[ \alpha = \frac{1.5}{2} = 0.75 \]

### Step 8: Calculate the Percentage of Degree of Dissociation
To find the percentage degree of dissociation:
\[ \text{Percentage of } \alpha = \alpha \times 100 = 0.75 \times 100 = 75\% \]

### Final Answer
The apparent degree of dissociation of Na₂SO₄ is **75%**.

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