When `0.004 M Na_(2)SO_(4)` is an isotonic acid with `0.01 M `glucose, the degree of dissociation of `Na_(2)SO_(4)` is

Options

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

Explanation

To find the degree of dissociation of \( \text{Na}_2\text{SO}_4 \) when it is isotonic with \( 0.01 \, M \) glucose, we can follow these steps:

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. For two solutions to be isotonic, the product of the van 't Hoff factor (\( i \)), concentration (\( C \)), and temperature (\( T \)) must be equal for both solutions.

### Step 2: Identify the van 't Hoff factor for glucose and \( \text{Na}_2\text{SO}_4 \)
- Glucose is a non-electrolyte, so its van 't Hoff factor \( i = 1 \).
- \( \text{Na}_2\text{SO}_4 \) dissociates into 3 ions: \( 2 \, \text{Na}^+ \) and \( \text{SO}_4^{2-} \). Thus, the van 't Hoff factor \( i = 3 \) if fully dissociated.

### Step 3: Write the expression for the van 't Hoff factor for \( \text{Na}_2\text{SO}_4 \)
Let \( \alpha \) be the degree of dissociation of \( \text{Na}_2\text{SO}_4 \). The concentration of \( \text{Na}_2\text{SO}_4 \) is \( 0.004 \, M \). The effective concentration of particles after dissociation can be expressed as:
\[ i = 1 + 2\alpha \]
where \( 1 \) accounts for the \( \text{Na}_2\text{SO}_4 \) molecule itself, and \( 2\alpha \) accounts for the two sodium ions produced per formula unit.

### Step 4: Set up the isotonic condition
Since the solutions are isotonic, we can equate the osmotic pressures:
\[ i_1 C_1 = i_2 C_2 \]
Substituting the known values:
\[ (1 + 2\alpha)(0.004) = (1)(0.01) \]

### Step 5: Solve for \( \alpha \)
Expanding the equation:
\[ (1 + 2\alpha)(0.004) = 0.01 \]
\[ 0.004 + 0.008\alpha = 0.01 \]
Rearranging gives:
\[ 0.008\alpha = 0.01 - 0.004 \]
\[ 0.008\alpha = 0.006 \]
\[ \alpha = \frac{0.006}{0.008} = 0.75 \]

### Step 6: Convert to percentage
To find the percentage degree of dissociation:
\[ \text{Percentage of dissociation} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

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

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