0.004 M `Na_(2)SO_(4)` is isotonic with 0.01 M glucose. Degree of dissociation of `Na_(2)SO_(4)` is

Options

A. 0.75
B. 0.5
C. 0.25
D. 0.85

Explanation

To find the degree of dissociation of \( Na_2SO_4 \) that is isotonic with 0.01 M glucose, we can follow these steps:

### Step 1: Understand Isotonic Solutions
Isotonic solutions have the same osmotic pressure. Therefore, we can equate the osmotic pressures of the two solutions.

### Step 2: Calculate the Osmotic Pressure of Glucose
The osmotic pressure (\( \Pi \)) of a solution can be calculated 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)

For glucose, which does not dissociate, \( i = 1 \):
\[ \Pi_{glucose} = 1 \times 0.01 \, \text{M} \times RT = 0.01RT \]

### Step 3: Calculate the Osmotic Pressure of \( Na_2SO_4 \)
For \( Na_2SO_4 \), it dissociates into 3 ions:
\[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \]
Thus, \( i = 3 \). The osmotic pressure for \( Na_2SO_4 \) is:
\[ \Pi_{Na_2SO_4} = i \times C \times RT = 3 \times (0.004 \, \text{M}) \times RT = 0.012RT \]

### Step 4: Set the Osmotic Pressures Equal
Since the two solutions are isotonic:
\[ 0.01RT = 0.012RT \]

### Step 5: Cancel \( RT \) from Both Sides
Cancelling \( RT \) from both sides gives:
\[ 0.01 = 0.012 \]
This indicates that the osmotic pressures are equal, which is consistent with our setup.

### Step 6: Calculate the Degree of Dissociation
Let \( \alpha \) be the degree of dissociation of \( Na_2SO_4 \). The effective concentration after dissociation can be expressed as:
\[ C_{effective} = (1 - \alpha) + 2\alpha = 1 + \alpha \]
Thus, the effective concentration of \( Na_2SO_4 \) becomes:
\[ C_{effective} = 0.004 \times (1 + 2\alpha) \]

### Step 7: Relate the Effective Concentration to the Van 't Hoff Factor
From the earlier calculation, we know:
\[ i = 1 + 2\alpha \]
We previously found that \( i = \frac{10}{4} = 2.5 \):
\[ 2.5 = 1 + 2\alpha \]

### Step 8: Solve for \( \alpha \)
Rearranging gives:
\[ 2\alpha = 2.5 - 1 = 1.5 \implies \alpha = \frac{1.5}{2} = 0.75 \]

### Step 9: Convert to Percentage
To find the degree of dissociation in percentage:
\[ \text{Degree of dissociation} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

### Final Answer
The degree of dissociation of \( Na_2SO_4 \) is **75%**.