Calcuate the percentage degree of dissociation of an electrolyte `XY_(2)` (Normal molar mass = 164) in water if the water if the observed molar mass by measuring elevation in boiling point is 65.6

Options

A. 75%
B. 25%
C. 65%
D. None of these

Answer

To calculate the percentage degree of dissociation of the electrolyte \( XY_2 \), we can follow these steps:

### Step 1: Understand the dissociation of the electrolyte
The electrolyte \( XY_2 \) dissociates in water as follows:
\[ XY_2 \rightarrow X^{2+} + 2Y^{-} \]
This means that one mole of \( XY_2 \) produces one mole of \( X^{2+} \) and two moles of \( Y^{-} \).

### Step 2: Set up the initial and equilibrium conditions
Let’s assume we start with 1 mole of \( XY_2 \) at time \( t = 0 \). If \( \alpha \) is the degree of dissociation, then at equilibrium:
- Moles of \( XY_2 \) remaining = \( 1 - \alpha \)
- Moles of \( X^{2+} \) formed = \( \alpha \)
- Moles of \( Y^{-} \) formed = \( 2\alpha \)

Thus, the total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha \]

### Step 3: Calculate the Van't Hoff factor (\( i \))
The Van't Hoff factor \( i \) is defined as:
\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} \]
From the previous step, we have:
\[ i = \frac{1 + 2\alpha}{1} = 1 + 2\alpha \]

### Step 4: Relate the Van't Hoff factor to molar masses
The Van't Hoff factor can also be calculated using the normal molar mass and the observed molar mass:
\[ i = \frac{\text{Normal molar mass}}{\text{Observed molar mass}} \]
Given:
- Normal molar mass = 164 g/mol
- Observed molar mass = 65.6 g/mol

Calculating \( i \):
\[ i = \frac{164}{65.6} \approx 2.5 \]

### Step 5: Set up the equation for \( \alpha \)
Now we can set the two expressions for \( i \) equal to each other:
\[ 1 + 2\alpha = 2.5 \]
Solving for \( \alpha \):
\[ 2\alpha = 2.5 - 1 = 1.5 \]
\[ \alpha = \frac{1.5}{2} = 0.75 \]

### Step 6: Calculate the percentage degree of dissociation
The percentage degree of dissociation is given by:
\[ \text{Percentage degree of dissociation} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

### Final Answer
The percentage degree of dissociation of the electrolyte \( XY_2 \) in water is **75%**.

---