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

Explanation

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

### Step 1: Write the dissociation equation
The dissociation of the electrolyte \( XY_2 \) can be represented as:
\[ XY_2 \rightleftharpoons X^{2+} + 2Y^{-} \]

### Step 2: Set up the initial and equilibrium concentrations
Initially, we have:
- Moles of \( XY_2 \) = 1
- Moles of \( X^{2+} \) = 0
- Moles of \( Y^{-} \) = 0

At equilibrium, if \( \alpha \) is the degree of dissociation, we have:
- Moles of \( XY_2 \) = \( 1 - \alpha \)
- Moles of \( X^{2+} \) = \( \alpha \)
- Moles of \( Y^{-} \) = \( 2\alpha \)

### Step 3: Calculate the total number of moles at equilibrium
The total number of moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha \]

### Step 4: Relate the normal molar mass and observed molar mass
The normal molar mass of the electrolyte \( XY_2 \) is given as \( 164 \, g/mol \) and the observed molar mass is \( 65.6 \, g/mol \).

Using the formula for the van 't Hoff factor \( i \):
\[ i = \frac{\text{Normal molar mass}}{\text{Observed molar mass}} = \frac{164}{65.6} \]

Calculating \( i \):
\[ i \approx 2.5 \]

### Step 5: Relate the van 't Hoff factor to degree of dissociation
The van 't Hoff factor \( i \) for the dissociation of \( XY_2 \) can also be expressed as:
\[ i = 1 + 2\alpha \]
Setting the two expressions for \( i \) equal gives:
\[ 2.5 = 1 + 2\alpha \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation:
\[ 2\alpha = 2.5 - 1 = 1.5 \]
\[ \alpha = \frac{1.5}{2} = 0.75 \]

### Step 7: Calculate the percentage degree of dissociation
To find the percentage degree of dissociation:
\[ \text{Percentage degree of dissociation} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

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

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