A compound `MX_(2)`has observed and normal molar masses `65.6 and 164`respectively .Calculated the apparent degree of ionization of `MX_(2)`:

Options

A. `75%`
B. `85%`
C. `65%`
D. `25%`

Answer

To solve the problem of calculating the apparent degree of ionization of the compound \( MX_2 \), we will follow these steps:

### Step 1: Identify the given data
- Observed molar mass (\( M_{obs} \)) = 65.6 g/mol
- Normal molar mass (\( M_{normal} \)) = 164 g/mol

### Step 2: Calculate the Van't Hoff factor (\( i \))
The Van't Hoff factor can be calculated using the formula:
\[ i = \frac{M_{normal}}{M_{obs}} \]
Substituting the values:
\[ i = \frac{164}{65.6} \approx 2.5 \]

### Step 3: Determine the number of ions produced (n)
For the compound \( MX_2 \), when it ionizes, it produces:
- 1 ion of \( M^{2+} \)
- 2 ions of \( X^{-} \)

Thus, the total number of ions (\( n \)) produced is:
\[ n = 1 + 2 = 3 \]

### Step 4: Relate the Van't Hoff factor to the degree of ionization (\( \alpha \))
The relationship between the Van't Hoff factor and the degree of ionization is given by:
\[ i = 1 + \alpha(n - 1) \]
Rearranging this equation to solve for \( \alpha \):
\[ \alpha = \frac{i - 1}{n - 1} \]

### Step 5: Substitute the values into the equation
Now substituting the values of \( i \) and \( n \):
\[ \alpha = \frac{2.5 - 1}{3 - 1} = \frac{1.5}{2} = 0.75 \]

### Step 6: Convert the degree of ionization to percentage
To express the degree of ionization as a percentage, multiply by 100:
\[ \text{Degree of ionization} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

### Final Answer
The apparent degree of ionization of \( MX_2 \) is **75%**.

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