`20`g of a binary electrolyte(mol.wt.=`100`)are dissolved in `500`g of water.The freezing point of the solution is `-0.74^(@)CK_(f)=1.86K "molality"^(-1)`.the degree of ionization of the electrolyte is

Options

A. `50%`
B. `75%`
C. `10%`
D. `0%`

Explanation

To find the degree of ionization of the binary electrolyte, we can follow these steps:

### Step 1: Calculate the change in freezing point (ΔTf)
The change in freezing point is given as:
\[ \Delta T_f = T_f^0 - T_f \]
Where \(T_f^0\) is the normal freezing point of the solvent (water, which is 0°C) and \(T_f\) is the freezing point of the solution (-0.74°C).
\[ \Delta T_f = 0 - (-0.74) = 0.74°C \]

**Hint:** Remember that the change in freezing point is the difference between the normal freezing point and the freezing point of the solution.

### Step 2: Use the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \(K_f\) is the cryoscopic constant (1.86 K kg/mol for water),
- \(m\) is the molality of the solution.

### Step 3: Calculate the molality (m)
First, we need to calculate the molality using the formula:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Given:
- Mass of solute (electrolyte) = 20 g
- Molecular weight of the electrolyte = 100 g/mol
- Mass of solvent (water) = 500 g = 0.5 kg

Calculate the moles of solute:
\[ \text{Moles of solute} = \frac{20 \, \text{g}}{100 \, \text{g/mol}} = 0.2 \, \text{mol} \]
Now, calculate the molality:
\[ m = \frac{0.2 \, \text{mol}}{0.5 \, \text{kg}} = 0.4 \, \text{mol/kg} \]

**Hint:** Make sure to convert the mass of the solvent from grams to kilograms when calculating molality.

### Step 4: Substitute values into the freezing point depression formula
Now substitute the values into the freezing point depression formula:
\[ 0.74 = 1.86 \cdot m \]
Substituting \(m = 0.4\):
\[ 0.74 = 1.86 \cdot 0.4 \]
Calculating the right side:
\[ 1.86 \cdot 0.4 = 0.744 \]
This confirms our calculations are consistent.

### Step 5: Calculate the degree of ionization (α)
The degree of ionization can be calculated using the formula:
\[ \frac{\text{Normal molecular weight}}{\text{Experimental molecular weight}} = 1 + \alpha \]
Where:
- Normal molecular weight = 100 g/mol (for the binary electrolyte),
- Experimental molecular weight can be calculated using the formula:
\[ \text{Experimental molecular weight} = \frac{M}{i} \]
Where \(i\) is the van 't Hoff factor, which is related to the degree of ionization:
\[ i = 1 + \alpha \]

From the freezing point depression, we can find \(i\):
\[ \Delta T_f = K_f \cdot m \cdot i \]
Substituting the known values:
\[ 0.74 = 1.86 \cdot 0.4 \cdot i \]
Solving for \(i\):
\[ i = \frac{0.74}{0.744} \approx 1 \]
Now substituting back into the degree of ionization formula:
\[ \frac{100}{100} = 1 + \alpha \]
This simplifies to:
\[ 1 = 1 + \alpha \implies \alpha = 0 \]

### Conclusion
The degree of ionization of the electrolyte is 0, which means it does not ionize at all in the solution.

**Final Answer:** The degree of ionization of the electrolyte is 0% (or α = 0).