Osmotic Pressure of 0.1 M aqueous solution of `MgCl_(2)`at 300k is 4.92 atm. What will be the percentage ionization of the salt ?

Options

A. `79%`
B. `59%`
C. `49%`
D. `69%`

Explanation

To solve the problem, we will follow a systematic approach using the formula for osmotic pressure and the concept of ionization of the solute.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Molarity (C) of the solution = 0.1 M
- Osmotic Pressure (π) = 4.92 atm
- Temperature (T) = 300 K
- The solute is magnesium chloride (MgCl₂).

2. **Identify the Formula for Osmotic Pressure**:
The formula for osmotic pressure is given by:
\[ \pi = iCRT \]
where:
- π = osmotic pressure
- i = Van't Hoff factor
- C = concentration of the solution
- R = universal gas constant (0.0821 L·atm/(K·mol))
- T = temperature in Kelvin

3. **Calculate the Van't Hoff Factor (i)**:
Magnesium chloride dissociates in water as follows:
\[ \text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^- \]
This means that for every 1 mole of MgCl₂, we get 3 moles of ions (1 Mg²⁺ and 2 Cl⁻). Hence, if α is the degree of ionization, the total number of particles after dissociation can be expressed as:
\[ n_f = 1 - \alpha + \alpha + 2\alpha = 1 + 2\alpha \]
The initial number of particles (before dissociation) is:
\[ n_i = 1 \]
Therefore, the Van't Hoff factor (i) is:
\[ i = \frac{n_f}{n_i} = 1 + 2\alpha \]

4. **Substituting Values into the Osmotic Pressure Equation**:
Substitute the known values into the osmotic pressure equation:
\[ 4.92 = (1 + 2\alpha)(0.1)(0.0821)(300) \]

5. **Simplifying the Equation**:
Calculate the right-hand side:
\[ 0.1 \times 0.0821 \times 300 = 2.463 \]
Thus, the equation becomes:
\[ 4.92 = (1 + 2\alpha) \times 2.463 \]

6. **Solving for α**:
Rearranging gives:
\[ 1 + 2\alpha = \frac{4.92}{2.463} \approx 1.996 \]
Therefore:
\[ 2\alpha = 1.996 - 1 = 0.996 \]
\[ \alpha = \frac{0.996}{2} \approx 0.498 \]

7. **Calculating Percentage Ionization**:
To find the percentage ionization, multiply α by 100:
\[ \text{Percentage Ionization} = \alpha \times 100 \approx 0.498 \times 100 \approx 49.8\% \]

8. **Final Answer**:
The percentage ionization of the salt (MgCl₂) is approximately **49.8%**.