What will be the degree of dissociation of 0.1 M `Mg(NO_3)_2` solution if van't Hoff factor is 2.74 ?

Options

A. `75%`
B. `87%`
C. `100%`
D. `92%`

Explanation

To find the degree of dissociation (α) of a 0.1 M solution of magnesium nitrate (Mg(NO₃)₂) given that the van't Hoff factor (i) is 2.74, we can follow these steps:

### Step 1: Understand the dissociation of magnesium nitrate
Magnesium nitrate dissociates in solution as follows:
\[ \text{Mg(NO}_3\text{)}_2 \rightarrow \text{Mg}^{2+} + 2 \text{NO}_3^{-} \]
This means that one mole of magnesium nitrate produces one mole of magnesium ions and two moles of nitrate ions.

### Step 2: Set up the initial and final moles
Let:
- \( n \) = initial moles of Mg(NO₃)₂ = 0.1 M
- \( \alpha \) = degree of dissociation

Initially, we have:
- Moles of Mg(NO₃)₂ = \( n \)
- Moles of Mg²⁺ = 0
- Moles of NO₃⁻ = 0

After dissociation:
- Moles of Mg(NO₃)₂ remaining = \( n(1 - \alpha) \)
- Moles of Mg²⁺ produced = \( n\alpha \)
- Moles of NO₃⁻ produced = \( 2n\alpha \)

### Step 3: Calculate the total number of particles after dissociation
The total number of particles after dissociation can be calculated as:
\[ \text{Total particles} = \text{Moles of Mg(NO}_3\text{)}_2 + \text{Moles of Mg}^{2+} + \text{Moles of NO}_3^{-} \]
\[ = n(1 - \alpha) + n\alpha + 2n\alpha \]
\[ = n(1 - \alpha + \alpha + 2\alpha) \]
\[ = n(1 + 2\alpha) \]

### Step 4: Relate van't Hoff factor to the degree of dissociation
The van't Hoff factor \( i \) is defined as the ratio of the total number of particles in solution after dissociation to the number of formula units initially present:
\[ i = \frac{\text{Total particles}}{\text{Initial particles}} = \frac{n(1 + 2\alpha)}{n} = 1 + 2\alpha \]

Given that \( i = 2.74 \):
\[ 1 + 2\alpha = 2.74 \]

### Step 5: Solve for α
Now, we can solve for α:
\[ 2\alpha = 2.74 - 1 \]
\[ 2\alpha = 1.74 \]
\[ \alpha = \frac{1.74}{2} \]
\[ \alpha = 0.87 \]

### Step 6: Convert to percentage
To convert the degree of dissociation into percentage:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.87 \times 100 = 87\% \]

### Final Answer
The degree of dissociation of the 0.1 M Mg(NO₃)₂ solution is 0.87 or 87%.

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