`0.96 g` of HI were, heated to attain equilibrium `2HI hArr H_(2)+I_(2)`. The reaction mixture on titration requires `15.7 mL` of `N//10` hypo solution. Calculate the degree of dissociation of `HI`.

Explanation

To calculate the degree of dissociation of HI in the reaction \(2HI \rightleftharpoons H_2 + I_2\), we will follow these steps:

### Step 1: Calculate the number of moles of HI
Given:
- Mass of HI = 0.96 g
- Molar mass of HI = 128 g/mol

Using the formula for number of moles:
\[ \text{Number of moles of HI} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.96 \, \text{g}}{128 \, \text{g/mol}} = 0.0075 \, \text{mol} \]

### Step 2: Set up the equilibrium expression
The reaction is:
\[ 2HI \rightleftharpoons H_2 + I_2 \]
Let \(x\) be the amount of HI that dissociates at equilibrium. The initial and equilibrium concentrations will be:
- Initial moles of HI = \(0.0075\)
- At equilibrium:
- Moles of HI = \(0.0075 - x\)
- Moles of \(H_2\) = \(x/2\)
- Moles of \(I_2\) = \(x/2\)

### Step 3: Relate the amount of \(I_2\) formed to the titration data
From the titration, we know that \(15.7 \, \text{mL}\) of \(N/10\) hypo solution was used. The equivalent weight of \(I_2\) is \(127\) (for iodine) and the moles of \(I_2\) formed can be calculated as follows:
\[ \text{mEq of } I_2 = \text{Volume (mL)} \times \text{Normality} = 15.7 \times \frac{1}{10} = 1.57 \, \text{mEq} \]
Since \(1 \, \text{mEq} = \frac{1 \, \text{mol}}{2}\) for \(I_2\):
\[ \text{Moles of } I_2 = \frac{1.57 \, \text{mEq}}{2} = 0.000785 \, \text{mol} \]

### Step 4: Relate \(I_2\) formed to \(x\)
From the equilibrium expression:
\[ \frac{x}{2} = 0.000785 \implies x = 0.00157 \, \text{mol} \]

### Step 5: Calculate the degree of dissociation \(\alpha\)
The degree of dissociation \(\alpha\) is given by:
\[ \alpha = \frac{\text{moles dissociated}}{\text{initial moles}} = \frac{x}{0.0075} \]
Substituting the value of \(x\):
\[ \alpha = \frac{0.00157}{0.0075} \approx 0.209 \text{ or } 20.9\% \]

### Final Answer
The degree of dissociation of HI is approximately \(20.9\%\).

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