If the observed and theoretical molecular mass of `NaCI` is found to be `31.80` and `58.50`, then the degree of dissociation of `NaCI` is:

Options

A. `83.96%`
B. `8.39%`
C. `90%`
D. `100%`

Explanation

To find the degree of dissociation of NaCl given the observed and theoretical molecular masses, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Theoretical molecular mass of NaCl = 58.50 g/mol
- Observed molecular mass of NaCl = 31.80 g/mol

2. **Calculate the Van't Hoff factor (i):**
The Van't Hoff factor (i) can be calculated using the formula:
\[ i = \frac{\text{Theoretical molecular mass}}{\text{Observed molecular mass}} \]
Substituting the given values:
\[ i = \frac{58.50}{31.80} \]

3. **Perform the calculation:**
\[ i = 1.838 \]

4. **Relate the Van't Hoff factor to the degree of dissociation (α):**
For NaCl, which dissociates into two ions (Na⁺ and Cl⁻), the relationship is given by:
\[ i = 1 + (n - 1) \cdot \alpha \]
where \( n \) is the number of ions produced upon dissociation. For NaCl, \( n = 2 \):
\[ i = 1 + (2 - 1) \cdot \alpha \]
Simplifying this gives:
\[ i = 1 + \alpha \]

5. **Substitute the value of i:**
\[ 1.838 = 1 + \alpha \]

6. **Solve for α:**
\[ \alpha = 1.838 - 1 = 0.838 \]

7. **Convert α to percentage:**
To express the degree of dissociation as a percentage:
\[ \alpha \text{ (in percentage)} = 0.838 \times 100 = 83.8\% \]

### Final Answer:
The degree of dissociation of NaCl is approximately **83.8%**.

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