The mole .weight of NaCl determined by studying freezing point depression of its `0.5%` aqueous solution is 30. The apparent degree dissociation of NaCl is

Options

A. `0.95`
B. `0.5`
C. `0.6`
D. `0.3`

Answer

To solve the problem of determining the apparent degree of dissociation of NaCl based on the given information, we can follow these steps:

### Step 1: Understand the given data
We are given:
- The observed molecular weight of NaCl from freezing point depression = 30 g/mol
- The actual molar mass of NaCl = 58.5 g/mol
- The concentration of the solution = 0.5%

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) is defined as the ratio of the actual molar mass to the observed molar mass. It can be calculated using the formula:

\[ i = \frac{\text{Actual molar mass}}{\text{Observed molar mass}} \]

Substituting the values:

\[ i = \frac{58.5 \, \text{g/mol}}{30 \, \text{g/mol}} = 1.95 \]

### Step 3: Relate the van 't Hoff factor to degree of dissociation
For NaCl, which dissociates into Na⁺ and Cl⁻ ions, the relationship between the van 't Hoff factor (i) and the degree of dissociation (α) is given by:

\[ i = 1 + \alpha \]

Where:
- i = van 't Hoff factor
- α = degree of dissociation

### Step 4: Solve for the degree of dissociation (α)
We can rearrange the equation to find α:

\[ \alpha = i - 1 \]

Substituting the value of i we calculated:

\[ \alpha = 1.95 - 1 = 0.95 \]

### Step 5: Conclusion
The apparent degree of dissociation of NaCl is 0.95, which means that 95% of NaCl is dissociated in the solution.

### Final Answer
The apparent degree of dissociation of NaCl is **0.95**.

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