0.1mole aqueous solution of NaBr freezes at -`0.335^@C` at atmospheric pressure , `k_f` for water is `1.86^@C` . The percentage of dissociation of the salt in solution is

Options

A. 90
B. 80
C. 58
D. 98

Answer

To solve the problem, we need to calculate the percentage of dissociation of NaBr in the aqueous solution given the freezing point depression. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Molarity of NaBr solution (m) = 0.1 moles/kg
- Freezing point of the solution (Tf) = -0.335 °C
- Freezing point depression constant for water (Kf) = 1.86 °C kg/mol

### Step 2: Calculate the depression in freezing point (ΔTf)
The depression in freezing point (ΔTf) can be calculated as:
\[ \Delta T_f = T_f^{\text{pure solvent}} - T_f^{\text{solution}} = 0 - (-0.335) = 0.335 \, °C \]

### Step 3: Write the formula for freezing point depression
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = freezing point depression constant
- \( m \) = molality of the solution

### Step 4: Determine the van 't Hoff factor (i)
For NaBr, which dissociates into Na⁺ and Br⁻ ions:
\[ \text{NaBr} \rightarrow \text{Na}^+ + \text{Br}^- \]
If α is the degree of dissociation, then:
- The amount of NaBr that dissociates = α
- Remaining NaBr = 1 - α
- Total particles = \( (1 - \alpha) + \alpha + \alpha = 1 + \alpha \)

Thus, the van 't Hoff factor \( i \) is:
\[ i = 1 + \alpha \]

### Step 5: Substitute values into the freezing point depression formula
Substituting the known values into the formula:
\[ 0.335 = (1 + \alpha) \cdot 1.86 \cdot 0.1 \]

### Step 6: Simplify and solve for α
\[ 0.335 = (1 + \alpha) \cdot 0.186 \]
\[ 1 + \alpha = \frac{0.335}{0.186} \approx 1.801 \]
\[ \alpha = 1.801 - 1 = 0.801 \]

### Step 7: Calculate the percentage of dissociation
The percentage of dissociation is given by:
\[ \text{Percentage of dissociation} = \alpha \times 100 = 0.801 \times 100 = 80.1\% \]

### Final Answer
The percentage of dissociation of NaBr in the solution is **80.1%**.

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