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 step by step, we will use the formula for freezing point depression and the concept of dissociation of electrolytes.

### Step 1: Understand the Freezing Point Depression Formula
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) = change in freezing point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = cryoscopic constant of the solvent (for water, \( K_f = 1.86 \, ^\circ C \))
- \( m \) = molality of the solution

### Step 2: Calculate the Change in Freezing Point
The freezing point of pure water is \( 0 \, ^\circ C \) and the freezing point of the solution is given as \( -0.335 \, ^\circ C \). Therefore, the change in freezing point is:
\[ \Delta T_f = 0 - (-0.335) = 0.335 \, ^\circ C \]

### Step 3: Substitute Values into the Freezing Point Depression Formula
We know:
- \( \Delta T_f = 0.335 \, ^\circ C \)
- \( K_f = 1.86 \, ^\circ C \)
- \( m = 0.1 \, \text{mol/kg} \)

Substituting these values into the formula:
\[ 0.335 = i \cdot 1.86 \cdot 0.1 \]

### Step 4: Solve for the Van 't Hoff Factor \( i \)
Rearranging the equation to solve for \( i \):
\[ i = \frac{0.335}{1.86 \cdot 0.1} = \frac{0.335}{0.186} \approx 1.80 \]

### Step 5: Determine the Percentage of Dissociation
For NaBr, which dissociates into two ions (Na\(^+\) and Br\(^-\)), the van 't Hoff factor \( i \) can be expressed as:
\[ i = 1 + \alpha \]
Where \( \alpha \) is the degree of dissociation. For NaBr:
\[ n = 2 \quad (\text{since it dissociates into 2 ions}) \]
Thus, we can use:
\[ \alpha = \frac{i - 1}{n - 1} \]
Substituting the values:
\[ \alpha = \frac{1.80 - 1}{2 - 1} = 0.80 \]

### Step 6: Calculate the Percentage of Dissociation
To find the percentage of dissociation:
\[ \text{Percentage of dissociation} = \alpha \times 100 = 0.80 \times 100 = 80\% \]

### Final Answer
The percentage of dissociation of NaBr in the solution is **80%**.

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