The freezing point of 0.08 molal aq `NaHSO_(4)` solution is `-0.372`. Calculate `alpha` (degree of dessociation) of `HSO_(4)^(-)` given `alpha` (degree of dissociation) of `NaHSO_(4)` is `100%` & `K_(f)` for `H_(2)O=1.86` `(K)/(m)`. Give answer after multiplying by 10.

Answer

To solve the problem, we need to calculate the degree of dissociation (α) of the bisulfate ion (HSO₄⁻) in a 0.08 molal aqueous solution of sodium bisulfate (NaHSO₄). We are given that the freezing point depression is -0.372°C and the cryoscopic constant (K_f) of water is 1.86 K·kg/mol.

### Step-by-Step Solution:

1. **Identify the Dissociation of NaHSO₄:**
Sodium bisulfate (NaHSO₄) dissociates in water as follows:
\[ \text{NaHSO}_4 \rightarrow \text{Na}^+ + \text{HSO}_4^- \]
The bisulfate ion (HSO₄⁻) can further dissociate:
\[ \text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-} \]

2. **Define the Degree of Dissociation (α):**
Let α be the degree of dissociation of HSO₄⁻. After dissociation:
- The concentration of Na⁺ ions = 1 (from NaHSO₄)
- The concentration of HSO₄⁻ ions = 1 - α
- The concentration of H⁺ ions = α
- The concentration of SO₄²⁻ ions = α

3. **Calculate the Total Number of Ions (i):**
The total number of ions after dissociation is:
\[ i = 1 + (1 - α) + α + α = 2 + α \]

4. **Use the Freezing Point Depression Formula:**
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(\Delta T_f = 0 - (-0.372) = 0.372\)
- \(K_f = 1.86 \, \text{K·kg/mol}\)
- \(m = 0.08 \, \text{mol/kg}\)

5. **Plug in the Values:**
Substitute the known values into the freezing point depression equation:
\[ 0.372 = (2 + α) \cdot 1.86 \cdot 0.08 \]

6. **Solve for α:**
Rearranging the equation:
\[ 0.372 = (2 + α) \cdot 0.1488 \]
\[ 2 + α = \frac{0.372}{0.1488} \approx 2.5 \]
\[ α = 2.5 - 2 = 0.5 \]

7. **Multiply α by 10:**
The final step is to multiply α by 10:
\[ 10α = 10 \times 0.5 = 5 \]

### Final Answer:
The answer after multiplying by 10 is **5**.