An aqueous solution freezes at
`-0.186^(@)C (K_(f)=1.86^(@)` ,`K_(b)=0.512^(@)`. What is the elevation in boiling point?

Options

A. 0.186
B. 0.512
C. `0.512/1.86`
D. 0.0512

Explanation

To find the elevation in boiling point (ΔTb) of the solution, we can use the relationship between freezing point depression (ΔTf) and boiling point elevation (ΔTb) based on the molality of the solution.

### Step-by-Step Solution:

1. **Identify the given values:**
- Freezing point depression (ΔTf) = -0.186°C
- Freezing point depression constant (Kf) = 1.86°C kg/mol
- Boiling point elevation constant (Kb) = 0.512°C kg/mol

2. **Calculate the molality (m) of the solution using ΔTf:**
The formula for freezing point depression is:
\[ \Delta T_f = K_f \times m \]
Rearranging this gives:
\[ m = \frac{\Delta T_f}{K_f} \]
Substituting the values:
\[ m = \frac{-0.186}{1.86} = -0.1 \text{ mol/kg} \]

3. **Calculate the elevation in boiling point (ΔTb):**
The formula for boiling point elevation is:
\[ \Delta T_b = K_b \times m \]
Substituting the values we have:
\[ \Delta T_b = 0.512 \times (-0.1) = -0.0512 \text{ °C} \]

4. **Final result:**
The elevation in boiling point is:
\[ \Delta T_b = 0.0512 \text{ °C} \]

### Summary:
The elevation in boiling point (ΔTb) for the given solution is 0.0512 °C.