An aqueous solution at `-2.55^(@)C`. What is its boiling point `(K_(b)^(H_(2)O)=0.52 K m^(-1)`,`K_(f)^(H_(2)O)=1.86 K m^(-1)`?

Options

A. `107.0^(@)C`
B. `100.6^(@)C`
C. `100.1^(@)C`
D. `100.7^(@)C`

Explanation

To find the boiling point of the aqueous solution given its freezing point, we can use the formulas for freezing point depression and boiling point elevation. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Freezing point of the solution, \( T_f = -2.55^\circ C \)
- Boiling point of pure water, \( T_b = 100^\circ C \)
- Freezing point depression constant, \( K_f^{H_2O} = 1.86 \, K \, m^{-1} \)
- Boiling point elevation constant, \( K_b^{H_2O} = 0.52 \, K \, m^{-1} \)

### Step 2: Calculate the freezing point depression (\( \Delta T_f \))
The freezing point depression is calculated as:
\[ \Delta T_f = 0 - T_f = 0 - (-2.55) = 2.55^\circ C \]

### Step 3: Relate freezing point depression to molality
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \cdot m \]
Where \( m \) is the molality of the solution. Rearranging gives:
\[ m = \frac{\Delta T_f}{K_f} = \frac{2.55}{1.86} \approx 1.3704 \, m \]

### Step 4: Calculate the boiling point elevation (\( \Delta T_b \))
Using the formula for boiling point elevation:
\[ \Delta T_b = K_b \cdot m \]
Substituting the values we found:
\[ \Delta T_b = K_b \cdot m = 0.52 \cdot 1.3704 \approx 0.7129^\circ C \]

### Step 5: Calculate the new boiling point
The boiling point of the solution is given by:
\[ T_b' = T_b + \Delta T_b = 100 + 0.7129 \approx 100.7129^\circ C \]

### Final Answer
The boiling point of the aqueous solution is approximately \( 100.71^\circ C \).

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