The boiling point of an aqueous solution of a non-volatile solute is `100.15^(@)C`. What is the freezing point of an aqueous solution obtained by dilute the above solution with an equal volume of water. The values of `K_(b)` and `K_(f)` for water are `0.512` and `1.86^(@)C mol^(-1)`:

Options

A. `-0.544^(@)C`
B. `-0.512^(@)C`
C. `-0.272^(@)C`
D. `-1.86^(@)C`

Explanation

To solve the problem, we need to determine the freezing point of the diluted solution after we have the boiling point elevation of the original solution. Here’s a step-by-step solution:

### Step 1: Determine the boiling point elevation
The boiling point of the solution is given as \(100.15^\circ C\). The normal boiling point of water is \(100^\circ C\). Therefore, the boiling point elevation (\(\Delta T_b\)) can be calculated as:

\[ \Delta T_b = T_b - T_{b0} = 100.15^\circ C - 100^\circ C = 0.15^\circ C \]

### Step 2: Calculate the molality of the solution
Using the formula for boiling point elevation:

\[ \Delta T_b = K_b \cdot m \]

where:
- \(\Delta T_b = 0.15^\circ C\)
- \(K_b = 0.512^\circ C \cdot \text{kg/mol}\)

We can rearrange the formula to find the molality (\(m\)):

\[ m = \frac{\Delta T_b}{K_b} = \frac{0.15^\circ C}{0.512^\circ C \cdot \text{kg/mol}} \approx 0.29297 \text{ mol/kg} \]

### Step 3: Determine the effect of dilution
When we dilute the solution with an equal volume of water, the molality will be halved because the amount of solute remains the same while the volume of the solvent doubles.

Thus, the new molality (\(m'\)) after dilution is:

\[ m' = \frac{m}{2} = \frac{0.29297}{2} \approx 0.146485 \text{ mol/kg} \]

### Step 4: Calculate the freezing point depression
Using the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m' \]

where:
- \(K_f = 1.86^\circ C \cdot \text{kg/mol}\)

Now we can calculate \(\Delta T_f\):

\[ \Delta T_f = 1.86^\circ C \cdot \text{kg/mol} \cdot 0.146485 \text{ mol/kg} \approx 0.272 \text{°C} \]

### Step 5: Determine the new freezing point
The normal freezing point of water is \(0^\circ C\). The freezing point of the solution will be lowered by \(\Delta T_f\):

\[ \text{Freezing Point} = 0^\circ C - \Delta T_f = 0^\circ C - 0.272^\circ C \approx -0.272^\circ C \]

### Final Answer
The freezing point of the diluted aqueous solution is approximately \(-0.272^\circ C\).

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