Which solute produces the highest boiling point in a `0.15` m aqueous solution ?

Options

A. `CaCl_(2)`
B. NaBr
C. `CuSO_(4)`
D. `CH_(3)OH`

Answer

To determine which solute produces the highest boiling point in a 0.15 molal aqueous solution, we will use the concept of boiling point elevation. The boiling point elevation can be calculated using the formula:

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

Where:
- \(\Delta T_b\) = boiling point elevation
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_b\) = ebullioscopic constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 1: Identify the van 't Hoff factor (i) for each solute
1. **CaCl₂**: Dissociates into 3 ions (1 Ca²⁺ and 2 Cl⁻), so \(i = 3\).
2. **NaBr**: Dissociates into 2 ions (1 Na⁺ and 1 Br⁻), so \(i = 2\).
3. **CuSO₄**: Dissociates into 2 ions (1 Cu²⁺ and 1 SO₄²⁻), so \(i = 2\).
4. **CH₃OH**: Does not dissociate (non-electrolyte), so \(i = 1\).

### Step 2: Calculate the effective molality for each solute
Since all solutions are 0.15 m, we can calculate the effective molality by multiplying the molality by the van 't Hoff factor:

- For **CaCl₂**:
\[ \text{Effective molality} = 0.15 \, \text{m} \times 3 = 0.45 \, \text{m} \]

- For **NaBr**:
\[ \text{Effective molality} = 0.15 \, \text{m} \times 2 = 0.30 \, \text{m} \]

- For **CuSO₄**:
\[ \text{Effective molality} = 0.15 \, \text{m} \times 2 = 0.30 \, \text{m} \]

- For **CH₃OH**:
\[ \text{Effective molality} = 0.15 \, \text{m} \times 1 = 0.15 \, \text{m} \]

### Step 3: Compare the effective molalities
Now we compare the effective molalities:

- CaCl₂: 0.45 m
- NaBr: 0.30 m
- CuSO₄: 0.30 m
- CH₃OH: 0.15 m

### Conclusion
The solute that produces the highest boiling point in a 0.15 m aqueous solution is **CaCl₂**, as it has the highest effective molality due to its van 't Hoff factor of 3.

### Final Answer
**CaCl₂ produces the highest boiling point in a 0.15 m aqueous solution.**

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