Which aqueous will have the highest boiling point?

Options

A. 1% glucose in water
B. 1% sucrose in water
C. 1% NaCl in water
D. `1% CaCl_(2)` in water

Explanation

To determine which aqueous solution has the highest boiling point, we will analyze the given options based on the colligative properties of solutions, specifically the 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 (which is the same for all solutions since the solvent is water)
- \(m\) = molality of the solution

Since \(K_B\) is constant for all solutions, we can focus on maximizing the product \(i \cdot m\).

### Step-by-Step Solution:

1. **Identify the Options**:
- A: 1% Glucose in water
- B: 1% Sucrose in water
- C: 1% NaCl in water
- D: 1% CaCl2 in water

2. **Calculate the van 't Hoff Factor (\(i\)) for Each Solute**:
- **Glucose (C6H12O6)**: Does not dissociate, so \(i = 1\).
- **Sucrose (C12H22O11)**: Does not dissociate, so \(i = 1\).
- **NaCl**: Dissociates into Na\(^+\) and Cl\(^-\), so \(i = 2\).
- **CaCl2**: Dissociates into Ca\(^{2+}\) and 2 Cl\(^-\), so \(i = 3\).

3. **Calculate the Molar Mass of Each Solute**:
- **Glucose**: Molar mass = 180 g/mol
- **Sucrose**: Molar mass = 342 g/mol
- **NaCl**: Molar mass = 58.5 g/mol
- **CaCl2**: Molar mass = 110.98 g/mol (40.08 for Ca + 35.45 for Cl \(\times 2\))

4. **Calculate the Molality (\(m\))**:
- For a 1% solution, we assume 1 g of solute in 100 g of solution, which means:
- Mass of solvent = 100 g - 1 g = 99 g = 0.099 kg
- Moles of solute = mass of solute / molar mass
- Therefore, molality \(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)

- **Glucose**:
\[ m = \frac{1 \text{ g}}{180 \text{ g/mol}} \div 0.099 \text{ kg} \approx 0.056 \text{ mol/kg} \]

- **Sucrose**:
\[ m = \frac{1 \text{ g}}{342 \text{ g/mol}} \div 0.099 \text{ kg} \approx 0.030 \text{ mol/kg} \]

- **NaCl**:
\[ m = \frac{1 \text{ g}}{58.5 \text{ g/mol}} \div 0.099 \text{ kg} \approx 0.171 \text{ mol/kg} \]

- **CaCl2**:
\[ m = \frac{1 \text{ g}}{110.98 \text{ g/mol}} \div 0.099 \text{ kg} \approx 0.091 \text{ mol/kg} \]

5. **Calculate \(i \cdot m\) for Each Solution**:
- **Glucose**: \(i \cdot m = 1 \cdot 0.056 = 0.056\)
- **Sucrose**: \(i \cdot m = 1 \cdot 0.030 = 0.030\)
- **NaCl**: \(i \cdot m = 2 \cdot 0.171 = 0.342\)
- **CaCl2**: \(i \cdot m = 3 \cdot 0.091 = 0.273\)

6. **Determine the Highest Boiling Point**:
- The solution with the highest \(i \cdot m\) value will have the highest boiling point.
- From our calculations, NaCl has the highest value of \(0.342\).

### Conclusion:
The aqueous solution with the highest boiling point is **1% NaCl in water** (Option C).