Which aqueous solution would process the lowest boiling point?

Options

A. `1%NaCl`solution
B. 1 % urea solution
C. 1% Glucose solution
D. 1 % sucrose solution

Explanation

To determine which aqueous solution has the lowest boiling point, we need to calculate the boiling point elevation (ΔTb) for each solution. The boiling point elevation can be calculated using the formula:

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

Where:
- \( \Delta T_b \) = elevation in boiling point
- \( K_b \) = ebullioscopic constant (a property of the solvent, which is constant for all solutions in this case)
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( m \) = molality of the solution (moles of solute per kg of solvent)

### Step-by-Step Solution:

1. **Identify the Solutions**:
Let's assume we have four different solutes in aqueous solutions: urea, glucose, sucrose, and a salt (like NaCl).

2. **Determine the Van 't Hoff Factor (i)**:
- Urea (non-electrolyte): \( i = 1 \)
- Glucose (non-electrolyte): \( i = 1 \)
- Sucrose (non-electrolyte): \( i = 1 \)
- NaCl (electrolyte): \( i = 2 \) (dissociates into Na⁺ and Cl⁻)

3. **Calculate the Molar Mass of Each Solute**:
- Urea: 60 g/mol
- Glucose: 180 g/mol
- Sucrose: 342 g/mol
- NaCl: 58.5 g/mol

4. **Assume a Constant Weight of Solvent**:
For simplicity, assume we are using 1 kg of water as the solvent for all solutions.

5. **Calculate the Molality (m)**:
Using a 1% solution:
- For 1 kg of water, 1% means 10 g of solute.
- Moles of solute = weight (g) / molar mass (g/mol)

- Urea:
\[ m = \frac{10 \text{ g}}{60 \text{ g/mol}} = 0.1667 \text{ mol/kg} \]
- Glucose:
\[ m = \frac{10 \text{ g}}{180 \text{ g/mol}} = 0.0556 \text{ mol/kg} \]
- Sucrose:
\[ m = \frac{10 \text{ g}}{342 \text{ g/mol}} = 0.0292 \text{ mol/kg} \]
- NaCl:
\[ m = \frac{10 \text{ g}}{58.5 \text{ g/mol}} = 0.1712 \text{ mol/kg} \]

6. **Calculate ΔTb for Each Solution**:
Since \( K_b \) is constant, we can compare the products \( i \cdot m \):
- Urea: \( \Delta T_b = K_b \cdot 1 \cdot 0.1667 \)
- Glucose: \( \Delta T_b = K_b \cdot 1 \cdot 0.0556 \)
- Sucrose: \( \Delta T_b = K_b \cdot 1 \cdot 0.0292 \)
- NaCl: \( \Delta T_b = K_b \cdot 2 \cdot 0.1712 \)

7. **Compare the Elevations**:
- Urea: \( 0.1667 K_b \)
- Glucose: \( 0.0556 K_b \)
- Sucrose: \( 0.0292 K_b \)
- NaCl: \( 0.3424 K_b \)

8. **Determine the Lowest Boiling Point**:
The solution with the lowest boiling point will be the one with the smallest ΔTb. Thus, the order from lowest to highest boiling point is:
- Sucrose < Glucose < Urea < NaCl

### Conclusion:
The aqueous solution of sucrose will have the lowest boiling point.