58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point (bp) of the resulting solutions.

Options

A. NaCl solution will show higher elevation of b.p.
B. Glucose solution will show higher elevation of b.p.
C. Both the solution will show equal elevation of b.p.
D. The b.p. elevation will be shown by either of the solutions.

Answer

To solve the problem of determining the elevation of boiling point for the solutions of NaCl and glucose, we will follow these steps:

### Step 1: Understand the Concept of Boiling Point Elevation
The elevation of boiling point (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = K_b \cdot m \cdot i \]
where:
- \( K_b \) is the ebullioscopic constant of the solvent (water in this case),
- \( m \) is the molality of the solution,
- \( i \) is the van 't Hoff factor (number of particles the solute breaks into).

### Step 2: Calculate the Molality of NaCl
1. **Find the molecular weight of NaCl**:
- Sodium (Na) = 23 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molecular weight of NaCl = 23 + 35.5 = 58.5 g/mol

2. **Calculate the molality of NaCl**:
- Mass of NaCl = 58.5 g
- Mass of water = 1000 g (since 1000 mL of water has a mass of 1000 g)
- Convert mass of water to kg: \( 1000 \, \text{g} = 1 \, \text{kg} \)
- Molality (\( m \)) = \(\frac{\text{mass of solute (g)}}{\text{molecular weight (g/mol)} \times \text{mass of solvent (kg)}}\)
\[ m_{NaCl} = \frac{58.5}{58.5 \times 1} = 1 \, \text{mol/kg} \]

### Step 3: Calculate the Molality of Glucose
1. **Find the molecular weight of glucose (C6H12O6)**:
- Carbon (C) = 12 g/mol × 6 = 72 g/mol
- Hydrogen (H) = 1 g/mol × 12 = 12 g/mol
- Oxygen (O) = 16 g/mol × 6 = 96 g/mol
- Molecular weight of glucose = 72 + 12 + 96 = 180 g/mol

2. **Calculate the molality of glucose**:
- Mass of glucose = 180 g
- Mass of water = 1000 g = 1 kg
- Molality (\( m \)) = \(\frac{180}{180 \times 1} = 1 \, \text{mol/kg}\)

### Step 4: Determine the Van 't Hoff Factor
1. **For NaCl**:
- NaCl dissociates into 2 ions: Na⁺ and Cl⁻.
- Therefore, \( i_{NaCl} = 2 \).

2. **For Glucose**:
- Glucose does not dissociate in solution.
- Therefore, \( i_{glucose} = 1 \).

### Step 5: Calculate the Elevation of Boiling Point for Both Solutions
1. **For NaCl**:
\[ \Delta T_{b, NaCl} = K_b \cdot m_{NaCl} \cdot i_{NaCl} = 0.52 \cdot 1 \cdot 2 = 1.04 \, \text{°C} \]

2. **For Glucose**:
\[ \Delta T_{b, glucose} = K_b \cdot m_{glucose} \cdot i_{glucose} = 0.52 \cdot 1 \cdot 1 = 0.52 \, \text{°C} \]

### Step 6: Compare the Elevation of Boiling Points
- The elevation of boiling point for NaCl is 1.04 °C.
- The elevation of boiling point for glucose is 0.52 °C.
- Therefore, the NaCl solution shows a higher elevation in boiling point than the glucose solution.

### Conclusion
The correct statement regarding the elevation of boiling point of the resulting solutions is:
- **NaCl solution will show higher elevation in boiling point.**