18 g of glucose, `C_(2)H_(12)O_(6)` is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil at (1.013 bar pressure). `K_(b)` for water is 0.52 K kg `"mol"^(-1)`.

Explanation

To solve the problem of determining the boiling point of water when 18 g of glucose is dissolved in 1 kg of water, we will follow these steps:

### Step 1: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. **Calculate the number of moles of glucose (C₆H₁₂O₆)**:
- Molecular weight of glucose = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
- Moles of glucose = mass of glucose / molecular weight = 18 g / 180 g/mol = 0.1 mol

2. **Calculate the molality**:
- Weight of solvent (water) = 1 kg
- Molality (m) = moles of solute / kg of solvent = 0.1 mol / 1 kg = 0.1 mol/kg

### Step 2: Calculate the elevation in boiling point (ΔTb)
The elevation in boiling point can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
Where:
- \( K_b \) = boiling point elevation constant for water = 0.52 K kg/mol
- \( m \) = molality of the solution = 0.1 mol/kg

Now, substituting the values:
\[ \Delta T_b = 0.52 \, \text{K kg/mol} \times 0.1 \, \text{mol/kg} = 0.052 \, \text{K} \]

### Step 3: Calculate the boiling point of the solution
The boiling point of pure water at 1.013 bar pressure is approximately 373.15 K (or 100 °C).

Now, we can find the boiling point of the solution:
\[ T_b = T_{b,\text{pure}} + \Delta T_b \]
\[ T_b = 373.15 \, \text{K} + 0.052 \, \text{K} = 373.202 \, \text{K} \]

### Final Answer
The boiling point of the solution is approximately **373.202 K**.

---