The boiling point of pure water is 373 K. calculate the boiling point of an aqueous solution containing 18 g of glucose (M.W. = 180) in 100g of water. Molal elevation constant of water is 0.52 K kg `"mol"^(-1)`.

Explanation

To solve the problem, we need to calculate the boiling point elevation of the aqueous solution containing glucose. Here are the steps to find the solution:

### Step 1: Identify the given data
- Boiling point of pure water, \( T_b^0 = 373 \, \text{K} \)
- Mass of glucose (solute), \( m_{solute} = 18 \, \text{g} \)
- Molar mass of glucose, \( M_{glucose} = 180 \, \text{g/mol} \)
- Mass of water (solvent), \( m_{solvent} = 100 \, \text{g} \)
- Molal elevation constant of water, \( K_b = 0.52 \, \text{K kg/mol} \)
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