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

Explanation

To find the boiling point of the solution, we need to calculate the elevation in boiling point using the given data. Let's go through the steps one by one.

### Step 1: Write down the given information
- Weight of glucose (\( W_{\text{solute}} \)) = 18 g
- Weight of water (\( W_{\text{solvent}} \)) = 1 kg = 1000 g
- \( K_b \) for water = 0.52 K kg mol\(^{-1}\)
- Molecular weight of glucose (\( M_{\text{solute}} \)) = 180 g/mol

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