`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 water when 18 g of glucose is dissolved in 1 kg of water, we will use the formula for boiling point elevation:

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

Where:
- \(\Delta T_b\) = boiling point elevation
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