Problem 1: Boiling Point Elevation with Glucose

Given: 18 g glucose (M = 180 g/mol) dissolved in 500 g water. Find the boiling point. (Kb = 0.52 K·kg/mol)

Step 1: Find moles of solute. $n_{glucose} = \frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ mol}$

Step 2: Convert solvent mass to kg. $w_1 = 500 \text{ g} = 0.5 \text{ kg}$

Step 3: Calculate molality. $m = \frac{n_{solute}}{w_{solvent}(\text{kg})} = \frac{0.1}{0.5} = 0.2 \text{ mol/kg}$

Step 4: Apply $\Delta Tb$ formula (glucose is non-electrolyte, i = 1). $\Delta T_b = K_b \times m = 0.52 \times 0.2 = 0.104 \text{ K}$

Answer: Boiling point = 100 + 0.104 = 100.104°C