boiling point elevation

Question

The boiling point elevation constant (Kb) for water is 0.52 K kg mol⁻¹. What is the boiling point of a solution containing 18 g of glucose (C₆H₁₂O₆) in 100 g of water?

RankGuru · Accepted Answer

The elevation in boiling point is given by:
$$\Delta T_b = K_b 	imes m$$
1. **Molality ($m$)**: Moles of Glucose $= \frac{18}{180} = 0.1 	ext{ mol}$.
   - Mass of solvent $= 100 	ext{ g} = 0.1 	ext{ kg}$.
   - $m = \frac{0.1 	ext{ mol}}{0.1 	ext{ kg}} = 1.0 	ext{ molal}$.
2. **Elevation**: $\Delta T_b = 0.52 	imes 1.0 = 0.52 	ext{ K}$.
3. **Boiling Point**: $T_b = T^0_b + \Delta T_b = 100^\circ C + 0.52 = 100.52^\circ C$.

*Note: The previous solution had a calculation error in the explanation text suggesting $100.052$.*

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