What is the boiling point of a solution of 0.1 mole of glucose in 200 mL of water? (K = 0.512C/m)

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BlackTom AI · Accepted Answer

To tackle boiling point elevation, I’ll use the formula ΔTb = i Kb m.
Option 1: 100.06°C. This would imply ΔTb = 0.06°C, which would correspond to a very small molality. Given the setup (0.1 mol in 0.200 kg solvent), the calculated molality is 0.5 m, leading to a larger shift than 0.06°C. So this value is too low.
Option 2: 100.13°C. Here ΔTb ≈ 0.13°C. While closer than option 1, the actual molality of 0.5 m with i = 1 and Kb = 0.512°C/m gives ΔTb ≈ 0.256°C, which is about twice 0.13°C. Therefore this option underestimates the boiling point rise.
Option 3: 100.26°C. This corresponds to ΔTb ≈ 0.26°C. Using ΔTb = i Kb m with i = 1, Kb = 0.512°C/m, and m = moles solute / kg solvent = 0.1 mol / 0.200 kg = 0.5 m, we get ΔTb = 0.512 × 0.5 ≈ 0.256°C, which rounds to 0.26°C. Adding to the pure water boiling point (100.0°C) yields about 100.26°C, matching this option. This is the correct calculation.
Option 4: 100.5°C. This would require ΔTb = 0.5°C, which would need either a much larger molality or a higher Kb than given. With the provided values, ΔTb is about 0.26°C, not 0.5°C, so this is inconsistent with the data.
In summary, the calculation follows from ΔTb = i Kb m with i = 1 for glucose (a non-electrolyte), m = 0.1 mol / 0.200 kg = 0.5 m, and Kb = 0.512°C/m, giving ΔTb ≈ 0.256°C and a boiling point of ≈ 100.26°C, which aligns with option 3.

What is the boiling point of a solution of 0.1 mole of glucose in 200 mL of water? (K = 0.512C/m)

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