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At a certain temperature, a glucose solution shows a lowering of vapour pressure of 0.6 mm Hg. What would be the vapour pressure of the pure solvent at the same temperature, if a glucose solution of molality (1/18) mol/kg gives a lowering of 0.6 mm Hg? (Assume dilute solution behaviour.)

1. 200 mm Hg
2. 300 mm Hg
3. 600 mm Hg
4. 720 mm Hg

Correct answer: 200 mm Hg

Solution

Molality = 1/18 mol/kg means 1/18 mol glucose per kg water. Moles of water per kg = 1000/18. Mole fraction of glucose = (1/18)/((1/18) + 1000/18) = (1/18)/(1001/18) = 1/1001 approximately 1/1000. Raoult's law: delta_P = x_solute * P⁰ = 0.6 mm Hg => P⁰ = 0.6 * 1000 = 600 mm Hg.

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