CBSE Class 12 · 2002 · Delhi Compartment · SolutionsShort Answer

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How is relative lowering of vapour pressure defined for a solution consisting of a volatile solvent and a non-volatile solute? How is this function related to the mole fraction of the solvent and of the solute?

Answer: Relative lowering = (p° − p)/p°; by Raoult's law it equals the mole fraction of the solute, x₂ = 1 − x₁.

Solution

Relative lowering of vapour pressure is the lowering of vapour pressure divided by the vapour pressure of the pure solvent:

\frac{p ^{\circ} - p}{p ^{\circ}}

By Raoult's law,

p = x_{1} p^{\circ}

, so

\frac{p ^{\circ} - p}{p ^{\circ}} = 1 - x_{1} = x_{2}

It equals the mole fraction of the solute and is

1 -

(mole fraction of the solvent).

CBSE Class 12 Board · 2002 · Delhi Compartment · Subjective

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