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Show that the relative lowering of vapour pressure is a colligative property.

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Question

Show that the relative lowering of vapour pressure is a colligative property.

Very Long Answer
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Solution

If p° represents the vapour pressure of a pure solvent and p represents the vapour pressure of the solution, we have

Lowering of vapour pressure = p° − p

Relative lowering of vapour pressure =  `(p^\circ - p)/p^\circ`

The relative lowering of vapour pressure as well as the lowering of vapour pressure are colligative properties and depend upon the number of solute particles but not upon its nature.

The relative lowering of vapour pressure is equal to the mole fraction `chi_"solute"` of solute, i.e.,

`(p^\circ - p)/p^\circ = chi_"solute"`

From the above equation, it is clear that the relative lowering of vapour pressure of a solution depends only upon the mole fraction of the solute and is independent of its nature. Since mole fraction represents the number of moles of solute dissolved in a definite number of moles of the solvent, the relative lowering in vapour pressure is a colligative property.

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Chapter 1: Solutions - REVIEW EXERCISES [Page 98]

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Nootan Chemistry [English] Class 12 ISC
Chapter 1 Solutions
REVIEW EXERCISES | Q 2.56 | Page 98
Nootan Chemistry [English] Class 12 ISC
Chapter 1 Solutions
VERY SHORT ANSWER TYPE QUESTIONS | Q 27. | Page 110
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