why does the vapour pressure gets lowered after addition of a non volatile solute?

Asked by joshika saraf | 20th Aug, 2011, 12:00: AM

Expert Answer:

The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction. This is the statement of the Raoult's law and teh lowering of vapour pressure is due to this only.

In equation form, this law reads:

In this equation, Po is the vapour pressure of the pure solvent at a particular temperature.

xsolv is the mole fraction of the solvent. That is exactly what it says it is - the fraction of the total number of moles present which is solvent.

We calculate this using:

Answered by  | 23rd Aug, 2011, 03:53: PM

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