RAOULT’S LAW

Lowering of vapour pressure-

The molecules of liquid are in a state of constant motion in all directions, some of these molecules have higher kinetic energy than others. Some molecules which are present at the surface of a liquid may leave the surface & go into vapour state. The pressure exerted by the vapours of a liquid on its surface is known as vapour pressure of that liquid.

Vapour pressure of a liquid depends upon temperature & is independent of amount of liquid & vapours present in the system.

When a nonvolatile solute is dissolved in a liquid, the escaping tendency of liquid molecules is lowered hence vapour pressure of liquid is lowered. This difference of vapour pressure is called lowering of vapour pressure.

Lowering of vapour pressure= p-ps

p=vapour pressure of pure solvent

ps= vapour pressure of pure solution

Relative lowering of vapour pressure = p-ps/p

Raoult’s Law-

According to Raoult’s law,

“The relative lowering of vapour pressure of solvent is equal to the mole fraction of the solute.”

Let, No. of moles of solute = n

No. of moles of solvent = N

Total moles = n+N

Vapour pressure of pure solvent= p

Vapour pressure of solution =Ps

lowering of vapour pressure= p-ps

Relative lowering of vapour pressure= p- ps/p

So, According to Raoult’s Law,

p- ps/p =n / n+N

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