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RAOULT’S LAW

RAOULT’S LAW

source : chemwiki.ucdavis.edu

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

Limitation of Raoult’s Law-

Raoult’s law is followed by dilute solutions

1- The mole fraction of the solute in a given solution remains constant at all temperatures

2- Lowering of vapour pressure is a colligative property.

For solute that dissociates or associated in solution, Vant Hoff factor (i) is defined as-

Vant Hoff factor (i) = no.of solute particles after dissociation or association / no. of solute particles dissolved

The term ‘n’ is replaced by ( i * n).

  1. Molecular weight of solute can be calculated by using Raoult’s law as follows-

p- ps/p =n / n+N

n =w/m ; N =W/M

p /p-ps =n+N/n

(p /p-ps) -1 = (n+N/n ) -1

ps/p-ps =N/n

p-ps /ps =n/N

n=w/m : N =W/M

p-ps /ps =wM/mW

m =w. M.ps /W(p-ps)

When dissociation or association takes peace, then ‘n’ is replaced by i × n

m=i.w.M.ps /W(p-ps)

  1. Consider a solution of two liquids A & B, their vapour  pressures in pure state be p1 & p2, their vapour pressures in solution be ps1 & ps2, their no. of moles in solution be n1 & n2.

According to Raoult’s law

p- ps/p =n / n+N

1 – (ps /p) =n / n+N

ps/p =1 – (n / n+N )

ps/p =N/n+N

ps = p (N/n+N )

Vapour pressure of solution = vapour pressure  of pure solvent × mole fraction of solvent

ps1 = p1 ( n1/n1+n2)

ps2 = p2 ( n2/n1+n2)

Vapour pressure of solution = ps1+ps2

=  p1 [ n1/(n1+n2)] +  p2[ n2/(n1+n2)]

= ( p1.n1 + p2.n2) /(n1+n2)

=  p1 [n1 +( p2 .n2/p1)]/  (n1 + n2)

If p2 >p1, the vapour pressure of solution will be greater than the vapour pressure of A & lesser than vapour pressure of  B.

If p2<p1, the vapour pressure of solution will be less than the vapour pressure of A & greater than the vapour pressure of B.

Saroj Bhatia: Dr. Saroj Bhatia is an Ph.D in chemistry who has been teaching chemistry for over a decade. Currently she is a respected principal of a renowned college in her hometown. She took this medium for online users. Her proudest achievement is helping people learn chemistry.
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