Chemistry, asked by binod036, 3 months ago

derive expression for Raoult's law. ​

Answers

Answered by TeraBhaii
3

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Answered by Anonymous
2

\bold\Answer

Raoult's Law: According to this law 'The vapour pressure of solution containing non-volatile solute is directly proportional to the mole fraction of the solvent'.

For a solution of two components A (Volatile solvent) and B (non-volatile solute)

Vapour pressure of solution = Vapour pressure of solvent ∝ Mole fraction of solvent

Or p=p

A

∝X

a

Pr P

A

=KX

A

....(i)

Where, k= Proportionality constant

Apply equation (i) for pure solvent for this purpose, put (i) X

a

=1 and instead of p

A

put p

A

0

where P

A

0

= vapour pressure of pure solvent

∴ P

A

0

=KX

1

Or P

A

0

=K ....(ii)

According to equation (ii) the value of proportionality constant k is equal to the vapour pressure of pure solvent by putting the value of k in equation (i) we get

P

A

=P

A

0

X

A

Or p solution =p pure solvent × mole fraction of solvent.

Derive it mathematically: Mathematically, if P is the vapour pressure of the pure solvent and P

S

, that of the solution, X

S

is the mole fraction of the solvent in the solution, n are number of moles of the solute and N are the number of moles of the solvent in the solution, then

P

S

∝X

S

or P

S

=KX

S

(Where X

S

=

n+N

N

and K is constant ....(i)

In case of pure solvent n=0 hence

Mole fraction of solvent,

X

S

=

n+N

N

=

0+N

N

=1 and P

S

=p

P=k×1=k

From equation (i) P

S

=P×X

S

...(ii)

Thus, the vapour pressure of the solvent in the solution is equal to the product of its mole fraction and its vapour pressure in the pure state at the same temperature.

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