Theory of dilute solutions

•Here we are deal with solutions of solid solutes dissolved in liquid solvents.
•Colligative properties:
1.Lowering of vapor pressure
2.Elevation of boiling point
3.Depression of freezing point
4.Osmotic pressure
•These properties depend on the number of the particles of the dissolved substance

and independent on their size or nature, therefore they have been grouped together as Colligative properties, these properties are important for determination the molecular weight of the unknown dissolved substance.

Lowering of vapor pressure, Raoult”s law:
Vapor pressure of a pure solvent is decreased when a non – volatile solute dissolved in it.
If P is the vapor pressure of the solvent,and

Ps that of the solution, then the lowering is:

P – Ps and P – Ps/P is called the relative lowering.

Raoult”s law states that” the relative lowering is equal to the mole fraction of the solute,

Thus P – Ps /P = n / n + N
Where n the number of moles of solute, and N the number of moles of solvent.

Derivation of Raoult”s law

•The vapor pressure of the solution is proportional to the mole fraction of the solvent, since the solute is non-volatile.

Thus Ps α N / n + N …………(1) 

In case of pure solvent n = 0 

Hence mol fraction of solvent = N / N + 0 = 1 

Thus from (1) P = k 

Ps = P . N /n + N 

Ps / P = N / n + N 

1 - Ps / P = 1 - N / n + N 

P – Ps / P = n / n + N 

Ideal solutions and deviation from Raoult”s law
•Suppose that γAB the attractive force between A and B molecules and γAA between A and A molecules. Then;

if γAB = γAA the solution is ideal

if γAB > γAA negative deviation is observed

if γAB < γAA positive deviation is observed

In very dilute solutions of non-electrolytes, the solvent and solute molecules are very much alike, such solutions approach the ideal behavior and obey Raoult”s law.


Determination of molecular weight from vapor pressure lowering.

if w gram of solute is dissolved in W gram of solvent, and m & M are molecular weight of solute and solvent respectively, then moles of solute (n) = w /m

moles of solvent (N) = W/M

substituting in Raoult”s law we get:

P- Ps/ P = n / n + N = (w/m) / W /M + w/m

Since for very dilute solution, n is very small it can be neglected in the denominator so

P – Ps / = w . M /mW



Measurement of lowering of vapor pressure by gas saturation method

• if w1 and w2 be the loss of weight in set of solution and in set of solvent respectively, 

then w1 α ps ………… (1) 

w2 α P – Ps …….(2) 

adding (1) and (2) we have 

w1 + w2 α ps+ P – Ps 

α P …. (3) 

Dividing (2) by (3) we get: 

P – Ps / P = w2 / w1 + w2 ………(4) 

Boiling point elevation
•As a result of vapor pressure lowering, the boiling point is elevated.
•If Tb is the boiling point of pure solvent, and T is that of the solution, the difference in boiling point (ΔT) is:

• ΔT = T - Tb 

•AB /AC = AD /AE 

•T1 – Tb / T2 – Tb = 

• P – P1/ P - P2 

•Hence the elevation of boiling point is directly proportional to the lowering of vapor pressure.

ΔT α P - Ps ………….(1) 

But since P is constant for the same solvent at fixed temperature, thus,

ΔT α P - Ps/ P ………..(2) 

But for Raoult”s law P - Ps/ P α w M /mW

M of solvent is constant, thus:

P - Ps/ P α w /mW……….(4) 

From (2) and (4) we get: 

ΔT α w /mW 

ΔT= kb . w/m . 1/W……….(5) 

•Kb is called molal elevation constant, if W of the solvent is expressed in kilograms, but if the mass of the solvent W is given in grames, it has to be converted to kilograms. Thus :
• ΔT = kb . w/m . 1/ W/1000………..(6)

From which we get: m = 1000 x kb x w / ΔT W

If the value of kb is given in k0 per 0.1 kg (100g) the last equation becomes:

m = 100 x kb x w / ΔT W

Thermodynamically kb can be calculated from

kb = RTb2 / 1000 Lv


•By Cottrell”s method: