Colligative Properties

These are the properties that depend on the concentration of solute molecule or ions in solution but not on the chemical identity of the solute. For example, addition of ethylene glycol to water lowers the freezing point of water below 0°C.

The magnitude of freezing point lowering is directly proportional to the number of solute molecules added to a particular quantity of solvent.

If 0.01 mole of ethylene glycol is added to 1 kg of water, the freezing point is lowered to ¾0.019°C while on adding 0.020 mole of ethylene glycol to 1 kg of water, the freezing point is lowered to ¾0.038°C.

The various colligative properties are

(i) Relative lowering of vapour pressure

(ii) osmotic pressure

(iii) elevation of boiling point

(iv) depression of freezing point

RELATIVE LOWERING OF VAPOUR PRESSURE

The addition of a non – volatile solute to a solvent at a given temperature and pressure results in the lowering of the vapour pressure of the solvent.

If a solution is formed by dissolving n moles of a non volatile solute in N moles of a volatile solvent. Then

Mole fraction of solvent, X₁ = $\frac{N}{n+N}$

Mole fraction of solute X₂ = $\frac{n}{N+n}$

Since the solute is non – volatile, so it would have negligible vapour pressure and thus the vapour pressure of the solution, is approximately same as the vapour pressure of the solvent.

According to Raoult's law

$P_1=X_1{P}_1^0...(i)$

Since X₁ + X₂ = 1 so X₁ = 1 - X₂

Now P₁ = $\left(1-X_2\right)P_1^0$

Or

$\frac{P_1^0-P_1}{P_1^0}=X_2...(ii)$

in equation (ii) $P_1^0-P_1$ is the vapour pressure lowering and $\left(P_1^0-P_1\right)/P_1^0$ is the relative vapour pressure lowering. From equation (ii) it is clear that the relative vapour pressure lowering is proportional to the mole fraction of the non – volatile solute in solution. It is independent of the nature of the solute. The relative vapour pressure lowering is, therefore a colligative property.

The mole fraction, X₂ = $\frac{n}{n+N}$

For a dilute solution, the number of moles of the solute (n) can be neglected as compared to the number of moles of the solvent (N). Hence

$X_2=\frac{n}{N}=\frac{W_1}{M_1}/\frac{W_2}{M_2}=\frac{W_2{M}_1}{W_1{M}_2}$

Where W₁ is the amount of the solute dissolved in W₁ amount of solvent, M₁ is the molar mass of the solvent and M₂ is the molar mass of the solute.

$\frac{P_1^0-P_1}{P_1^0}=X_2=\frac{W_2{M}_1}{W_1{M}_2}$

Vapour pressure in terms of molality of the solution:

$\frac{P_1^0-P_1}{P_1^0}=X_2=\frac{m{M}_1}{1000gm},wheremisthemolality$

Illustration 1.

The vapour pressure of a solvent decreased by 10 mm Hg when a non volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg.

(A) 0.8

(B) 0.6

(C) 0.4

(D) 0.2

Solution:

P_­o – P_s = P_o x mole fraction of solute; 10 = P_o x 0.2

Again, 20 = P_o x mole fraction solute

Hence, n = 0.4, so, mole fraction of solvent = 1 – 0.4 = 0.6

Hence, (B) is correct.

There is a natural tendency of solutes in a solution to diffuse from a higher concentration to a lower concentration so as to bring about a uniform distribution throughout.

Certain membranes allow solvent molecules to pass through them but not solute molecules, particularly not those of large molecular weight. Such a membrane is called semipermeable and might be an animal bladder, a vegetable tissues or a piece of cellophone.

Osmosis is the phenomenon of solvent flow thorough a semi permeable membrane to equalize the solute concentrations on both sides of the membrane.

Osmotic pressure is a colligative property of a solution equal to the pressure that, when applied to the solution just stops osmosis. It is denoted by p. The osmotic pressure p of a solution is related to the molar concentration of solute M.

$\pi =MRT$, Where R is the gas constant and T is absolute temperature.

Van't Hoff equation for dilute solution is similar to ideal gas equation,

$\pi V=nRT$

Where p = osmotic pressure

R = gas constant

Illustration 2.

Calculate the molecular weight of cellulose acetate if its 0.5% (wt./vol) solution in acetone (sp. gr. = 0.9) shows an osmotic rise of 23 mm against pure acetone at 27°C.

Solution:

0.5% (wt. / vol) solution means 0.5 gm of cellulose acetate is dissolved in 100 ml solution.

Osmotic pressure = 23 mm of pure acetone

$\pi =2.3$cm of pure acetone = $\frac{2.3\times 0.9}{13.6}$cm of Hg = 0.1522 cm of Hg

$\pi$ = $\frac{0.1522}{76}$ atm = 0.002 atm

Let the molecular weight of the cellulose acetate be M

Here, number of mole of cellulose acetate (n) $=\frac{0.5}{m}$

Volume = V = 100 ml = 0.1 lit

R = 0.082 Lit atm mol^–1 K^–1, T = (27 + 273) = 300 K

Osmotic pressure (p) = $\frac{n}{V}$ xRT

$\Rightarrow$ 0.002 = $\frac{0.5}{M}\times \frac{1}{0.1}$x 0.0821 x 300

$\therefore$ m = 61575

ISOTONIC SOLUTIONS

A pair of solutions having same osmotic pressure is called isotonic solution. When two solutions having the same osmotic pressure are put into communication with each other through a semi permeable membrane, there will be net transference of solvent from one solution to the other, but there is a dynamic equilibrium between two solutions. According to Van't Hoff equation, it is evident that isotonic solutions must have the same molar concentration, if both are at same temperature.

For isotonic solution, ${\pi }_1={\pi }_2$

Or $C_1=C_2\Rightarrow \frac{n_1}{V_1}=\frac{n_2}{V_2}$

Or $\frac{W_1}{m_1{V}_1}=\frac{W_2}{m_2{V}_2}$

Illustration 3.

Calculate the osmotic pressure of sucrose solution containing 1.75 gms in 150 ml of solution at 17°C.

Solution: We know that

$\pi V$ = nRT = $\frac{W_B}{m_B}RT$

$\pi \times$ 0.15 = $\frac{1.75}{342}\times 0.0821\times 290$

$\pi$ = 0.812 atm

REVERSE OSMOSIS AND WATER PURIFICATION

The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solutions side. That is, now the pure solvent flows out of the solution through the semipermeable membrane (SPM). This phenomenon is called reverse osmosis and is of great practical utility. Reverse osmosis is used in the desalination of sea water. A schematic set up for the process is shown in figure. When pressure more than osmotic pressure is applied pure water is squeezed out of the sea water through the membrane. A variety of polymer membrane are available for this purpose.

The pressure required for reverse osmosis are quite high and a workable porous membrane is a film of cellulose acetate placed over a suitable support. Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water. These days many countries use desalination plants to meet their water requirement.

BOILING POINT ELEVATION BY NON VOLATILE SOLUTE

The boiling point T_b of a liquid is the temperature at which its pressure becomes equal to the atmospheric pressure. When a non – volatile solute is added to a liquid, the vapour pressure of the liquid is decreased. Hence it must be heated to a higher temperature in order that its vapour pressure becomes equal to that of the atmospheric pressure. This means that the addition of a non – volatile solute to a liquid raises its boiling point.

If $T_b^0$ is the boiling point of the solvent and T_b is the boiling point of the solution, then

$T_b-T_b^0=\Delta T_b$(Elevation of boiling point)

$\Delta T_b\alpha$ molality (m)

$\Delta T_b=K_b\times m$, k_b: molal elevation constant or Ebullioscopic constant

$m:$ molality of the solution

Elevation in boiling point can be calculated as

$\Delta T_b=\frac{R{T}_b^2{M}_{1}m}{\Delta H_{vap}}$ [M₁ is the molar mass of solvent, $\Delta H_{vap}$ is enthalpy of vaporization per gram of solvent.]

For a given solvent, $R{T}_b^2{M}_1/\Delta H_{vap}$is constant, it is denoted by K_b and called molal elevation constant of the solvent thus, $K_b=R{T}_b^2{M}_1/\Delta H_{vap}$

Hence $\Delta T_b=m.k_b$

If molality m = 1, i.e. 1 mole of the solute is dissolved in 1 kg of the of the solvent. Then $K_b=\Delta T_b$ Thus molal boiling point elevation constant or ebullioscopic constant of a solvent is defined as the elevation in boiling point of the solution which may be theoretically be produced when 1 mole of the non volatile, non electrolyte solute is dissolved in 1 kg of the solvent.

The elevation in boiling point depends only on the molality of the solute and is independent of the nature of the solute, it is therefore a colligative property.

$Unitof{K}_b=Kkgmo{l}^{-1}$

Determination of molar mass using boiling point elevation method

If W₂ grams of the solute of molar mass M₂ is dissolved in W₁ kg of the solvent, then the number of moles of the solute dissolved in 1 kg of the solvent would be given by:

$\frac{W_2}{M_2}\times \frac{1}{W_1},then$

$\Delta T_b=K_b\times \frac{W_2}{M_2{W}_1}or{M}_2=\frac{K_b{W}_2}{W_1\Delta T_b}$

Illustration 4.

Y g of non - volatile organic substance of molecular mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is K_b. Elevation in its boiling point is given by

(A)$\frac{M}{K_{b}Y}$ (B)$\frac{4K_{b}Y}{M}$

(C)$\frac{K_{b}Y}{4M}$ (D)$\frac{K_{b}Y}{M}$

Solution: DT = $\frac{1000\times K_b\times Y}{250\times M}$ = $\frac{4K_{b}Y}{M}$

Hence, (B) is correct.

DEPRESSION OF FREEZING POINT BY A NON – VOLATILE SOLUTE

The temperature at which solid and liquid states of a substance have the same vapour pressure is called freezing point. If a non volatile solute is mixed with a pure solvent, the freezing point of pure solvent is always greater than the impure one.

$T_f^0-T_f=\Delta T_f$, where T_f⁰ is the freezing point of the pure solvent and T_f is the freezing point of a non – volatile solute containing solvent and $\Delta T_f$ is the depression in freezing point.

$\Delta T_f\propto$ molality

or $\Delta T_f=K_f\times m$

K_f = molal freezing point depression constant of the solvent (cryoscopic constant)

m = molality of the solution

Molal freezing point depression constant of the solvent or cryoscopic constant is defined as the depression in freezing point which may be theoretically produced by dissolving 1 mole of any Non-volatile, non-electrolyte solute in 1000 gm of solvent.

$or\Delta T_f=\frac{1000\times K_f\times w}{m_1\times W}$

Where m₁ = molecular weight of solute

w = weight of solute

W = weight of solvent

Illustration 5.

The amount of ice that will separate on cooling a solution containing 50g of ethylene glycol in 200g water to –9.3°C is : [K_f = 1.86 K molality^–1]

(A) 38.71 g (B) 38.71 mg

(C) 42 g (D) 42 mg

Solution: $\Delta T$ = $\frac{1000\times K_f\times w}{W\times M}$

or 9.3 = $\frac{1000\times 1.86\times 50}{62\times W}$

$\therefore$ W = 161.29

$\therefore$ Ice separated = 200 – 161.29 = 38.71 g

Hence, (A) is correct.

Illustration 6.

The freezing point of aqueous solution contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is: (${K_f}_{H_{2}O}$ = 1.86 K molality^–1)

(A) 290.2 K (B) 285.5 K

(C) 269.93 K (D) 250 K

Solution: $\Delta T_f=$ $\Delta T_f$ for glucose + $\Delta T_f$ for KCl + $\Delta T_f$ for urea

= $\frac{1000\times 1.86\times 10}{100\times 180}$ + $\frac{1000\times 1.86\times 1\times 2}{74.5\times 100}$ + $\frac{1000\times 1.86\times 5}{100\times 60}$ = 3.069°

Freezing point = 273 – 3.069 = 269 .93 K

Hence, (C) is correct.

Illustration 7.

The molal freezing point constant for water is 1.86 K.molarity^–1. If 34.2 g of cane sugar (C₁₂H₂₂O₁₁) are dissolved in 1000g of water, the solution will freeze at

(A) –1.86°C (B) 1.86°C

(C) –3.92°C (D) 2.42°C

Solution: $\Delta T_f$ = $\frac{1000\times K_f\times w}{MW}=\frac{1000\times 1.86\times 34.2}{100\times 342}$ = 1.86°

$\therefore$ T_f = 0 – 1.86 = –1.86° C

Hence, (A) is correct.

Illustration 8.

An aqueous solution containing 0.25 g of a solute dissolved in 20g of water freezes at 0.4°C. Calculate the molar mass of the solute. K_f = 1.86 K kg mol^-1.

Solution: $\Delta T_f$ = $K_f\times m$

$M=\frac{K_f\times w\times 1000}{W\times \Delta T}=\frac{1.86\times 0.25\times 1000}{20\times 0.4}$= 58. 13g/mol