Notes-Part-2-Class-12-Chemistry-Chapter-2-Solutions-Maharashtra Board

Boiling point elevation as a consequence of vapour pressure lowering :

• The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 x 10³ Nm^-2). When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
• If the liquid has a low vapour pressure, it has a higher boiling point.

To understand the elevation of boiling point, let us compare the vapour pressures of solution and those of pure solvent. The vapour pressures of solution and of pure solvent are plotted as a function of temperature as shown in Fig.

When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.

If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,

ΔT_b = T — T₀

The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀—P, where P₀ and P are the vapour pressures of the pure solvent and the solution. [ΔT_b ∝ (P₀—P) or ΔT_b ∝ ΔP]

Relation between molar mass of the solute and boiling point elevation :

 The boiling point elevation, ΔT_b of a solution is directly proportional to molality (m) of the solution.  ΔT_b ∝ m

ΔT_b = K_bm

where K_b is a proportionality constant

If m = 1 molal, then

K ΔT_b = K_b.

where K is called molal elevation constant.

The molality of the solution is given by,

m = \(\frac{\text{ Number of moles of the solute}}{\text{Weight of the solvent in kg}}\)

Let W₁ = Weight (in gram) of a solvent,

W₂ = Weight (in gram) of a solute

M₂ = Molecular weight of the solute

Then the molality (m) of the solution is given by

m = \(\frac{W_2×1000}{W_1M_2}=\frac{W_2}{W_1M_2}\) mol kg⁻¹

∴ ΔT_b = K_b x \(\frac{W_2×1000}{W_1M_2}\)

∴ M₂ = \(\frac{K_bW_2×1000}{ΔT_bW_1M_2}\)  or M₂ = \(\frac{K_bW_2}{ΔT_bW_1M_2}\) mol kg⁻¹

If the weights and molecular weight are expressed in kg, then,

∴ ΔT_b = K_b x \(\frac{W_2}{W_1M_2}\)

Freezing point depression as a consequence of vapour pressure lowering :

The effect of dissolution of a nonvolatile solute into a solvent on freezing point of solvent can be understood in terms of the vapour pressure lowering.

The freezing point of a liquid is the temperature at which the liquid and the solid have the same vapour pressure.

Addition of a nonvolatile solute to a liquid decreases the freezing point, i.e. the freezing point of the solution is less than that of the pure solvent.

This is due to the lowering of the vapour pressure of the solvent by the addition of the nonvolatile solute.

When a liquid is cooled from the point A, its vapour pressure decreases and at the point B, it freezes (solidifies).

In case of a solution, since the vapour pressure is lowered, the freezing point decreases. Hence, if the solution is cooled from the point F, it freezes at lower temperature E, than the pure liquid. This is also due to separation of solvent molecules due to solute molecules decreasing their intermolecular attraction.

If  T₀ and T are the freezing points of a pure solvent and the solution, then, the depression in the freezing point is given by,

ΔT_f = T₀ — T.

This depression ΔT_f depends on the lowering of the vapour pressure (P₀ — P)

ΔT_f ∝ (P₀ — P),

where P₀ and P are the vapour pressures of the pure liquid and the solution.

Relationship between freezing point depression of a solution containing non-volatile nonelectrolyte and its molar mass :

The freezing point depression, ΔT_f of a solution is directly proportional to molality (m) of the solution.

 ΔT_f ∝ m

 ΔT_f = K_f m

where Kf is a molal depression constant.

The molality of a solution is given by,

m = \(\frac{\text{ Number of moles of the solute}}{\text{Weight of the solvent in kg}}\)

If W₁ grams of a solvent contain W₂ grams of a solute of the molar mass M₂, then the molality m of the solution is given by,

m = \(\frac{W_2×1000}{W_1M_2}=\frac{W_2}{W_1M_2}\) mol kg⁻¹

∴ ΔT_f = K_f x \(\frac{W_2×1000}{W_1M_2}\)

∴ M₂ = \(\frac{K_fW_2×1000}{ΔT_fW_1M_2}\) or M₂ = \(\frac{K_fW_2}{ΔT_fW_1M_2}\) mol kg⁻¹

If the weights and molecular weight are expressed in kg, then,

∴ ΔT_f = K_f x \(\frac{W_2}{W_1M_2}\)

The unit of K_f is K kg mol^-1 .

Hence, from the measurement of the depression in the freezing point of the solution, the molar mass of the substance can be determined.

Osmosis can be demonstrated with experimental set up shown in Fig.

• Consider an inverted thistle funnel on the mouth of which a semipermeable membrane is firmly fastened.
• It is filled with the experimental solution and immersed in a solvent like water.

• As a result, solvent molecules pass through the membrane into the solution in the funnel causing rising of level in the arm of thistle funnel.
• This increases the hydrostatic pressure.
• At a certain stage this rising level stops indicating an equilibrium between the rates of flow of solvent molecules from solvent to solution and from solution to solvent. The hydrostatic pressure at this stage represents osmotic pressure of the solution in the thistle funnel.

Remember : It is important to note that osmotic pressure is not the pressure produced by a solution. It exists only when the solution is separated from the solvent by a suitable kind of semipermeable membrane.

Relation between osmotic pressure and concentration of solution :

Consider V dm³ of a solution in which n₁ moles of a solvent contains n₂ moles of a nonvolatile solute at absolute temperature T.

The osmotic pressure, p of a solution is given by,

π = nRT/V

R is gas constant having value 0.08206 dm³ atm K^-1 mol^-1 (OR L atm K^-1 mol^-1). Since concentration, C of a solution is in mol dm_-3 or molarity is,

C = n/V mol dm^-3 or M

∴ π = CRT,  (If concentration C is expressed in mol m^-3 and R = 8.314 J K^-1 mol^-1, then π will be in SI units, pascals or Nm^-2.)

• Measurements of osmotic pressure are made at a certain constant temperature.
• Molality depends upon temperature but molality is independent of temperature.
• Hence in osmotic pressure measurements, concentration is expressed in molarity.

Relation between osmotic pressure and molar mass of a solute :

Consider V dm³ (litres) of a solution containing W₂ mass of a solute of molar mass M₂ at a temperature T.

Number of moles of solute, n₂ = W₂/M₂

The osmotic pressure p is given by,

π = \(\frac{W_2RT}{M_2V}\)

∴ \(M_2=\frac{W_2RT}{πV}\)

By measuring osmotic pressure of a solution, the molar mass of a solute can be calculated.

Since osmotic pressure can be measured more precisely, it is widely used to measure molar masses of the substances

Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,

C_A = n_A/V (in mol dm^-3) and C_B = n_B/V (in mol dm^-3)

If the absolute temperature of both the solutions is T, then by the van’t Hof equation,

π_A = C_ART and p_B = C_BRT, where  π_A and π_B are their osmotic pressures.

For the isotonic solutions,

π_A = π_B

∴ C_A = C_B

∴ n_A/V = n_B/V

∴ n_A = n_B

Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

Reverse osmosis : osmosis is a flow of solvent through a semipermeable membrane into the solution. The direction of osmosis can be reversed by applying a pressure larger than the osmotic pressure.

The phenomenon of the passage of solvent like water under high pressure from the concentrated aqueous solution like sea water into pure water through a semipermeable membrane is called reverse osmosis.

The osmotic pressure of sea water is about 30 atmospheres. Hence when pressure more than 30 atmospheres is applied on the solution side, regular osmosis stops and reverse osmosis starts. Hence pure Water from sea water enters the other side of pure water.

For this purpose a suitable semipermeable membrane is required which can withstand high pressure conditions over a long period.

This method is used successfully in Florida since 1981 producing more than 10 million litres of pure water per day.

Remember : Osmotic pressure is much larger and therefore more precisely measurable property than other colligative properties. It is therefore, useful to determine molar masses of very expensive substances and of the substances that can be prepared in small quantities.

Van’t Hoff factor : To obtain the colligative properties of electrolyte solutions by using relations for nonelectrolytes, van’t Hoff suggested a factor i.

It is defined as a ratio of the observed colligative property of the solution to the theoretically calculated colligative property of the solution without considering molecular change.

The van’t Hoff factor can be represented as,

This colligative property may be the lowering of vapour pressure of a solution, the osmotic pressure, the elevation in the boiling point or the depression in the freezing point of the solution. Hence,

\(i=\frac{\text{Observed lowering of vapour pressure}}{\text{Theoretical lowering of vapour pressure}}\)

= ΔP_(ob)/ΔP_(th)

\(i=\frac{\text{Observed elevation in boiling point}}{\text{Theoretical elevation in boiling point}}\)

= ΔT_b(ob)/ΔT_b(th)

\(i=\frac{\text{Observed depression in freezing point}}{\text{Theoretical depression in freezing point}}\)

= ΔT_f(ob)/ΔT_f(th)

\(i=\frac{\text{Observed osmotic pressure}}{\text{Theoretical osmotic pressure}}\)

= π_(ob)/ π_(th)

• When the solute neither undergoes dissociation or association in the solution, then, i = 1
• When the solute undergoes dissociation in the solution, then, i > l
• When the solute undergoes association in the solution, then i < 1

From the value of the van’t Hoff factor, the degree of dissociation of electrolytes, degree of association of nonelectrolytes can be obtained.

van’t Hoff factor gives the important information about the solute molecules in the solution and chemical bonding in them.

van’t Hoff factor related to molecular mass of the substance :

The van’t Hoff factor is also defined as,

\(i=\frac{\text{Actual moles of particles in solution after dissociation }}{\text{Moles of formula units in dissolved solution}}=\frac{M_{th}}{M_{ob}}\)

In case of nonelectrolytes, i < 1.

In case of electrolytes, i > 1. For example for KNO₃ and NaCl, i = 2, for Na₂SO₄, CaCl₂, i = 3, etc.

Modification of expressions of colligative properties : The expressions of colligative properties mentioned earlier for nonelectrolytes are to be modified so as to make them applicable for electrolyte solutions.

The modified equations are:

By Raoult’s law :

\(\frac{P_0-P}{P} = ix_2 = i× \frac{W_2M_1}{W_1M_1}\)

For elevation in boiling point :

ΔT_b = i x K_b x m = i x K_b x \(\frac{W_2x1000}{W_1M_1}\)

ΔT_b = i x K_b x \(\frac{W_2}{W_1M_1}\)

For depression in freezing point

ΔT_f = i x K_f x m = i x K_f x \(\frac{W_2×1000}{W_1M_1}\)

ΔT_f = i x K_f x \(\frac{W_2}{W_1M_1}\)

For Osmotic pressure :

π = iCRT =\(i×\frac{WRT}{MV}\)

Relationship between degree of dissociation of an electrolyte and van’t Hoff factor :

Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives x number of A^y+ ions and y number of B^x- ions.

Let a be the degree of dissociation.

At equilibrium,

AxBy ⇔ xA^y+ + yB^x-

For 1 mole of electrolyte : 1 − α, xα, yα

and For ‘m’ moles of an electrolyte : m(1 − α), mxα, myα are the number of particles.

Total number of moles at equilibrium, will be,

Total moles = m(1 − α) + mxα + myα

= m[(1 − α) + xα + yα]

= m[1 + xα + yα − α]

= m[1 + α(x + y − 1)]

The van’t Hoff factor i will be,

\(i=\frac{\text{Observed colligative property}}{\text{Theoretical colligative property}}\)

∴\(i=\frac{m[1 + α(x + y − 1)]}{m}\)

i = 1 + α(x + y − 1)

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n

∴ i = 1 + α(n − 1)

∴α(n − 1) = i − 1

∴ α = \(\frac{i-1}{n-1}\)  …..(1)

This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Relation between van’t Hoff factor and molar mass of electrolyte :

Consider an elctrolyte AxBy that dissociates in aqueous solution as

                      AxBy  ⇔ xA^y+ + yB^x-

Initially :            1          0         0

At equilibrium : (1-∝)    (x∝)     (y∝)

If i is van’t Hoff factor and n is total number of ions produced from dissociation of one mole of electrolyte then, the degree of dissociation ∝ is given by,

∝ = \(\frac{i-1}{n-1}\)

Now if M_th and M_ob are theoretical and observed molecular (or molar) masses respectively, then,

∴ ∝(n — 1) = i – 1 = \(\frac{M_{th}}{M_{ob}}-1\)

∴ ∝(n — 1) =\(\frac{M_{th}-M_{ob}}{M_{ob}}\)

∴ ∝ a = \(\frac{1}{n-1}[\frac{M_{th}-M_{ob}}{M_{ob}}]\) ….. (2)

This is a relation between degree of dissociation and molecular masses of the dissolved electrolyte AxBy