Solution-Class-12-Chemistry-Chapter-2-Solutions-Maharashtra Board

(d) 12 mm Hg

(c) osmotic pressure

(a) molarity

(a) 1m

(a) nature of solvent

(b) boiling point of solvent is lower than that of solution

(c) 4.92 atm

(a) 5.41 %

(d) directly proportional to the mole fraction of the solvent

(a) boiling point of water involved in cooking is increased

(c) 0.026 M

(c) they are isotonic solutions

Osmotic pressure : The osmotic pressure is defined as the excess mechanical pressure required to be applied to a solution separated by a semipermeable membrane from pure solvent or a dilute solution to prevent the osmosis or free passage of the solvent molecules at a given temperature. The osmotic pressure is a colligative property.

While calculating osmotic pressure by equation, p = CRT, the concentration is expressed in molarity but not in molality.

This is because the 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.

The boiling point elevation is directly proportional to the molality of the solution. Thus,

ΔT_b ∝ m or ΔT_b = K_bm

where m is the molality of solution. The proportionality constant K_b is called boiling point elevation constant or molal elevation constant or ebullioscopic constant.

If m = 1, ΔT_b = K_b

Thus, ebullioscopic constant is the boiling point elevation produced by 1 molal solution.

Given : m = 0.1 m, ΔT_f = 0 - (- 0.43) = 0.43 ⁰C, K_f = 1.86 K kg mol^-1, i=?

ΔT_f = i x K_f x m

∴ i = \(\frac{ΔT_f}{k_f×m}=\frac{0.43}{1.86×0.1}\)= 2.312

Answer is  van’t Hoff factor = i = 2.312.

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.

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.(degree of ionization)

• Freezing point depression is a colligative property, hence depends on the number of particles in the solution.
• More the number of particles in the solution, higher is the depression in freezing point.

The number of particles (ions) from electrolytes are,

a, 0.1 m NaCl

NaCl —> Na⁺ + Cl^- (Total ions = 0.1 + 0.1=0.2m)

b. 0.05 m Al₂(SO₄)₃

Al₂(SO₄)₃ --> 2Al₃⁺ + 3SO₄^1- , (Total ions = 2 x 0.05 + 3 x 0.05 =0.1 +0.15 = 0.25 m)

Therefore, Al₂(SO₄)₃ solution will have higher freezing point depression. .

Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

Explanation : Let P₀ and P be the vapour pressures of a pure solvent and a solution respectively. If x₁ is the mole fraction of the solvent then Raoult’s law can be represented as, P = x₁P₀

For a binary solution containing one solute, if x₁ and x₂ are mole fractions of a solvent and a solute respectively then,

x₁ + x₂ = 1

 x₁ = 1 - x₂

 P = x₁P₀

= (1 - x₂)P₀

= P₀ - x₂P₀

 P₀ - P = x₂P₀

∴ x₂ = \(\frac{P_0-P}{P_0}\)

P₀ - P = ΔP is the lowering of vapour pressure

∴ x₂  = \(\frac{ΔP}{P_0}\)

In the equation,  is called relative lowering of vapour pressure.

Hence Raoult’s law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute.

• The boiling point of water (or any liquid) depends on its vapour pressure.
• Higher the vapour pressure, lower is the boiling point.
• When 1 mole of volatile methyl alcohol is added to 1 dm³ of water, its vapour pressure is increased decreasing the boiling point of water.

• Since osmotic pressure has large values, it can be measured more precisely.
• The osmotic pressure can be measured at a suitable constant temperature.
• The molecular masses can be measured more accurately.
• Therefore, it is more useful to determine molecular masses of expensive substances by osmotic pressure.

• 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]

Hypotonic solutions : When two solutions have different osmotic pressures, then the solution having lower osmotic pressure is said to be a hypotonic solution with respect to the other solution.

Explanation : Consider two solutions of the substances A and B having osmotic pressures π_A and π_B.  If π_B is less than π_A, then the solution B is a hypotonic solution with respect to the solution A.

Hence, if C_A and C_B are their concentrations, then, C_B < C_A. Hence, for equal volumes of the solutions, n_B < n_A.

Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.

Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B.  If p_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.

Hence, if C_A and C_B are their concentrations, then, C_B > C_A. Hence, for equal volumes of the solutions, n_B > n_A.

• When a solvent and a solution containing a nonvolatile solute are separated by a semipenneable membrane, there arises a flow of solvent molecules from solvent to solution as well as from solution to solvent.
• Due to higher vapour pressure of solvent than solution, the rate of flow of solvent molecules from solvent to solution is higher.
• As more and more solvent passes into solution due to osmosis, the solvent content increases, and the rate of backward flow increases.
• At a certain stage an equilibrium is reached where both the opposing rates become equal attaining an equilibrium.

Given : π_cacl2 = 0.605 atm, π_urea = 0.245 atm

For urea solution, van’t Hoff factor,

i = 1

π_cacl2 = i x (CRT)_CaCl2

π_urea = (CRT) _urea

π_cacl2/π_urea = (i x (CRT)_CaCl2)/(CRT) _urea

∴ i = π_cacl2/π_urea = 0.605/0.245 = 2.47

Answer is : van’t Hoff factor = i = 2.47

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.

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

When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.

• If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ − P) is the lowering of the vapour pressure.

If x₁ and x₂ are the mole fractions of a solvent and a solute respectively,

then x₁ + x₂ = 1

By Raoult’s law, P = x₁P_0. 

where P₀ is the vapour pressure of a pure solvent and P is the vapour pressure of the solution at given temperature.

∴ P/P₀ = x₁

1− P/P₀ = 1 − x₁

 = x₂

If P₀ − P = ΔP

then ΔP/ P₀  = x₂.

∴ΔP = P₀x₂

• Boiling point elevation and freezing point depression involve temperature changes, (ΔT_b and ΔT_f).
• Since molarity depends on temperature but molality is independent of temperature we use molality and not molarity in considering boiling point elevation and freezing point depression.

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.

The solubility of a solid solute depends upon temperature.

Generally rise in temperature increases the solubility. This is due to expansion of holes or empty spaces in the liquid solvent. Generally 10 °C rise in temperature, increases the solubility of solids two fold.

Figure shows the result of experimental determination of solubilities of some ionic solids in water at various temperatures.

Following are some experimental observations from Fig

• Dissolution process may be endothermic Or exothermic.
• Solubilities of NaBr, NaCl and KCl change slightly with temperature.
• Solubilities of KNO₃, NaNO₃ and KBr increase appreciably with increasing temperature.
• Solubility of Na₂SO₄ decreases with increase of temperature.

This variation in solubility with temperature can be used to separate the salts from the mixture by fractional clystallisation.

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.

• 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]

Given : Required concentration of O₂ = 3.8 mg/L =

Solubility of O₂ = 2.2 x 10^-3 mol L^-1

P = 1 atm

Partial pressure of O₂ needed = Po₂ = ?

S = K_H x Po₂

∴ K_H = S/ Po₂ = (2.2 x 10^-3)/1 mol L^-1 atm^-1

= 2.2 x 10^-3 mol L^-1 atm^-1

Po₂ = S/K_H =

= 0.05397 atm

Ans. Pressure needed = Po₂ = 0.05397 aim.

Given : Vapour pressure of pure solvent (water) = P₀ = 17 mm Hg

Weight of solvent = W₁ = 50 g

Weight of solute (urea) = W₂ = 2.8 g

Molecular weight of a solvent = M₁ = 18

Molecular weight of a solute (urea) = M₂ = 60 g mol^-1

\(\frac{P_0-P}{P_0} =\frac{W_2/M_2}{W_1/M_1}\)

\(\frac{17-P}{17} =\frac{2.8×18}{50×60}=0.0168\)

17-P = 17 x 0.0168

∴ P = 17 − 0.2856 = 16.7144 mm Hg

Vapour pressure of solution = 16.7144 mm Hg

Given : W₂ = 5 g cane sugar, W₁ = 100 − 5 = 95 g, M₂ = 342 g mol^-1, T_f1 = 271 K;

ΔT_f1 =273 − 271 = 21K, T_f = ?

W₂’ = 5 g glucose, W₁’ = 100 − 5 = 95 g,

M₂’ = 180 g mol^-1, ΔT_f2 = ?

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

K_f = \(\frac{ΔT_fW_1M_2}{W_2×1000}\)

ΔT_f2 = K_f x \(\frac{W'_2×1000}{W'_1M'_2}=\frac{13×5×1000}{95×180}=3.801 K\)

Freezing point of solution = T_f = = 273 − 3.801 = 269.199 K

Freezing point of solution = 269.2 K.

Given : W₁ = 50 g acetic acid

ΔT_b = 1.76 K

K_b = 3.07 K kg mol^-1

m = ?

ΔT_b = K_b x m

m = ΔT_b/ K_b = 1.76/3.07

= 0.5733 m

Molality of solution = 0.5733 m

Given: Density of a solution d = 1.063 g mL^-1 = 1.063, Osmotic pressure of solution  π = 12.16 atm, Temperature T = 25 °C = 298.15 K,

Freezing point of solution = T_f = –1.03 ⁰C, ∴ ΔT_f = 0 – (−1.03) =1.03 ⁰C

R = 0.08206 dm³ atm K^-1 mol^-1, K_f of water = 1.86 K kg mol^-1

ΔT_f = K_f m

∴ m = ΔT_f/ K_f = 1.03/ 1.86 = 0.554 mol kg^-1 = 0.554 M

π = CRT

12.16 = C x 0.08206 x 298.15

∴ C = 0.497 mol dm^–3 = 0.497 M

Mass of solvent W₁ = C/m = (0.497/0.554) x 1 dm³ = 0.897 kg = 897 g

Mass of Solution = 1.063 x g mL^–1 x 1000 mL = 1063g = 1.063 kg =1063 g

∴ Mass of Solute W₂ = 1.063 – 0.897 = 0.166 kg =166 g

m = \(\frac{W_2×1000}{W_1M_2}\)

∴ M₂ = \(\frac{W_2×1000}{W_1m}= \frac{166g×1000g\,Kg^{-1}}{897g×0.554mol\,kg^{-1}}\) = 334 g mol^-1

Molar mass of the compound = 334 g mol^-1

Given : 30% by mass of toluene (T) and 70% by mass of benzene (B).

W_T = 30 g, W_B = 70 g, P⁰_T = 36.7 mm Hg, P⁰_B = 118.2 mmHg

M_T  = 92 g mol^-1, M_B = 78 g mol^-1

P_T = ?, P_B = ?, P_soln = ?

n_T = W_T/ M_T = 30/92 = 0.3260 mol

n_B = W_B/ M_B = 70/78 = 0.8974 mol

Total number of moles = n_Total = n_T + n_B

= 0.326 + 0.8974

= 1.2234 mol

Mole fractions :

x_T = n_T/ n_Total = 0.326/1.2234 = 0.2665

x_B = 1-0.2665 = 0.7335

P_soln = x_T x P⁰_T + x_B x P⁰_B

= 0.2665 x 36.7 + 0.7335 x 118.2

= 9.780 + 86.7

= 96.48 mm Hg

Partial pressures :

P_T = x_T x P_soln

= 0.2665 x 96.48

= 25.71 mm Hg

P_B = x_B X P_soln

= 0.7335 x 96.48

= 70.77 mm Hg

Answer is : Total pressure = P_soln = 96.48 mm Hg

Partial pressures Toluene : P_T = 25.71 mm Hg

Partial pressures Benzene : P_B = 70.77 mm Hg

Given: T= 273 + 25 = 298 K, C= 0.1 m ≅ 0.1 M, K_f = 1.86 K kg mol^-1,

Per cent dissociation = 1.35, Freezing point = t_f = ?, p = ?

α = 1.35/100 = 0.013

CH₃COOH ↔ CH₃COO + H⁺

1 − α                  α                 α

i = 1 − α + α + α = 1 + α = 1 + 0.0135 = 1.0135

(i) ΔT_f = i x K_f x m

= 1.0135 x 1.86 x 0.1

= 0.1885 °C

Freezing point of solution = 0 − 0.1885 = − 0.1885 °C

(ii) π = iCRT

= 1.035 x 0.1 x 0.08206 x 298

= 2.53 atm

Answer is : (i) Freezing point of solution = − 0.1885 °C

                   (ii) Osmotic pressure = π = 2.53 atm.

Given : c = 0.15 m KCl ≅ 0.15 M KCl, ΔT_f = 0 − T_f = 0 − (− 0.510) = 0.510 °C

T = 273 K, K_f = 1.86 K kg mol^-1

i = ?; π = ?

ΔT_f  = i x K_f x m

∴ i = \(\frac{ΔT_f}{K_fm}\)

π = iCRT

= 1.828 x 0.15 x 0.08206 x 273

= 6.143 atm

Answer is i = 1.828, Osmotic pressure = p = 6.143 atm