Class XII Chem

Class XIIth Chemistry Electrochemistry 1

SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

Notes: -

Solutions

26

Lecture - I

Pre requisites

Various types of solutions e.g., solid in liquid, gas

in liquid, liquid in liquid, Solid in solid etc.

Knowledge of Henry’s law. ___________________________________________

Slide 1

In every day life we deal with mixtures or solutions.

Some properties associated with solutions are of great practical importance.

In this chapter we study all aspects for example the colligative properties.

The ways in which concentration can be expressed are as follows:-

__________________________________________ Slide 2

(i) Mass percentage or percent by mass:. massof solute

100massof solution

massof solute 100

volumeof solution density of solution

(ii) Percent by volume volumeof solute 100

volumeof solution

Slide 3

(iii) Percent mass by volume = massof solute 100

volumeof solution

(iv) Strength or conc. =

–1massof solute gms

g Lvolume of solution inL.

(v) Parts per million = 6massof solute10

massof solution

(vi) Mole fraction:

mole fraction of solute (B) = B

BB

A

nX

n n

mole fraction of solvent (A) = B

AA

A

nX

n n

And XA + XB = 1. __________________________________________

Slide 4

(vii) Molality (m) = molesof solutekgof the solvent

B

B A

W 1000m

M W

WB and MB are wt. and molar mass of solute. WA is wt. in grams of solvent.

(viii) –1molesL

molesof soluteMolarity M

Volume of solutioninL.

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164 Solutions Chemistry (XII)


Notes: -

(ix) –1g equivL

gram equivalents or

equivalents of soluteNormality N

Volume of soluioninL.

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Slide 5 (A)

Solubility of Gases

The solubility of a gas at a given temperature is directly proportional to the pressure at which it is dissolved.

(xB) mole fraction is taken as a measure of its solubility and p is the partial pressure of gas in equilibrium with the solution.

xB

p or p

xB

or p = KHxB. __________________________________________

Slide 5 (B)

Here KH is Henry’s constant.

Different gases have different KH values at the same temperature so KH is a function of the nature of gas.

Higher the value of KH at a given pressure the lower is the solubility of gas in liquid.

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Slide 6

With rising temperature the solubility of gases decreases in liquids.

In solution the gas molecules are present in liquid phase and the process of dissolution can be considered as similar to condensation and heat is evolved in the process.

The process involves dynamic equilibrium.

Since dissolution is exothermic the solubility will decrease with increase in temperature (Le chatelier’s rule)

Slide 7

Illustration: 1

Henry’s law constant for the molality of methane in benzene at 298 K is 4. 27 × 105 mm Hg. Calculate the solubility of methane in benzene, at 298 K under a pressure of 760 mm Hg, in terms of its mole fraction in solution. __________________________________________

Slide 8

Vapour Pressure:

The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure of the liquid. __________________________________________

Slide 9 (A)

Raoult’s law

(A) For Non-volatile solute

The vap. Pressure of solution containing non-volatile solute is less than the vapour pressure of pure solvent.

It p1 is the vapour pressure of solvent in solution and 1p

is the vapour pressure of pure

solvent and x1 is the mole fraction of solvent in solution then lowering of vapour pressure is equal to 1 1 1 1 1p – p p – p x

1 1 1since p x p

1 1or p 1– x

__________________________________________

Slide 9 (B)

or 1 1 1 1 2p – p p p x

where x2 is the mole

fraction of solute in solution.




Notes: -

or relative lowering of vapour pressure

1 1 22

1 21

p – p nx

n np

Where n1 and n2 are the moles of solvent and solute respectively.

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Slide 10

For dilute solutions neglecting n2 in the denominator we get

2 2 11 1

1 1 2 1

n W Mp – pp n M W

Where W1 and W2 are

the masses and M1 and M2 are the molar masses of solvent and solute respectively.

Knowing other quantities the molar mass of solute can be calculated

This is known as Raoults law. That is for a solution containing non-volatile solute, the relative lowering of vapour pressure is equal to mole fraction of the solute.

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Slide 11

Illustration: 2

Vapour pressure of water at 293 K is 17.535 mm of Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water. ___________________________________________

Slide 12 (A)

(B) Solution of two volatile liquids Two volatile liquids A and B. PA

XA Or PA = PA°XA

PB

XB Or PB = PB°XB

PA° and PB° are vapour pressure of pure components A and B respectively.

__________________________________________

Slide 12 (B)

PA and PB are partial vapour pressures of A and B in solution

__________________________________________

Slide 13

According to Dalton’s law of Partial Pressure P = PA + PB

= PA°XA + PB°XB ( XA + XB = 1) = PA°(1 – XB) + PB°XB

= PA° – PA°XB + PB°XB

B A B AP P – P X P




Notes: -

Slide 14

Illustration: - 3

Heptane and octane form an ideal solution. At 373 k the vapour pressure of two liquids are 105.2 K Pa and 46.8 k Pa respectively. What will be the vapour pressure of a mixture having 26.0 g heptane and 35.0 g of octane? __________________________________________

Slide 15

Illustration:- 4

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively at 350 K. Find out the composition of the liquid mixture in terms of mole fractions if the total vapour pressure is 600 mm Hg. Also find out the mole fractions of the two liquids in vapour phase. __________________________________________

Slide 16

The solutions which obey Raoult’s law over the entire range of concentration are known are Ideal solutions.

Two more properties should hold true for ideal solutions. i.e., Hmix = 0 and Vmix = 0.

It means no heat should be evolved or absorbed and there should be no volume change on mixing the two components.

A third necessary condition for the solution to be ideal is, A – B = A – A and B – B.

Slide 17 (A)

A perfectly ideal solution is rare but a mixture of n-hexane and n-heptane; bromoethane and chloroethane; benzene and toluene etc, are solutions which are nearly ideal in behaviour.

If the vapour pressure of solution is more than the value predicted by Raoult’s law, it exhibits a positive deviation as depicted in the following figures:

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Slide 17 (B)

Vapour pressureof solution

Vap

our

pres

sure

p1 p2

Vapour pressureof solution

Vap

our

pres

sure

p1p2

x = 01

x = 12

Mole fraction x = 11

x = 02x1

x = 01

x = 12

Mole fraction x = 11

x = 02x1

x2 x2

(a) (b) __________________________________________

Slide 18

Molecular interactions are responsible for those deviations.

For positive deviation A – B < A – A or B – B.

Examples are ethanol and acetone or carbon disulphide and acetone.

For negative deviation A – B > A – A or B – B.

Phenol and aniline; chloroform and acetone are examples of this type.




Notes: -

Slide 19

Some liquid solutions form azeotropes which are binary mixtures.

These have same composition in liquid and vapour phase and boil at a constant temperature.

It is not possible to separate the components by fractional distillation.

Azeotropes can be minimum or maximum boiling.

__________________________________________

Slide 20

Ethanol-water mixtures show a large positive deviations and form a minimum boiling azeotrope.

This azeotrope contains approx 95% ethanol and 5% water,

Similarly nitric acid and water have less vapour pressure than the ideal value and show a negative deviation and form a maximum boiling azeotrope.

It contains almost 68 percent nitric acid and 32% water by mass.




Notes: -

Lecture Assignment

1 Mark Questions

Q1. Give one example of

(i) Gas in liquid type solution (ii) Solid solution

Q2. Define molarity and molality of solutions. Q3. Define the term mole fraction giving one

example. Q4. What do you understand by mass percent-

age? Q5. Why does the molality of a solution remain

unchanged with temperature?

2 Mark Questions

Q6. State Henry’s law. Q7. List the factors on which the Henry’s

constant depends. Q8. What are ideal liquid solutions?

3 Mark Questions

Q9. The vapour pressure of a pure liquid A is 30

mmHg at 320 K. The vapour pressure of this liquid in a solution with liquid B is 28 mmHg at the same temperature. What is the mole fraction of B in the solution if it obey’s Raoult’s law.

Q10. What properties depend on the number of particles dissolved in solution rather than their chemical identity? Name two such properties.

Q11. Define osmosis. Q12. Henry’s law constant for the molality of

methane in benzene is 4.27 × 105 mmHg. Calculate the solubility of methane in benzene at 298 K under 760 mmHg.




Notes: -

Lecture - II

Pre requisites

Raoults law.

Deviations from Raoults law.

__________________________________________

Slide 21

Elevation in boiling point:

The temperature at which the vapour pressure becomes equal to the atmospheric pressure is called boiling point.

On dissolving a non-volatile solute in the solvent the boiling temperature of the solution goes up.

The following figure depicts the boiling point of pure solvent and solution.

__________________________________________

Slide 22

1 atm.Atmospheric pressure

A

C

Solvent

Solution

B D

v.p.

Temp

Tb

T0 T1

Slide 23 (A)

Experiments have shown that for dilute solution

Tb

m

or Tb = Kb.m

2b

bvap.

R.M.TK

1000. H

is called molal elevation

constant or molal ebullioscopic constant.

Here R is gas constant, M is molar mass of solvent, Tb is boiling point of pure solvent, Hvap. is enthalpy of vapourisation.

__________________________________________

Slide 23 (B)

Calculation of Molecular Mass

B

B A

W 1000m

M WB

Here WB and MB and are weight and molar mass of solute and WA is weight of solvent

Bb b b

B A

b BB

A b

W 1000T K m K

M W

K W 1000or M

W T

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Slide 24

Depression in freezing point

Freezing point is defined as the temperature at which its solid and liquid phase have the same vapour pressure.




Notes: -

SolidSolven

t

A

B

A’D P’

P

C

C’

Tf

T1 T0

v.p.

Temp

Solvent

Solution

__________________________________________

Slide 25 (A)

Tf

m Tf = Kf × m

Kf = molal depression constant or cryoscopic constant. 2f

ffus.

R.M.TK

1000. H

Here R is gas constant, M is molar mass of solvent, Tf

is freezing point of solvent and Hf is enthalpy of fusion.

Slide 25 (B)

Calculation of Molecular Mass

B

B A

Bf f

B A

f BB

A f

W 1000m

M W

W 1000or T K

M W

K W 1000or M

W T

___________________________________________ Slide 26

Illustration: - 5

A very small amount of a non-volatile solute (that does not dissociate) is dissolved in 56.8 cm3 of benzene (density 0.889g cm-3). At room temperature vapour pressure of this solution is 98.88 mm Hg while that of benzene is 100 mm Hg. Find the Molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene?




Notes: -

Lecture Assignment

2 Mark Questions

Q1. Calculate the mole fraction of ethyl alcohol

(C2H5OH) and water in a solution in which 46 g of ethyl alcohol and 90 g of water have been mixed.

Q2. A 250 cm3 solution of sodium sulphate contains 3.01 × 1022 sodium ions. What is the molarity of the solution?

Q3. When does a solution deviate from the ideal behaviour?

Q4. Define azeotropic mixture. Q5. What are the different types of azeotropes?

Give one example of each. Q6. Which will have higher boiling point, 0.1 M

NaCl or 0.1 M BaCl2 solution in water? Q7. Which has the highest freezing point:

(a) 1 M glucose (b) 1 M NaCl (c) 1 M CaCl2

(d) 1 M AlF3?

3 Mark Questions

Q8. What happens when two solutions of the same solvent are separated by a semipermeable membrane?

Q9. The vapour pressure of pure liquids A and B are 450 and 700 mmHg respectively at 350 K. Find out the composition of the liquid mixture

if total vapour pressure is 600 mmHg. Also, find the composition of the vapour phase.

Q10. 100 g liquid A (molar mass 140 g/mol) was dissolved in 1000 g of liquid B (molar mass 180 g/mol). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the V.P. of pure liquid A and its vapour pressure in the solution if the total V.P. of solution is 475 torr.

Q11. A solution was prepared by dissolving 0.915 g of sulphur, S8, in 100.0 g of acetic acid, CH3COOH. Calculate the boiling point of the solution. The boiling point of pure acetic acid is 118.5°C. The value of kb is 3.08°C kg mol–1.

Q12. Solution A is obtained by dissolving 1g of urea in 100 g of water and solution B is obtained by dissolving 1g of glucose in 100 g of water. Which solution will have a higher boiling point and why?

5 Mark Questions

Q13. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 gram of polymer of molar mass 185,000 in 450 ml of water at 37°C.

Q14. 0.6 mL of acetic acid (CH3COOH) having density 1.06 g mL–1 is dissolved in 1 L of H2O. The depression in freezing point was observed to be 0.0205°C. Calculate the Van’t Hoff factor and the dissociation constant of acid. (kf = 1.86 K kg mol–1)




Notes: -

Lecture - III

Pre requisites

Knowledge of various colligative properties.

Application in determining various constants.

__________________________________________

Slide 27

Osmosis and Osmotic Pressure

The osmotic pressure of the solution at a particular temperature may be defined as the excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semi-permeable membrane.( )

Isotonic: Two solutions having same osmotic pressure.

Hypertonic solution: Solution having higher osmotic pressure than some other is said to be hypertonic w.r.t. other.

__________________________________________

Slide 28

Hypotonic solution: Solution having lower osmotic pressure relative to some other. Reverse osmosis: Purification of sea water. Vant Hoff’s equation:

V nRT

V

nRT CRT

Calculation of molecular mass

VBn

RT

BM VBW RT

or V

BB

W RTM

Slide 29

Abnormal Colligative Properties

Colligative properties depend on the number of particles present in solution.

If the solute undergoes dissociation or association in solution, the number of particles change and the observed values of the colligative properties would also change.

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Slide 30

Vant Hoff introduced a factor i known as van’t Hoff factor to account for the extent of dissociation or association. Thus

Totalno.of particlesafter disso.or asso.i ;

Totalno.of particlesbeforedisso.or asso.

observedcolligativepropertyi ;

Calculatedcolligativeproperty

Normalmolarmassi

observedmolarmass.

__________________________________________

Slide 31

In the case of association or dissociation of solute particles the expressions are:

2

1

b b

f f

2

npi

p n

T iK .m

T iK .m

in RT / V

n2 = moles of solute and n1 = mole of solvent




Notes: -

Slide 32

Illustration: - 6

Calculate the mass of a non-volatile solute (M = 40) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. __________________________________________

Slide 33

Illustration: - 7

The elements ‘A’ and ‘B’ form compounds AB2 and AB4. When dissolved in 20g of benzene 1.0 g of AB2

lowers the freezing point by 2.3 K where as 1.0 g of AB4 lowers it by 1.3K. Kf for benzene is 5.1 K. Kg mol–1. Calculate atomic mass of ‘A’ and ‘B’. __________________________________________

Slide 34

Question 8

A 1.0 m aqueous solution of a monobasic acid (M = 20) freezes at –1.9°. Kf of water is 1.86 K.Kg. mol-1. If the density of solution is 1.12 g/mL, find out Ka of the acid

Slide 35

Illustration: - 9

Urea dissolved in 250 g of water freezes at –0.744°C. Solution was cooled when some ice was formed. Solution was decanted and heated to 100°C when the vap.pr. was found to be 752.7 mm of Hg. How much ice was formed and find out the temperature to which solution was cooled? Kf (H2O) = 1.86 __________________________________________

Slide 36

Illustration: - 10

Calculate the amount of NaCl which must be added to 100g water so that freezing point is depressed by 2 K. For water Kf = 1.86 K kg mol–1




Notes: -

Lecture Assignment

2 Mark Questions

Q1. Define formality. Q2. If

is the osmotic pressure and V is the

volume in litres of solution containing one gm mol of solute, what would be the value of V at 273 K.

Q3. What is relation between absolute temperature and osmotic pressure?

Q4. What do you understand by “colligative properties”?

Q5. Define osmotic pressure.

3 Mark Questions

Q6. Why the osmotic pressure measurement is preferred for the molar mass determination of macromolecules over other colligative properties?

Q7. Before giving intravenous injection what care is generally taken and why?

Q8. When fruits and vegetables that have dried are placed in water, they slowly swell and return to original form, why? Would a temperature increase accelerate the process? Explain.

Q9. The solubility of Ba(OH)2.8H2O in water at 288 K is 5.6 g per 100 g of water. What is the molality of the hydroxide ions in the saturated solution of barium hydroxide at 288 K? (Atomic Mass: Ba = 137, O = 16, H = 1).

5 Mark Questions

Q10. (i) What is osmotic pressure and how is it

related to the molecular mass of non- volatile substance?

(ii) What advantage the osmotic pressure method has over the elevation of boiling point method for determining molecular masses?

Q11. What is the molar concentration of solute particles in the human blood if the osmotic pressure is 7.2 atm at the body temperature of 37°C? (R = 0.0821 L atm K–1 mol–1)

Q12. How much glucose must be dissolved in one litre of an aqueous solution so that its osmotic pressure is 2.57 atm at 300 K? (R = 0.0821 L atm K–1 mol–1)

Q13. Calculate the molar concentration of urea solution if it exerts an osmotic pressure of 2.45 atm at 300 K. (R = 0.0821 L atm mol–1 K–1)




Notes: -

Chapter Assignment

2 Mark Questions

Q1. For which of the following van’t Hoff factor can

not be greater than unity: (a) K4[Fe(CN)6] (b) AlCl3

(c) NH2CONH2

(d) KNO3? Q2. Osmotic pressure of a solution containing 7g

of dissolved protein per 100 cm3 of the solution is 20 mm Hg at 310 K. calculate the molar mass of protein. [R = 0.082 L atom K–1 mol–1]

Q3. An electrolyte AB is 50% ionized in aqueous solution. Calculate the freezing point of 1 molal aqueous solution.

Q4. What is molar concentration of solute particles in a human blood if the osmotic pressure is 7.2 atm at the body temperature of 37°C? (R = 0.0821 L atm K–1 mol–1)

Q5. In aqueous solution, NaCl is completely ionized into Na+ and Cl– ions. Compute the osmotic pressure of 0.255 M solution of NaCl at 300 K.

3 Mark Questions

Q6. 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate the (i) Freezing point depression and (ii) The freezing point of the solution. (kf for

water = 1.86 K kg mol–1) Q7. In a solution of urea, 3.0 g of it is dissolved in

100 ml of water. What will be the freezing point of this solution? State the approximation made if any. [Kf for water = 1.86 K kg mol–1, molar mass of urea = 60 g mol–1]

Q8. A solution of 3.800 g of sulphur in 100 g of CS2 (boiling point = 46.30°C) boils at 46.66°C. What is the formula of sulphur molecules in this solution? (Atomic mass of sulphur = 32 g mol–1 and Kb for CS2 = 2.40 K kg mol–1)

Q9. Calculate the osmotic pressure of a solution obtained by mixing 100 cm3 of 0.25 M solution of urea and 100 cm3 of 0.1 M solution of cane sugar at 293 K. [R = 0.082 L atm mol–1 K–1]

Q10. An aqueous solution of glucose is made by dissolving 10 g of glucose (C6H12O6) in 90 g of water at 303 K be 32.8 mm Hg, what would be the vapour pressure of the solution?

Q11. The solubility of Ba(OH)2.8H2O in water at 288 K is 5.6 g per 100 g of water. What is the molality of the hydroxide ions in saturated solution of Ba(OH)2.8H2O at 288 K?

Q12. Calculate molarity and molality of a 13% solution (by weight) of sulphuric acid. Its density is 1.020 g cm–3. (Atomic mass H = 1, O = 16, S = 32 u)

Q13. 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.103 bar?

Q14. 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate: (a) Freezing point depression, (b) Freezing point of the solution (Kf for water

= 1.86 K kg mol–1; Atomic masses: C = 12, H = 1, O = 16 u)

Q15. A solution containing 18 g of non-volatile solute in 200 g of water freezes at 272.07 K. Calculate the molecular mass of solute. (Kf = 1.86 K/m)




Notes: -

5 Mark Questions

Q16. An antifreeze solution is prepared from 222.6

g of ethylene glycol (C2H6O2), and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g/mL, then what is the molarity of the solutions?

Q17. If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 L of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293 K is 76.48 K bar.

Q18. At 40°C, methanol and ethanol form nearly an ideal solution with the pressure given by p = 119 x + 135 Where x is the mole fraction of methanol and P is the vapour pressure in torr. What are the vapour pressures of the pure methanol and pure ethanol at this temperature?

Q19. What is meant by positive and negative deviations from Rault’s law and how is the sign of Hsol related to positive and negative deviations from Rault’s law?

Q20. Ethylene glycol (C2H6O2) is used as antifreeze and is non-volatile and has density of 1.109 g/cm3. How many litres of ethylene glycol should be used per litre of water to protect it from freezing at – 3.7°C?

Q21. The molal freezing point depression constant of benzene (C6H6) is 4.90 K kg mol–1. Selenium exists as a polymer of the type Sex. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower than for pure benzene. Deduce the molecular formula of selenium. (Atomic mass of Se = 78.8 g mol–1)

Q22. Assuming complete ionization, calculate the expected freezing point of solution prepared by dissolving 6.00 g of Glauber’s salt, Na2SO4.10H2O in 0.1 kg of H2O. (Kf for H2O = 1.86 K kg mol–1) [At. mass of Na = 23, S = 32, O = 16, H = 1 u]


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Class XII Chem

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