Class XII Chem

Class XIIth

Chemistry

Electrochemistry

1

Class XII

26 Solutions 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:. mass of solute 100 mass of solution mass of solute 100 volume of solution density of solution

(ii) Percent by volume

volume of solute 100 volume of solution

Slide 3 mass of solute 100 volume of solution

(iii) Percent mass by volume = (iv) Strength or conc. = mass of solute gms volume of solution inL.

(v) Parts per million =

g L–1

mass of solute 106 mass of solution

(vi) Mole fraction: mole fraction of solute (B) =

nB nA

mole fraction of solvent (A) =

nB nA

nA

nB

XB XA

And XA + XB = 1. __________________________________________ Slide 4 (vii) Molality (m) =

moles of solute kg of the solvent

WB 1000 MB WA WB and MB are wt. and molar mass of solute. WA is wt. in grams of solvent. m

(viii) Molarity M –1

moles L

moles of solute Volume of solution inL.

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(ix)

Normality N g equiv L–1

gram equivalents or equivalents of solute Volume of soluioninL.

__________________________________________ 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. __________________________________________ 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)

Chemistry (XII) 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 nonvolatile solute is less than the vapour pressure of pure solvent. It p1 is the vapour pressure of solvent in solution and p1 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 p1 – p1 p1 – p1 x1 since p1 x1p1 or

p1 1 – x1

__________________________________________ Slide 9 (B) or p1 – p1

p1

p1 x 2 where x2 is the mole

fraction of solute in solution.

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165

Solutions or relative lowering of vapour pressure p1 – p1 n2 x2 n1 n2 p1

Where n1 and n2 are the moles of solvent and solute respectively. __________________________________________ Slide 10

Chemistry (XII) 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

For dilute solutions neglecting n2 in the denominator we get n2 W2 M1 p1 – p1 Where W1 and W2 are p1 n1 M2 W1 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. __________________________________________ 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 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 P

PB – PA XB

PA

Slide 12 (A) (B) Solution of two volatile liquids Two volatile liquids A and B. PA XA Or PA = PA°XA

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Solutions

Chemistry (XII)

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

Slide 17 (A) A perfectly ideal solution is rare but a mixture of nhexane 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: __________________________________________ Slide 17 (B)

Illustration:- 4

Vapour pressure

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

Vapour pressure of solution

p1

x1 = 0 x2 = 1

p2

Mole fraction x1 = 1 x2 = 0 x1 x2

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.

(a)

Vapour pressure of solution Vapour pressure

166

p2

x1 = 0 x2 = 1

p1

Mole fraction x1 = 1 x2 = 0 x1 x2

(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.

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167

Solutions

Chemistry (XII)

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.

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Solutions

Chemistry (XII)

Lecture Assignment 1 Mark Questions

3 Mark Questions

Q1.

Q9.

Q2. Q3. Q4. Q5.

Give one example of (i) Gas in liquid type solution (ii) Solid solution Define molarity and molality of solutions. Define the term mole fraction giving one example. What do you understand by mass percentage? Why does the molality of a solution remain unchanged with temperature?

2 Mark Questions Q6. Q7. Q8.

State Henry’s law. List the factors on which the Henry’s constant depends. What are ideal liquid solutions?

Q10.

Q11. Q12.

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

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169

Solutions

Chemistry (XII)

Lecture - II Pre requisites

Slide 23 (A) Experiments have shown that for dilute solution Tb m or Tb = Kb.m R.M.Tb2 Kb is called molal elevation 1000. Hvap.

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.

v.p.

Atmospheric pressure B D nt e v l o S n A utio Tb So l C

Temp

T0 T1

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 m

WB MB

1000 B WA

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

Kb m Kb

Kb

WB MB

1000 WA

WB 1000 WA Tb

__________________________________________ 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.

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170

Solutions

Chemistry (XII)

C nt o lv e

v.p.

A

nt S lid olve B o tio n S S P So lu A’ D P’

Slide 25 (B)

C’

Tf T1 T0 Temp __________________________________________ Slide 25 (A) Tf m Tf = Kf × m Kf = molal depression constant or cryoscopic constant. R.M.Tf2 Kf 1000. Hfus. Here R is gas constant, M is molar mass of solvent, Tf is freezing point of solvent and Hf is enthalpy of fusion.

Calculation of Molecular Mass m

WB MB

or Tf

1000 WA Kf

WB MB

1000 WA

WB 1000 WA Tf ___________________________________________ Slide 26 or MB

Kf

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?

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Solutions

Chemistry (XII)

Lecture Assignment 2 Mark Questions Q1.

Q2.

Q3. Q4. Q5. Q6. Q7.

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. A 250 cm3 solution of sodium sulphate contains 3.01 × 1022 sodium ions. What is the molarity of the solution? When does a solution deviate from the ideal behaviour? Define azeotropic mixture. What are the different types of azeotropes? Give one example of each. Which will have higher boiling point, 0.1 M NaCl or 0.1 M BaCl2 solution in water? Which has the highest freezing point: (a) 1 M glucose (b) 1 M NaCl (c) 1 M CaCl2 (d) 1 M AlF3?

Q10.

Q11.

Q12.

5 Mark Questions Q13.

3 Mark Questions Q8.

Q9.

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

Q14.

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. 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)

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Chemistry (XII)

Lecture - III Slide 29

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 n RT CRT V Calculation of molecular mass nB RT V WB RT WB RT MB or V MB V

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. __________________________________________ 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 particlesbefore disso.or asso.

observedcolligativeproperty ; Calculatedcolligativeproperty Normalmolar mass i observedmolar mass. __________________________________________ Slide 31 i

In the case of association or dissociation of solute particles the expressions are: n p i 2 p n1 Tb

iK b .m

Tf

iK f .m in2RT / V

n2 = moles of solute and n1 = mole of solvent

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Chemistry (XII) Slide 35

Slide 32 Illustration: - 6

Illustration: - 9

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

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: - 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

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

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

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Solutions

Chemistry (XII)

Lecture Assignment 2 Mark Questions Q1. Q2.

Q3. Q4. Q5.

Define formality. 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. What is relation between absolute temperature and osmotic pressure? What do you understand by “colligative properties”? Define osmotic pressure.

5 Mark Questions Q10.

Q11.

3 Mark Questions Q6.

Q7. Q8.

Q9.

Why the osmotic pressure measurement is preferred for the molar mass determination of macromolecules over other colligative properties? Before giving intravenous injection what care is generally taken and why? 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. 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).

Q12.

Q13.

(i) What is osmotic pressure and how is it related to the molecular mass of nonvolatile substance? (ii) What advantage the osmotic pressure method has over the elevation of boiling point method for determining molecular masses? 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) 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) 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)

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175

Solutions

Chemistry (XII)

Chapter Assignment 2 Mark Questions Q1.

Q2.

Q3.

Q4.

Q5.

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

Q8.

Q9.

Q10.

Q11.

Q12.

Q13.

3 Mark Questions Q6.

Q7.

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) 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]

Q14.

Q15.

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) 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] 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? 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? 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) 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? 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) 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)

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176

Solutions

Chemistry (XII)

5 Mark Questions Q20. Q16.

Q17.

Q18.

Q19.

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? 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. 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? 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?

Q21.

Q22.

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? 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) 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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Notes: -

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