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SOLUTIONS

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SOLUTIONSSOLUTIONS:Homogeneous mixture of two or more components.

Binary solution:Consisting of only two components.

Solute & Solvent: If two components are in different physical states. The component which is in the same physical state as

that of solution is solvent & other is solute. e.g. solution of sugar in water is a liquid. Solution & water are in same physicalstate, so water is solvent and sugar is solute.

If two components are in same physical state, the component in larger quantities is solvent and one in smaller quantities is

solute.

TYPES OF SOLUTIONS1. Gas (solute) in gas (solvent): Air i.e. O2in N2.2. Liquid (solute) in gas (solvent): Chloroform mixed with N2gas.3. Solid (solute) in gas (solvent): Camphor in N2gas.4. Gas (solute) in liquid (solvent): CO2dissolved in water.5. Liquid (solute) in liquid (solvent): Ethanol in water.6. Solid (solute) in liquid (solvent): Sugar in water.7. Gas (solute) in solid (solvent): H2in Pd.8. Liquid (solute) in solid (solvent): Hg/Na9. Solid (solute) in solid (solvent): AlloysMethods of expressing concentration of a solution1. Mass percentage (w/w):

Mass% =of a component

100mass of solution

Mass

Total

2. Mass-Volume percentage (w/v)Mass-Volume% = 100

mlinsolutionofVolume

componentaofMass

3. Volume percentageVolume% = 100

solutionofvolumeTotal

componentofVolume

4. Parts per million (ppm) = 610solutionofpartsofnumberTotal

soluteofpartsofNumber

5. Normality of solution (N)N =

litresinVolume

soluteofmassequivalentgmofNumber

EQUIVALENT MASS OF A SOLUTE

(i) Eq. mass of acid =acidofBasicity

.M.M

(ii) Eq. mass of base =baseofAcidity.M.M

(iii) Eq. mass of oxidizing or reducing agent =moluecule1bylostorgainedelectronsof.No

.M.M

(iv) Eq. mass of simple salt =anionsorcationsalloverchargeTotal

.M.M

6. Molarity of a solution (M)


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M =litresinsolutionofVolume

soluteofmolesof.No

7. Molality of a solution (m)m =

KginsolventofassM

soluteofmolesof.No

8. Mole fraction of a component =21

1

nn

n

n1= Number of moles of one component

n2= Number of moles of 2ndcomponent

SOME IMPORTANT RELATIONSHIPS

1. N = x M : x =mass.Eq

.M.M

2. Normality equation: N1V1= N2V23. Molarity equation : M1V1= M2V2

x1M1V1= x2M2V2

4. Number of gm equivalents of solute = N volume in litres5. Number of mgm equivalents of solute = N volume in ml6. Number of gm moles of solute = M volume in litres7. Number of mgm mole of solute = M volume in ml8. Relation between molarity (M) and molality (m) of a solution

m =1000/MMd

M

2, m = molality

M = molarity

d = density of solution

M2= M.M. of solute

9. Relation between molality (m) and mole fraction (x2)m =

11

2

Mx

x1000 x1= mole fraction of solvent

x2= mole fraction of soluteM1= M.M. of solvent

10.Relation between molarity (M) and mole fraction of solute (x2)M =

2211

2

MxMx

xd1000

ASSIGNMENT

1. Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of ethylene glycol by mass.[0.067]

2. Calculate molarity of a solution containing 5gm of NaOH in 450 ml solution [0.28 M]3. Calculate molality of 2.5 gm of ethanoic acid (CH3COOH) in 75 gm of benzene [0.55 m]4. Calculate the mass % of benzene and carbon tetrachloride if 22 gm of benzene is dissolved in 122 gm of carbon

tetrachloride. [benzene =15.28%, CCl4=84.72%]

5. Calculate the mole fraction of benzene in solution containing 30% by mass of CCl4.[0.82]6. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 m aqueous solution[36.95 g]7. Calculate

(a) molality, [1.50]

(b) molarity [1.448]

(c) mole fraction [0.026]

of KI if the density of 20% (w/w) aqueous KI is 1.202 gm/ml. [c mol8. Concentrated HNO3used in the lab is 68% by mass in aqueous solution. What is the molarity of the acid if

density of solution is 1.504 gm/ml. [16.23 M]


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9. A solution of glucose in water is labeled as 10% (w/w). What would be the molality, molarity and mole fraction ifthe density of solution is 1.2 gm/ml [M=0.667,m 0.617 mole fraction= 0.011 ]

10. How many ml of 0.1 M HCl are required to react completely with 1 gm mixture of Na 2CO3and NaHCO3containing equimolar amounts of both. [157.5ml]

11. A solution is obtained by mixing 300 gm of 25% solution and 400 gm of 40% solution by mass. Calculate themass % of the resulting solution. [33.57%]

12. An antifreeze solution is prepared from 222.6 gm of ethylene glycol (C2H6O2) and 200 gm of water. Calculatemolality of solution. If the density of the solution is 1.072 gm/ml, then what shall be molarity of solution?[m=17.95 N 9.1]

13. A sample of drinking water was found to be severely contaminated with chloroform supposed to be a carcinogen.The level of contamination was 15 ppm (by mass)

(i) Express this in % by mass [% by mass=1.5x10-3, m = 1.255x 10-4](ii) Determine the molality of solution in water

14. A 6.90 M solution of KOH in water contains 30% by mass. Calculate density of KOH solution. [1.288g/ml]15. Define the term molarity and molality of a solution. How does a change in temperature effects the molarity &

molality of the solution.

SOLUBILITY:Maximum amount of solute that can be dissolved in the given amount of solvent to give a saturated solution

at a given temperature. Solubility depends upon nature of solute, nature of solvent, temperature & pressure.

Solubility of a solid in a liquid: Principle of solubility is like dissolves like i.e. a solute dis solves in a solvent if inter-molecular interactions are similar in the two. Polar solutes dissolve in polar solvents, e.g. Sodium chloride & barium chlorideare polar, so soluble in polar solvents like water. Sugar, glucose and alcohol dissolve in water due to hydrogen bonding.

Naphthalene and anthracene are non-polar, so soluble in non-polar solvents like benzene.Effect of temperature on solubility: This follows Le-Chatelier principle. If the dissolution of a solid in a liquid is

exothermic then increase in temperature will decrease the solubility. If the dissolution is endothermic then increase in

temperature will increase the solubility.

Effect of pressure: There is no significant effect of change in pressure on the solubility of a solid in a liquid.

Solubility of a gas in a liquid

Factors effecting solubility of a gas in a liquid1.Pressure:

Henrys Law:At a given temperature the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Mole fraction of the gas in a solution is directly proportional to partial pressure of the gas over the solution.X (g) p (g)

= KH p (g) In this equation unit of KHin bar1or atm1.

Or

p (g) X (g)= KH X (g) In this equation units of KHin bar or atm.

X (g) = Mole fraction of gas in solution

p (g) = Partial pressure of gas over solution

KH= Henrys constant

If value of KHis in bar or atm higher is the value of KHlower will be solubility of gas and with increase in temperature value

of KHwill increase.

Applications of Henrys law(i) To increase the solubility of CO2in soft drinks the bottles are sealed under high pressure.(ii) For respiration of deep sea divers air is diluted with helium in place of air.(iii) Anoxia is created at high altitudes as partial pressure of oxygen is less, this leads to low concentration of

oxygen in blood and tissues.

2. Effect of temperatureWith increase in temperature the solubility of a gas in a liquid decreases.

ASSIGNMENT


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1. If N2gas is bubbled through water at 293 K, how many milimoles of N 2gas would dissolve in 1 litre of water.Assume that N2exerts a partial pressure of 0.987 bar. Given that Henrys law constant for N2gas at 293 K is 76.48

kbar. [0.716 m moles]

2. H2S is a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H2S gas in water atSTP is 0.195 m. Calculate the Henrys law constant.[285.71 atm]

3. Henrys law constant for CO2in water is 1.67 108Pa. at 298 K. Calculate the quantity CO2in 500 ml of sodawater when packed under 2.5 atm, CO2pressure at 298 K. [27.78 moles]

4. The partial pressure of ethane over a solution containing 6.56 103gm of ethane is 1 bar. If the solution contains5.00 102gm of ethane, then what shall be partial pressure of the gas? [7.62 bar]

5. Henrys law constant for the molarity of methane in benzene at 298 K is 4.27 10 5mm Hg. Calculate the solubilityof methane in benzene at 298 K under 760 mm Hg. [mol fraction of CH 4=1.78 x 10

-3]

6. State Henrys law correlating the pressure of a gas and its solubility in a solvent and mention two applications of thelaw.

7. At the same temperature hydrogen is more soluble in water than helium. Which of them will have higher value ofKH and Y.

Vapour pressure of a liquidEquilibrium pressure exerted by the vapours of the liquid over the liquid in equilibrium with the liquid.

Raoults law:At a given temperature the vapour pressure due to a volatile component of a solution is directly proportional to mole fraction

of this component in the solution.In case of a binary solution of two volatile liquids A and B if partial vapour pressure of A is p 1and partial vapour pressure of

B is p2and vapour pressures of pure liquids A and B are p1oand p2

oand respective mole fractions are x1and x2.

For component A p1 x1

= p1o x1

For component B

p2= p2o x2

so total pressure will be equal to p1+ p2

p total = x1p1o+ x2p2

o

= (1x2) p1o+ x2p2

o

= p1o+ (p2

op1

o) x2

Raoults law for the solution of a non-volatile solute in a liquid

At a given temperature the vapour pressure of a solution of a non-volatile solute is directly proportional to the mole fractionof solvent in the solution.

p1 x2

p1= po x2 p1is vapour pressure of solution,

pois vapour pressure of pure liquid

x2is mole fraction of liquid

Let the mole fraction of solute in the solvent is x1

So mole fraction of solvent x2= 1x1

So p1= po (1x1)

o

1

p

p= (1x1)

x1= 1o

1

p

p=

o

1o

p

pp

i.e. relative lowering of vapour pressure of the solution of a non-volatile solute is equal to mole fraction of solute.

IDEAL SOLUTIONS The solutions which obey Raoults law at entire range of concentration are known as ideal solutions. When such solutions

are formed there is neither any heat change nor any volume change.mixH = 0, mixV = 0 mixS > 0Ideal solutions are formed when inter molecular attractions between the molecules of two liquids are of the same order.Examples: (i) Benzene + Toluene

(ii) Cyclohexane + Cycloheptane (iii) Hexane + Heptane


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(iv) C6H5Cl + C6H5Br (v) C2H5Cl + C2H5Br

ASSIGNMENT1. Vapour pressure of Chloroform (CHCl3) and dichloromethane (CH2Cl2) at 298 K are 200 mm Hg and 415 mm Hg

respectively. Calculate (i) the vapour pressure of solution prepared by mixing 25.5 gm of CHCl 3and 40 gm ofCH2Cl2at 298 K (ii) mole fraction of each component in vapour phase. [3947.9 mm] .[XCHCl3 = 0.78],

2. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzeneand toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in

vapour phase if 80 gm of benzene is mixed with 100 gm of toluene.[X benzene=0.598 X toluene= 0.402]

3. 100 gm of liquid A (molar mass 140 gm / mole) was dissolved in 1000 gm of liquid B (molar mass 180 gm / mole)the vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and itsvapour pressure in the solution if the total vapour pressure of the solution 475 torr. [280.7 mm]

Non-Ideal solutionsThe solutions which do not obey Raoults law over the entire range of concentration are known as non-ideal solutions. In this

solutions new intermolecular interactions (A vs B) are different from old intermolecular forces of interactions (A vs A and B

vs B). So escaping tendency of molecules is different from expected so vapour pressure of the solution is different from

expected.

mixH 0, mixV 0

Non-Ideal solution with positive deviationIn these solutions new intermolecular forces of interaction (A vs B) are lower than old intermolecular forces of interaction (A

vs A and B vs B). So escaping tendency of molecules is more than expected and vapour pressure is more than expected.


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In this solutions mixH = +ve, mixV = +ve

The partial vapour pressure of both components is more than expected as per Raoults law.The solution of composition corresponding to maxima of the curve forms a constant boiling mixture. i.e. solution of this

composition boils at a constant temperature & distils over without change in composition as if it is a pure liquid. Such a

mixture is known as azeotropic mixture or azeotrope. Since the partial pressures of the two components is more than

expected. So boiling point of the mixture is lower than both the components of the mixture. So such an azeotrope is known as

low boiling azeotrope.

(i) Acetone + Carbon disulphide

(ii) Acetone + Ethyl alcohol

(iii) Acetone + Benzene(iv) Methyl alcohol + water

(v) Ethyl alcohol + Water

(vi) Carbon tetrachloride + Chloroform(vii) Carbon telrachloride + Benzene

(viii) Carbon tetrachloride + Toluene

Non-Ideal solution with negative deviationIn this type of solution the new intermolecular interactions (A vs B) are stronger than old intermolecular interactions (A vs A

and B vsB). So escaping tendency of molecules is less than expected. So partial vapour pressures of the two components are

less than expected.

In this solution mixH = ve, mixV = ve


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The partial vapour pressures of both the components of the mixture are less than expected as from Raoults law. The solution

of composition corresponding to minima of the curve forms a constant boiling mixture, i.e. azeotrope. Since the partial

vapour pressures of both the components is less than expected, so boiling point of the mixture will be more than both the

components. Such a mixture is known as high boiling azeotrope.

(i) Chloroform + benzene(ii) Chloroform + diethylether

(iii) Acetone + aniline(iv) HCl + water(v) HNO3 + water

(vi) Acetic acid + pyridine

COLLIGATIVE PROPERTIESThe properties of dilute solutions of non-volatile solutes which depend upon number of solute particles in the solution and are

independent of the nature of solute are known as colligative properties.

The important colligative properties are

(i) Relative lowering of vapour pressure(ii) Elevation in boiling point

(iii) Depression in freezing point

(iv) Osmotic pressure

Relative lowering of vapour pressureAccording to Raoults law for the solution of a non-volatile solute.

21

1

o

so

nn

n

p

pp

, where po= vapour pressure of pure solvent

ps= vapour pressure of solutionn1= number of mole of solute

n2= number of moles of solvent

From this expression it is clear that the relative lowering of vapour pressure depends only on the number of moles of solute

dissolved in a definite number of moles solvent, so it is a colligative property.

Determination of molecular masses of solute from relative lowering of vapour pressureAccording to Raoults law for the solution of a non-volatile solute.

21

1o

s

o

nnn

ppp

,

2

22

1

11

Mwn,

Mwn

w1= mass of solute

w2= mass of solvent

M1= M.M. of soluteM2= M.M. of solvent

Substituting these values

2211

11

o

so

M/wM/w

M/w

p

pp

For a very dilute solution n1is negligible as compared to n2, so n1can be neglected in comparision with n2. So expression wil

be

2

1

o

so

n

n

p

pp

And

22

11

o

so

M/w

M/w

p

pp

In case of both dilute as well as concentrated solution following modified formula is used


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12

21

s

so

Mw

Mw

p

pp

ASSIGNMENTS1. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid

weighing 0.5 gm when added to 39.0 gm of benzene (M.M. 78 gm/mole), vapour pressure of the solution then is0.845 bar. What is the molecular mass of the solid?

2. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K, when 25 gmof glucose is dissolve in 450 gm of water.

3. A solution containing 30 gm of non-volatile solute exactly in 90 gm of water has a vapour pressure of 2.8 KPa at298 K. Further 18 gm of water is then added to this solution and new vapour pressure becomes 2.9 KPa at 298 K.

Calculate (i) Molar mass of solute (ii) Vapour pressure of water at 298 K.

4. Calculate the mass of a non-volatile solute (M.M = 40 gm/mole) which should be dissolved in 114 gm of octane toreduce its vapour pressure to 80%.

5. Vapour pressure of water is 12.3 KPa at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatilesolute in it.

6. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of thesolvent. What is the molar mass of solute? (vapour pressure of pure water at 100

oC is 1.013 bar)

7. Two liquids A and B boil at 145oC and 190oC respectively. Which of them has higher vapour pressure at 80oC.8. Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution containing 10% bymass of urea at 40oC. (vapour pressure of water at 40oC = 55.3 mm Hg).

9. A solution is made by dissolving 30 gm of non-volatile solute in 90 gm of water. It has a vapour pressure of 2.8 KPaat 298 K. At 298 K vapour pressure of pure water is 3.64 KPa. Calculate molecular mass of solute.

10. How much urea (M.M = 60gm/mole) should be dissolved in 50 gm of water, so that its vapour pressure at roomtemperature is reduced by 25%. Calculate molality of the solution.

11. Expain with a suitable diagram and appropriate example why some non-ideal solutions show negative deviation.12. What are ideal and non-ideal solutions? Explain with a suitable diagram the behaviour of ideal solutions.Elevation in boili ng point

The boiling point of a liquid is the temperature at which vapour pressure of the liquid becomes equal to atmospheric

pressure.

By addition of a non-volatile solute to a solvent the mole fraction of solvent becomes less than 1 in the solution and as

the vapour pressure of a solution is directly proportional to mole fraction of the solvent. So vapour pressure of a solventdecreases on addition of non-volatile solute to it. i.e. vapour pressure of the solution of a non-volatile solute is always

less than vapour pressure of pure solvent. This may be explained by taking of plot between vapour pressures of solvent

& solutions & temperature.

At point E (Tob) vapour pressure of solvent is equal to atmospheric pressure

So Tob is the boiling point of solvent. At point F (Tb) the vapour pressure of solution becomes equal to atmospheric

pressure. So Tb is the boiling point of solution as Tb is greater than T ob, i.e. boiling point of solution of non-volatile

solute is higher than boiling point of pure solvent. The difference is known as elevation in boiling point.

Tb = Tb Tob


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Tb m

Or Tb = Kb m [ Kb = molal elevation constant or Ebullioscopic constant]

Molal elevation constant is equal to elevation in boiling point shown by 1 molal solution.

As m =21

1

2

11

wM

w1000

1000/w

M/w

w1= weight of solute

w2= weight of solventM1= molecular mass of solute

Or Tb =21

1

wM

w1000Kb

M1=2

1

wTb

w1000Kb

Form the above expression it is clear that elevation in boiling point is directly proportional to molality of the solution, i.e

depends upon on the concentration of the solute. So it is a colligative property.

Kb=v

2o2

v

2o

H1000

RTM

l1000

RT

To= boiling point pure solvent

lv= latent heat of vaporization per gram of solventHv= latent heat of vaporization per mole of the solvent

M2 = M.M. of solvent

R = Molar gas constant

ASSIGNMENT1. 18 gm of glucose (C6H12O6) is dissolved in 1 Kg of water. At what temperature will water boil at 1.013 bar? Kb for

water is 0.52 K Kg / mole.

2. The boiling point of benzene is 353.23 K. When 1.80 gm of a non-volatile solute was dissolved in 90 gm ofbenzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute. Kb for benzene is 2.53 KKg/mole.

3. A solution containing 8 gm of a compound in 100 gm of diethylether boils at 36.86oC, whereas the pure ether boilsat 35.60oC. Determine the molecular mass of solute. (For ether Kb = 2.02 K Kg/mole)

4. On dissolving 3.24 gm of sulphur in 40 gm of benzene boiling point of solution was higher than that of benzene by0.81 K. kb value for benzene is 2.53 K Kg/mole. What is molecular formula of sulphur? (At. Mass of sulphur is 32gm/mole)

5. A solution containing 12.5 gm of non-electrolytic substance in 175 gm of water gave a boiling point elevation of0.70 K. Calculate molar mass of the substance. (Kb = 0.52 K Kg/mole)

6. 0.90 gm of a non-electrolyte was dissolved in 87.90 gm of benzene. This raised the boiling point of benzene by0.25oC. If the molecular mass of the non-electrolyte is 103.0 gm/mole. Calculate molal elevation constant for

benzene.

Depression i n f reezing point

At freezing point the solid phase is in equilibrium with the liquid phase. So freezing point is the temperature at which

vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the liquid phase. On addition of non

volatile solute, the freezing point of a liquid decreases. This can be shown by plotting curves between vapour pressures

of liquid solvent, solution and solid solvent.


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In this plot at point B i.e. at ofT , the vapour pressure of solvent becomes equal to vapour pressure in solid phase. SoofT

is the freezing point of solvent. At point E i.e. at Tf, the vapour pressure of solution becomes equal to vapour pressure in

solid phase, so Tfis the freezing point of solution as T fis smaller thanofT i.e. freezing point of solution is always less

than freezing point of solvent. This decrease in freezing point is known as depression in freezing point.

fT =

o

fT - Tf, where Tf= depression in freezing pointofT = freezing point of solvent

Tf= freezing point of solution

f

T m

Or, fT = Kf m, Kf= molal depression constant or Cryoscopic constant

It is the depression in freezing point shown by 1 molal solution.

m =1000/w

M/w

2

11 w1= wt. of solute,

w2= wt. of solvent

M1= M.M. of solute

So Tf=12

1f

Mw

1000wK

Or M1=f2

1f

Tw

1000wK

From this expression, it is clear that depression in freezing depends only on the molality i.e. concentration of solute, so it is a

colligative property.

Kf=H1000

TMR

fusion

of2

2

R = molar gas constant,

M2= M.M. of solvent,

Tfo=freezing point of pure solvent

fusionH = enthalpy of fusion of solvent

ASSIGNMENT1. 45 gm of ethylene glycol (C2H6O2) is mixed with 600 gm of water. Calculate (a) freezing point depression ( b) The

freezing point of solution. (Kffor water = 1.86 K Kg/mole)

2. 1.00 gm of non-electrolyte solute is dissolved in 50 gm of benzene lowered the freezing point of benzene by 0.40 KThe freezing point depression constant of benzene is 5.12 K Kg/mole. Find the molar mass of solute.

3. A 5% solution by mass of cane sugar in water has a freezing point 271 K. Calculate the freezing point of 5% glucosein water if freezing point of pure water is 273.15 K.


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4. Two elements A and B form compounds having molecular formula AB2 and AB4. When dissolved in 20 gm obenzene 1 gm of AB2 lowers the freezing point by 2.3 K, whereas 1.0 gm of AB4lowers it by 1.3 K. The molar

depression constant for benzene is 5.1 K Kg/mole. Calculate atomic masses of A and B.

5. Calculate the temperature at which a solution containing 54 gm of glucose (C6H12O6) in 250 gm of water will freeze(Kffor water = 1.86 K Kg/mole)

6. A 0.1539 molal aqueous solution of cane sugar (M.M. = 342 gm/mole) has a freezing point of 271 K. While thefreezing point of pure water is 273.15 K, what will be the freezing point of an aqueous solution containing 5 gm ofglucose per 100 gm of solution?

7. Calculate the freezing point of a solution containing 0.52 gm of glucose dissolved in 80.2 gm of water.8. Show graphically that freezing point of a liquid will be depressed when a non-volatile solute is dissolved in it.9. The molal freezing point depression constant of benzene is 4.90 K Kg/mole. Selenium exists as a polymer Se x

When 3.26 gm of Selenium is dissolved in 226 gm of benzene. The observed freezing point is 0.112 o lower than

pure benzene. Determine the molecular formula of Selenium (At. Mass of Se = 78.8 gm/mole)

10. An aqueous solution freezes at 272.4 K, while pure water freezes at 273 K. Determine (i) Molality of solution (iiBoiling point of solution (iii) Lowering of vapour pressure at 298 K (Kf= 1.86 K Kg/mole, Kb= 0.512 K Kg/mole

and vapour pressure of pure water at 298 K = 23.756 mm Hg)

Osmosis & Osmoti c pressure

OsmosisWhen a solution is separated from solvent or two solutions of different concentration are separated by a semi permeable

membrane, then solvent molecules flow from solvent to solution or from solution of lower concentration of solute tosolution of higher concentration of solute. This spontaneous flow of the solvent molecules from the solvent to solution or

from a less concentrated solution to a more concentrated solution through a semi permeable membrane is known as

osmosis.

Types of semi permeable membranes(a)Natural semi permeable membranes: Vegetable and animal membranes found just under the outer skin act as semi

permeable membrane. Most common animal membrane is pigs bladder.

(b)Ar ti fi cial semi permeable membranes : Parchment paper, cello phane, copper ferrocyanide, silicates of ironc cobaland nickel.

Exo-osmosis : When a cell is placed in a solution having higher concentration the cell save, there is flow of solvent from cel

to outside this process involves shrinkage of cell (plasmolysis). It is known as exo-osmosis. This is also known as crenation.

Endo-osmosis:When a cell is placed in less concentrated solution, solvent flows from outside to cell and swelling of cell

takes place. This process is known as endo-osmosis. This may also result in rupture of cell wall, this process is known ashemolysis.

Torgor :The pressure developed inside the shell due to osmosis is known as torgor.

Osmotic Pressure:(1) It is the equilibrium hydrostatic pressure developed due to osmosis.(2) It is the minimum pressure that has to be applied on solution to prevent the flow of the solvent to semi permeable

membrane is called osmotic pressure.Reverse osmosis: When pressure more than osmotic pressure is applied on the solution then solvent molecules flow from

solution to solvent. This process is known as reverse osmosis. This is used for desalination of water.

C

T

C T

= R C T

C = V

n

So, =V

nRT

As n = w/M

So =V

RT

M

w


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Or M =V

wRT

, [Vant Hoff equation for dilute solution]

Where M= M.M. of solutew = wt. of solute

R = molar gas constant

= osmotic pressureV = volume of solution in L

T = temperature in K

Method used for finding out the osmotic pressureBarkeley and Hartleys Method

The above expression shows that osmotic pressure depends only on the concentration of solute, so it is a colligative property.

The method use to find out the molecular masses of macro-molecules like proteins is osmotic pressure as (i) The osmotic pressure is determined at room temperature, so denaturation of proteins is avoided.(ii) The molecular mass of macro-molecules is so large that in the solution, the molar concentration is very small

and it is only osmotic pressure whose value is measurable.

Isotonic solution:Solutions having same osmotic pressure.

Hypotonic solution:Solution having lower osmotic pressure.

Hypertonic solutions:Solution having higher osmotic pressure.

0.91% solution of NaCl is isotonic with human red blood cell (RBC).

ASSIGNMENT1. 200 ml of an aqueous solution of a protein contains 1.26 gm of protein. The osmotic pressure of such a solution at

300 K is found to be 2.57 103

bar. Calculate the molar mass of protein. (R = 0.083 L bar mole1

K1

)

2. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 gm of polymer of molarmass 185000 in 450 ml of water at 37

oC.

3. At 300 K 36 gm of glucose present in 1 L of its solution has an osmotic pressure of 4.98 bars. If the osmoticpressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

4. Define osmotic pressure. Describe how the molecular mass of a substance; can be determined by a method based onmeasurement of osmotic pressure?

5. What are conditions which result in reverse osmosis?6. A 4% solution of sucrose is isotonic with 3% solution of an unknown organic substance. Calculate molecular mass

of unknown substance.7. 1 L aqueous solution of sucrose (M.M. = 342 gm/mole) weighing 1015 gm is found to record an osmotic pressure o

4.82 atm at 293 K. What is the molarity of the solution?

8. Calculate the osmotic pressure of a solution obtained by mixing 100 ml of 1.5% solution of urea and 100 ml of3.42% solution of cane sugar at 20oC.

ABNORMAL MOLECULAR MASSESThe colligative properties depend upon actual concentration of solute particles in the solution. In some solutions, the

solute may undergo dissociation or association. So effective concentration of solute particles is not as calculated and the

calculated molecular mass using these properties will be different from the actual molecular mass. So the molecular

masses determinate are known as abnormal molecular masses.

If the solute dissociates in the solution then the effective concentration of solute particles will be more than calculatedSo observed colligative properties will be higher than calculated and observed molecular mass will have lower value

than calculated.If the solute molecules associate in solution then effective concentration of solute particles will be less than calculated

and colligative properties will be less than calculated and observed molecular masses will be higher than calculated. In

some cases on addition of a solute to a solution, due to complex formation the concentration of solute particles decreases(instead of increasing).

e.g. when HgI2is added to KI solution following complex is formed

2KI + HgI2 K2[HgI4]

2KI 2K++ 2I

K2[HgI4] 2K++ HgI4

2


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So on addition of HgI2to KI solution the effective concentration of solute particle decreases, so magnitude of colligative

properties decreases.

Vant Hoff factor (i)This factor gives the extent of association or dissociation of solute molecules in the solution.

i =ondissociatinornassociationeitherishen thereproperty wecolligativofvalueCalculated

propertyecolligativofvaluealExperiment.

i =f

f

TofvalueNormal

TofvalueObserved

or i =

b

b

TofvalueNormal

TofvalueObserved

i =

ofvalueNormal

ofvalueObserved

i =pressurevapourweringofrelativeloofvalueNormal

pressurevapourofloweringrelativeofvalueObserved

i =M.M.Observed

M.M.Normal

Calculation of the degree of dissociation of solute in a solutionDegree of dissociation is defined as defraction of the total number of molecules that ionizes. Let the degree of dissociation be

.

An nA

1- nSo total number of moles of solute in solution equal to 1 + n = 1 + (n 1)

i =1

)1n(1

=1n

1i

Calculation of degree of association of solute molecules in solutionDegree of association is fraction of total number of molecules of solute which have undergone association in the solution. Le

degree of association be . Let n molecules of solute undergo association.

nA An

1- /n

So total number of moles of solute particles in solution = 1 + /n

i =1

/n-1

So =1-n

ni)-(1

ASSIGNMENT1. 2 gm of benzoic acid (C6H5COOH) dissolved in 25 gm of benzene shows a depression in freezing point equal to

1.62 K. Molal depression constant for benzene is 4.9 K Kg/mole. What is the percentage association of the acid if iforms a dimmer in the solution?

2. 0.6 ml of acetic acid (CH3COOH) having density 1.06 gm/ml is dissolved in 1 L of water. The depression infreezing point observed was 0.0205oC. Calculate the Vant Hoff factor and the dissociation constant of acid.

3. Determine the osmotic pressure of a solution prepared by dissolving 25 mgm of K2SO4 in 2 L of water at 25oCassuming that it is completely dissociated.

4. Determine the amount of CaCl2(i = 2.47) dissolved in 2.5 L of water, such that its osmotic pressure is 0.75 atm at27oC.

5. 19.5 gm of CH2FCOH is dissolved in 500 gm of water. The depression in the freezing point of water observed is1.0oC. Calculate the Vant Hoff factor and dissociation constant of acid.

6. Calculate the depression in freezing point of water when 10 gm of CH3CH2CHClCOOH is added to 250 gm owater. (Ka= 1.4 10

3, Kf= 1.86 K Kg/mole)

7. Of 0.1 m solution of glucose and NaCl respectively which one will have higher boiling point and why?


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8. On dissolving 3.1 gm of BaCl2in 250 gm of water, the solution boils at 100.083oC. Calculate Vant Hoff factor andmolality of solution. (Kb= 0.52 K Kg/mole, M.M of BaCl2= 208.3 gm/mole)

9. The depression in freezing point of water observed for same molar concentration of acetic acid, trichloro acetic acidand trifluoro acetic acid increases as stated about. Explain.

10. Why is the freezing point of 0.1 M solution of NaCl nearly twice that of 0.1 M glucose solution?11. The freezing point of a solution containing 0.3 gm of acetic acid in 30.0 gm of benzene is lowered by 0.45oC

Calculate Vant Hoff factor. (Kf = 5.13 K Kg/mole)12. What is meant by Vant Hoff factor? The osmotic pressure of 0.0103 M solution of an electrolytic is found to be

0.70 atm at 27oC. Calculate the Vant Hoff factor. R = 0.082 L atm mole1K1.)

Multiple Choice Questions with one Answer1. The ratio of the value of any colligative property for KCl solution to that for sugar solution is nearly

(a) 1.0 (b) 0.5 (c) 2.0 (d) 2.5

2. Which of the following solutions will have maximum freezing point ?(a) 0.01 M urea (b) 0.01 M KCl (c) 0.01 M BaCl2(d) 0.01 M NaCl

3. The osmotic pressure observed when benzoic acid is dissolved in benzene is less than expected.This may be due to thereason that

(a) Benzoic acid is an organic solute(b) Benzoic acid has higher molar mass than benzene(c) Benzoic acid gets associated in benzene(d) Benzoic acid gets dissociated in benzene .

4. The relative lowering of vapour pressure of a solution on the addition of a non- volatile sloute(a) Is equal to mole fraction of the solute(b) Is equal to the mole fraction of the solvent(c) Depends upon the nature of the solute(d) Depends upon the nature of the solute and solvent.

5. What happens when an egg is kept in a saturated solution of NaCl after removing hard shell of the egg in dilutehydrochloric acid ?(a) Egg will shrink (b) Egg will grow

(c) Egg will remain same (d) Egg will first shrink and then grow

6. The normality of 1 M H3PO4is(a) 1N (b) 3N (c) 0.33N (d) 0.5 N

7.

Osmotic pressure is measured by(a) Ostwald's method (b) Berkeley and Hartley method

(c) Pfeffer's method (d) Beckmann's method

8. When glycerine is added to a litre of water, which of the following is observed ?(a) Water evaporates more readily(b) The viscosity of water decreases(c) The freezing point of water is lowered(d) The temperature of water increases

9. What will happen if a thin slice of sugar beet is placed in a concentrated solution of sodium chloride?(a) Sugar beet will lose water from its cells(b) Sugar beet will absorb water from solution(c) Sugar beet will neither absorb nor lose water(d) Sugar beet will dissolve in solution.

10. A solution contains 16 g of methanol and 90 g of water. The mole fraction of methanol in the solution is(a) 0.909 (b) 0.826 (c) 0.091 (d) 0.182

11. What volume of 36N and 1N sulphuric acid should be mixed to get 1L of 6N acid solution(a) 96.7 ml of 36N +903.3 ml of 1N (b) 142.8ml of 36N +857.2ml of 1N

(b) 903.3ml of 36N +96.7ml of 1N (d) 100ml of 36N +900ml of 1N

12. An aqueous solution of urea having 6 g in 500ml has a density of 1.05. If the molar mass of urea is 60 , its molalitywill be

(a) 0.193m (b) 0.190m (c) 0.200m (d) 0.10m

13. The weight of (COOH)2 2H2O needed to prepare 500ml of 0.1M solution is(a) 12.6 g (b) 9.0 g (c) 4.5 g (d) 6.3 g


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14. The vapour pressure of ethanol at 298K is 40mm of mercury. Its mole fraction in a solution with methanol is 0.80. Itvapour pressure in the solution is

(a) 8 mm Hg (b) 32 mm Hg (c) 48 mm Hg (d) 80 mm Hg

15. The osmotic pressure of a 0.2 m solution of non- electrolyte at 270C (R= 0.082 L K-1mol-1) is :(a) 4.92 atm (b) 1 atm (c) 0.2 atm (d) 27 atm

16. Two liquids A and B have vapour pressures 500 and 200 torr respectively at a certain temperature. In an ideal solution ofthe two, the mole fraction of A at which two liquids have equal partial pressures is(a) 0.50 (b) 0.20 (c) 0.286 (d) 0.714

17. Phenol associates in benzene to form the dimers, (C6H5OH)2. The normal mass and the observed molar mass of phenol inbenzene have been found to be 94 and 146.3 respectively.The degree of association of phenol in benzene is :

(a) 72.7% (b) 75% (c) 32.6% (d) 71.5%

18. A litre of public water contains 5 mg of chlorine. The concentration of chlorine in ppm is :(a) 5 10-4 (b) 5 (c) 50 (d) 0.5

19. The molality of 1 litre solution of 98% sulphuric acid (w/v)having density 1.88g/ml is(a) 10.1 (b) 11.1 (c) 5.55 (d) 2.6

20. The Van't Hoff factor for calcium nitrate Ca(NO3)2is 2.5 .Using the dissociation equilibrium,Ca(NO3)2Ca

2++ 2NO3- the degree of dissociation is :

(a) 0.25 (b) 0.33 (c) 0.66 (d) 0.7521. 100 ml of 0.3 N HCl are mixed with 200 ml of 0.6 N H2SO4. The final normality of the resulting solution will be

(a) 0.1 N (b) 0.2 N (c) 0.3 N (d) 0.5 N22. How many grams of a dibasic acid (molar mass = 200) should be present in 100 ml of its aqueous solution to give

decinormal strength?

(a) 1 g (b) 2g (c) 10 g (d) 20 g

23. 0.01 M solution each of urea, common salt and Na2SO4are taken, the ratio of depression of freezing point is(a) 1 : 1 : 1 (b) 1 : 2 : 1 (c) 1 : 2 : 3 (d) 2 : 2 : 3

24. The molecular weight of benzoic acid in benzene as determined by depression in freezing point method corresponds to(a) Ionization of benzoic acid (b) Dimerization of benzoic acid

(c) Trimerization of benzoic acid (d) Saturation of benzoic acid25. 6.3g of hydrated oxalic acid is dissolved to get 500ml solution. Its normality will be:

(a) 0.1N (b) 0.2N (c) 0.25N (d) 0.4N26. Out of molarity (M), molality(m), mole fraction (x) and normality (N), those depenedent on temperature are:

(a) M. m (b) m. x (c) M, N (d) N, x

27.

100 ml of decinormal HCl is mixed with 400ml of decimolar HCl. The normality of the resultant solution will be:(a) 0.1 (b) 0.01 (c) 0.05 (d) 0.2

28. The molecular mass of KCl as determined by cryoscopic method is nearly:(a) 74.5 (b) 149.0 (c) 37.25 (d) None

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