SOLUTIONS
Solution:?It is the homogeneous mixture, in which conc. of solute can be varied with in the limit.
e.g.sugar in water (before saturation point)
Concentration:?Amount of solute present in a known amount of solvent or solution .
Units of concentration:?1.Percentage
a.Percentage by weight:?Weight of solute in grams present in 100g of solution.
e.g. 10% NaCl soln by wt. means solute + solvent ? soln
or 10%NaCl soln(w/w) 10g 10010=90g 100g
b.Percentage by volume(i. For liquid solute):?The vol. of solute in ml present in 100 ml of solution.
e.g. 10% HCl soln by vol. or solute + solvent ? soln
10%HCl soln(v/v) means 10 ml 10010=90ml 100ml
(ii. For solid solute)The wt. of solute in grams present in 100 ml of solution.
e.g. 10%NaCl soln by vol.or solute + solvent ? soln
10% NaCl soln(w/v) means 10g ? 100ml
2.Strength:?Amount of solute in grams present per litre of soln.
i.e. strength =wt. of solute in grams/vol. of soln in litre = w*1000/vol. of soln in ml
e.g. 10g of NaCl is present in 250 ml of soln.Find strength of soln
soln:? w = 10g
vol. of soln = 250 ml
strength = 10*1000/250 =  g/L (solve it )
3.Molarity:?Number of moles of solute present per litre of solution. It is denoted by “M”.
M =No.of moles of solute /vol. of soln in lit. = (Given wt. of solute *1000) / ( M.Mass of solute*vol. of soln in ml)
i.e.M = w * 1000/GMM * vol.
e.g.Find molarity of the solution obtained by dissolving10g of NaCl per 250 ml of soln ?
soln:? w =10g
GMM = 23+35.5=58.5g
Vol. of soln = 250ml
M =10*1000/58.5*250 =  M (solve it)
4.Normality:?Number of gram equivalents of solute present per litre of soln.It is denoted by “N”.
N = No. of gm eq. of solute/vol. of soln in litre = w*1000/GEW*Vol. of soln (ml)
e.g. 0.98g of H2SO4 has been dissolved per 250 ml of soln.Find molarity & normality of soln.
soln :? w = 0.98 g
GMM =2*1+32*1+16*4=98 g
GEW =98/2 =49 g (as H2SO4 is Dibasic acid , so n = 2 & we know GEW =GMM / n)
Vol. of soln = 250 ml
N = 0.98*1000 / 49*250 =  N (?)
M = 0.98*1000 / 98*250 =  M(?)
NOTE:?From here N = 2M In general N = nM ( Name = naren Malik) n is acidity or basicity
5.Molality:?Number of moles of solute present in 1 kg of solvent.It is denoted by “m”.
m = No. of moles of solute / wt. of solvent in kg = w2*1000 / GMM*W1 here W1 =wt. of solvent, w2 = wt. of solute
e.g. 0.4g of NaOH has been dissolved in 250 g of solvent.Find the molality of soln.
soln:? w2 = 0.4 g
w1 = 250 g
GMM = 23+16+1=40g
m = 0.4*1000/40*250 =  m (?)
6.Mole fraction:?Mole fraction of a component in a soln is defined as the number of moles of that component divided by the total number of moles of all the component.It is denoted by “X”
X1= n1/n1+n2 and X2 = n2 / n1+n2 ? X1 + X2 = 1
7.ppm ( parts per million ):?Mass of solute present in one million parts by mass of solution
Ppm = mass of solute*106/mass of solution
?? NUMERICALS
1.Find molarity, molality ,normality and mole fraction of each component of 10% aq. soln of H2SO4 (w/w).d=1.10g/cm3
2.Find M,N ,X1.X2.m of 10%aq. soln of H2SO4 (v/v) d=1.10 g/cm3.
3.How many ml of a 0.1 M HCl are required to react completely with 1g mixture of Na2CO3 and NaHCO3containg equimolar amount of the two ?
Henry’s Law:? The mass of the gas dissolved in a given volume of solvent is directly proportional to the pressure of the gas present in equilibrium with the liquid.
i.e. m ? p => m = KH p here KH is Henry’s const.
or
The solubility of a gas in a liquid at a particular temperature is directly proportional to the partial pressure of that gas in the mixture.
i.e. pA ? ?A => pA = KH ?A
• More the Henry’s const. , lesser will be solubility of gas.
Solution of solids in solids( solid solution )?Those solutions in which both the component are solids. These are of two types:?1.Substitutional solid solutions ?If the atoms , ions or molecules of one solid present at the lattice site are replaced by the atoms ,ions or molecules of another similar solid, having similar sizes ,the solids thus obtained are called substitutional solid
2.Interstitial solid solutions?If in the lattice of a solid , the atoms of some other solid ,small in size ,occupy the interstitial sites ,the solids obtained are called interstitial solid solutions.
Vapour pressure ?Vapour pressure of a liquid is the pressure exerted by the vapours in equilibrium with the liquid at a particular temperature.
Effect of Adding a nonvolatile solute on vapour pressure of a liquid (solvent):?Vapour pressure decreases because some of the surface is occupied by nonvolatile solute particles which reduces the escaping tendency of solvent molecules into the vapour phase.
Raoult’s Law (for volatile solute):?The partial vapour pressure of a component is equal to the product of mole fraction of that component and.vapour pressure of that component in pure state.
i.e. pA = xA p0A and pB = xB p0B
Let us consider a mix. of two completely miscible volatile liquids A & B ,having mole fraction xA& xB
p0B Then according to Raoult’s Law :
pA = xA p0A & pB = xB p0B
P = pA+pB Total pressure of soln is :
 p0A pB=xB p0B P = pA + pB = xA p0A + xB p0B
Vapour pressure  P = xA p0A + (1 xA ) p0B
 P = ( p0A – p0B ) xA + p0B
 pA=xA p0A

xA=1 Mole fraction
xB=0 ____________
Raoult’s as a special case of Henry’s law:?According to Henry’s law :? p = KH x …………….( 1 )
For very very dilute soln x ~ 1 & p = p0 Putting these values in eqn. ( 1 ) we get
P0 = KH . 1 => KH = p0 Putting this value of KH in eqn. ( 1 ) we get
p = p0 x
Raoult’s Law (for the solution containing nonvolatile solute) or Relative lowering in Vapour Pressure ?
Vapour pressure of solution(pS) = Vapour pressure of solvent( pA) + vapour pressure of solute( pB)
ps = pA ( as solute is nonvolatile,so pB = 0) ps = p0A XA ( acc. to Raoult’s law ) ps = p0A ( 1 XB ) ( as X A+XB =1 )
ps = p0A – p0A XB
p0A  ps = p0A XB poA – ps is called lowering in vapour pressure
(poA  ps)/poA = XB (poA  ps )/poA is called relative lowering in vapor
 Pressure
This shows that relative lowering in vapour pressure is equal to the mole fraction of solute.
? Relative lowering in vapour pressure is colligative property . Comment.
Ans. Yes, Relative lowering in vapour pressure is colligative property because it depends on no. of moles of solute according to following relation: ( poA  ps ) /poA = XB = n2 /(n1+ n2)
Determination of molecular masses of solute from lowering in vapour pressure:?
(poA  ps)/poA = XB = n2 /(n1+ n2) Putting n2 = w2/M2 & n1 = w1/M1
(poA  ps)/poA = (w2/M2 ) /( w1/M1 + w2/M2 ) For dilute soln. ,we can write:
(poA  ps)/poA = (w2/M2 ) /( w1/M1 )
Or (poA  ps)/poA =w2 .M1 / M2 .w1 From here we can find mol. Mass of solute ,M2
NUMERICALS
1.Heptane and octane form ideal solution .At 373 K,the vapour pressure of the two liquid components are 105.2 k Pa and 46.8 k Pa resp.What will be the vapour pressure in k Pa of a mixture of 25g of heptane and 35g of octane ?
2.The vapour pressure of water is 12.3 kPa at 300 K.Calculate the vapour pressure of 1 molal solution in it .
3.Calculate the mass of a nonvolatile solute( m .mass 40 ) that should be dissolved in 114g of octane to reduce its pressure to 80% .
4.The vapour pressure of a 5% aq. soln of a nonvolatile organic substance at 373 K is 745 mm .Calculate the molar mass of the solute .
Ideal Solutions:?The soln which obeys Raoult’s law at all conc. and temperatures.The formation of ideal soln neither involve any change in enthalpy nor in volume . Thus condition of ideal soln are :
(i) It must obey Raoult’s law. i.e. pA = xA p0A and pB = xB p0B
(ii) ?Hmixing = 0
(iii) ?Vmixing = 0
?The soln in which solventsolvent and solutesolute interactions are of the same type as solventsolute interactions behave as ideal soln.
NonIdeal solutions:?The soln which do not obey Raoult’s law and accompanied by change in enthalpy and change in volume during their formation.
(i) It does not obey Raoult’s law. i.e. pA ? XA p0A and pB ? XB p0B
(ii) ?Hmixing ? 0
(iii) ?Vmixing ? 0
The nonideal solutions are further classified into two types:
(a)Solution showing positive deviations:?The soln in which the forces of interaction b/n solutesolvent is less than the solventsolvent and solutesolute interactions.As a result, the partial vapour pressure of each component and the total pressure of the solution will be greater as compared to an ideal solution.
(i) pA > xA p0A and pB > xB p0B
(ii) ?Hmixing > 0
(iii) ?Vmixing > 0
(b)Solution showing negative deviations:? The soln in which the forces of interaction b/n solutesolvent is more than the solventsolvent and solutesolute interactions.As a result, the partial vapour pressure of each component and the total pressure of the solution will be less as compared to an ideal solution.
(i) pA < XA p0A and pB < XB p0B
(ii) ?Hmixing < 0
(iii) ?Vmixing < 0
Example of Ideal soln:?( a) Benzene and Toluene (b) nHexane and nheptane (c) Ethyl bromide and ethyl iodide
Example of Solution showing positive deviations:? (i) Ethyl alc.and cyclohexane (ii) Ethyl alc.and water (iii) Ethyl alc.and acetone (iv) CCl4 and CHCl3 (v) Acetone & benzene (vi) acetone & CS2
Example of Solution showing negative deviations?: (i)Acetone & chloroform (ii)HCl &H2O(iii)chloroform & diethyl ether (iv)HNO3 & H2O
Azeotropic Mixture :?The soln having definite composition ,and boils like a pure liquid is called azeotropic mix. or azeotrope .
Minimum boiling azeotrope:?The composition of the mix. corresponding to which it have maximum vapour pressure and hence boil at a lower temp.
Maximum boiling azeotrope:?The composition of the mix. corresponding to which it have minimum vapour pressure and hence boil at a higher temp.
Azeotropic mix. can’t be seprated into their component by fractional distillation .
Graph showing +ve &  ve deviation are as follows:
Colligative Properties:?The properties of ideal soln which depend only on no. of moles of solute and do not depend on the nature of solute.
Osmosis:?The flow of solvent from a less concentrated soln to a more concentrated soln through a SPM .
Osmotic pressure:?The minimum pressure required to prevent the entry of the solvent into the solution through
SPM.
Expression for the osmotic pressure:? Osmotic pressure( ? ) of a soln is found to be directly proportional to the molar conc. of the soln & its temp.
i.e. ? ? C
? T
So ? ? C.T
? =R.C.T
? = CRT
? =n/v RT        (2)
?.V = nRT This eqn. is called vant’s Hoff eqn. for dilute solutions
Also ? = (w/M).RT/V
? = wRT / MV
M = wRT / ?V From here mass of solute can be calculated
From eqn (2) ,as ? ? n So osmotic pressure is colligative property.
Isotonic solutions :?The solutions which have same osmotic pressure are called isotonic solutions .Isotonic soln have same conc.
Hypotonic :?If one soln is of lower osmotic pressure, it is called hypotonic with respect to the more concentrated
solution.
Hypertonic:?The soln which have higher osmotic pressure is called hypertonic w.r.t. dilute soln.
NUMERICALS
1.Calculate the conc, of that soln. of sugar which has osmotic pressure of 2.46 atm. At 300 K.
2.Calculate the osmotic pressure at 273 Kof a 5% soln of urea by volume.( R=0.0821 L atm/K/mole)
3.A 4% soln of sucrose is isotonic with 3% soln of an unknown organic substance .Calculate the molar mass of the unknown substance.
ELEVATION IN BOILING POINT
It is found that the boiling point of the solution is always higher than
that of the pure solvent.The increase is called elevation in boiling point.
Expression for the elevation in boiling point:?
?Tb = Kb m here Kb is called the molal elevation const.
or ebullioscopic const.
m is the molality of soln.
If m = 1 ,then ?Tb = Kb ;Hence,
Molal elevation const. or ebullioscopic const may be defined as the
Elevation in boiling point when the molality of the solution is unity
Also Kb = RT2o /1000 lv = M1RT2o / 1000 ?Hv
Derivation of ?Tb = Kb m :? As elevation in boiling point is
directly proportional to the lowering in vapour pressure .
i.e. ?Tb ? ?p According to Raoult;s Law ; ?p ? x2
?Tb ? x2
?Tb = k x2 But x2 = n2/(n1+n2) ? n2 / n1 ( for dilute soln)
?Tb = k n2 /n1      (3)
?Tb = k M1 n2/w1 as n1 = w1 / M1
?Tb = k M1 m m = molality = no. of moles of solute /wt. of solvent in kg = n2/w1
?Tb = Kb m taking kM1 = Kb
• From eqn. (3) ?Tb ? n2 Hence elevation in boiling point is colligative property
Calculation of molecular mass of the solute :?
As ?Tb = Kb m or ?Tb = Kb w2.1000 / M2.w 1
M2 = Kb w2.1000 / ?Tb .w 1
NUMERICALS
1. 10g of nonvolatile solute when dissolved in 100g of benzene raises its boiling point by 1o .What is the
m.mass of the solute ? ( Kb for benzene = 2.53 K m1 )
2.Calculate the b.pt. of a solution containing 0.456gof camphor(m.mass = 152 ) dissolved in 31.4g of acetone( b.pt. = 56.30oC ) ,if the molecular elevation const. per 100g of acetone is 17.2oC .
( Hint Kb = 17.2 / 10 )
Depression in Freezing Point :?
It is found that the Freezing point of the solution is always lower than
that of the pure solvent.The decrease in freezing point is
called. Depression in Freezing Point
Expression for the Depression in Freezing Point:?
?Tf = Kf m here Kf is called the molal depression const.
or cryoscopic const.
& m is the molality of soln.
If m = 1 ,then ?Tf = Kf ;Hence,
Molal depression const. or cryoscopic const may be defined as the
Depression in Freezing Point when the molality of the solution is unity
Also Kf = RT2o /1000 lf = M1RT2o / 1000 ?Hf
Derivation of ?Tf = Kf m :? As Depression in Freezing Point is
directly proportional to the lowering in vapour pressure .
i.e. ?Tf ? ?p According to Raoult;s Law ; ?p ? x2
?Tf ? x2
?Tf = k x2 But x2 = n2/(n1+n2) ? n2 / n1 ( for dilute soln)
?Tf = k n2 /n1      (4)
?Tf = k M1 n2/w1 as n1 = w1 / M1
?Tf = k M1 m m = molality = no. of moles of solute /wt. of solvent in kg = n2/w1
?Tf = Kf m taking kM1 = Kf
• From eqn. (4) ?Tf ? n2 Hence depression in freezing point is colligative property
Calculation of molecular mass of the solute :?
As ?Tf = Kf m or ?Tf = Kf w2.1000 / M2.w 1
M2 = Kf w2.1000 / ?Tf .w 1
NUMERICALS
1. A 5 % solution ( by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.1 K .
2. Two elements A & B forms compounds having molecular formula AB2 & AB4 . When dissolved in 20 g of benzene , 1 g of AB2 lower the freezing point by 2.3 K whereas 1.0 g of AB4 lower it by 1.3 K .The molal molal depression constant for benzene is 5.1 K kg /mol .Calculate atomic mass of A &B
ABNORMAL MOLECULAR MASS ? If the molecular mass calculated from colligative properties is different from actual value ,then it is called ABNORMAL MOLECULAR MASS.
The abnormal molecular masses are observed in the following cases:
1. When the solution is non ideal i.e. the soln is not dilute .
2. When the solute undergoes association in the solution .
3. When the solute undergoes dissociation in the solution .
Abnormal mol. Mass due to association ? Due to association , no. of moles decreases so C.P. also decreases.
As C.P.is inversely proportional to m.mass ,hence m.mass inceases
e.g. Acetic acid has m.mass equal to 60. But its m.mass is found to be 120 by using C.P.
It is due to association of acetic acid in the form of dimmer because of H – bonding
O       H  O
CH3 –C O H     O = CCH3
Abnormal mol. Mass due to dissociation ? Due to dissociation , no. of moles increases so C.P. also increases.
As C.P.is inversely proportional to m.mass ,hence m.mass decreases
e.g .KCl ? K + Cl due to dissociation no. of moles becomes double , so C.P. also doubles,Hence m.mass is half of the actual value.
van’t Hoff factor? It is the ratio of observed value of colligative property to the actual (normal) value of the colligative property .It is denoted by ‘i’ .
i = observed (experimental) value of colligative property / actual (normal or calculated) value of the colligative property
or i = actual (normal or calculated) value of m.mass / observed (experimental) value of m.mass
or i = Mc / Mo
Modified eqn of colligative properties :? (poA  ps)/poA = i XB
? = i CRT
?Tb = i Kb m
?Tf = i Kf m
Calculation of degree of dissociation ? It is defined as the fraction of molecules get dissociated . It is denoted by ‘? ’ .
Let us consider the rexn. An ? n A
Initially 1mol 0
After dissociation 1  ? n ?
Total moles before dissociation = 1
Total moles after dissociation = 1 – ? + n ?
i = observed no. of moles / normal no. of moles
i = [1 – ? + n ? ] / 1
i = 1 + ? ( n  1 )
i – 1 = ? ( n – 1 )
? = ( i  1 ) / ( n – 1 )
? = [ Mc / Mo  1 ] / (n – 1)
? = [ Mc  Mo ] / Mo (n – 1 )
Calculation of degree of association ? It is defined as the fraction of molecules get associated . It is denoted by ‘?’
Let us consider the rexn. n A ? An
Initially 1mol 0
After dissociation 1  ? ? / n
Total moles before dissociation = 1
Total moles after dissociation = 1 – ? + ? / n
i = observed no. of moles / normal no. of moles
i = [1 – ? + ? / n ] / 1
i = 1 + ? ( 1  n ) / n
i – 1 = ? ( 1  n ) / n
? = ( i  1 )n / ( 1 – n )
? = (1  i )n / ( 1 – n )
? = [ 1  Mc / Mo ] / (n – 1)
? = [ Mo – Mc ] n / Mo (n – 1 )
NUMERICALS
What is the value of van’t Hoff factor for (i) KCl (ii) K4[Fe(CN)6] (iii) 80% dissociated Al2(SO4)3
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