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1. Define the term solution. What kind of solutions are possible ?Write
briefly about each kind of solution with an example.
•
Solution:
Solutions are homogeneous mixtures of two or more substances in a single phase. Most of the
solutions can be considered as having a majority ingredient called the Solvent and one or more
minority ingredients called Solutes.
Different types of Solutions:
Type of Solutions
Common Example
Gaseous Solutions:Gas in gas
A mixture of oxygen and nitrogen gases.
Liquid in gas
Chloroform vapours mixed with nitrogen gas.
Solid in gas
Camphor vapours in nitrogen gas.
Liquid Solutions:Gas in liquid
Oxygen dissolved in water.
Liquid in liquid
Ethanol dissolved in water.
Solid in liquid
Sucrose dissolved in water
Solid Solutions:Gas in solid
Solution of hydrogen in Palladium.
Liquid in solid
Amalgam of mercury with Sodium.
Solid in solid
Copper dissolved in gold.
2. Suppose a solid solution is formed between two substances, one
whose particles are very large and the other whose particles are very
small. What type of solid solution is this likely to be ?
•
in
Solution:
These solid solutions are called Interstitial Solid Solutions constitute the planning of atoms of small
size into the voides that exist between atoms in the host lattice.
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3. Define the following terms (i) Mole fraction (ii) Molality (iii) Molarity
(iv) Mass percentage.
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Solution:
(i) Mole Fraction
Mole fraction of any component is the number of moles of that particular component divided by the
sum of moles of all the components present in a solution. It is denoted by the symbol c . Thus mole
fraction, c A of a component A in solution with another component B is defined by:
=
nA and nB number of moles of A and B respectively.
(ii) Molality
Molality of a solution is defined as the number of moles of the solute that are present in 1 kg of the
solvent.
Molality (m) =
(iii) Molarity
Molarity of a solution is expressed as the number of moles of a solute present in one litre of the
solution.
Molarity M =
(iv) Mass Percentage
It is defined as the mass of the solute expressed in gms per 100 gms of solution.
4. Concentrated HNO3 used in the laboratory work is 68% HNO3 by mass
in aqueous solution.What should be the Molarity of such a sample of the
acid if the density of the solution is 1.504 g ml –1 ?
•
Solution:
68% HNO3 means 68 gms HNO3 dissolved in 100 gms of the solution.
Volume of HNO3 =
= 66.489 mol.
Molecular mass of HNO3 = 63
in
Density = 1.504 gm ml-1 =
The number of moles of HNO3 =
= 1.079 mol.
1.079 moles of HNO3 present in 66.489 ml.
2
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1000 ml of solution contains =
Molarity = 16.22 M.
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5. A solution of glucose in water is labeled as 10%. W/w what would be
the molality and mole fraction of each component in the solution. If the
density of the solution is 1.2 g ml-1. Then what shall be the molarity of the
solution ?
•
Solution:
Calculation of Molality:
Percentage of glucose solution = 10%
i.e., 10 g of glucose present in = 90 g of water
Molecular mass of glucose = 180
No. of moles in 10 g of glucose =
90 g of water contain
moles of glucose.
Molality of the 10% glucose solution =
= 0.617 m.
Calculation of Mole Fraction:
No. of moles of glucose =
=
No. of moles of water =
=
Mole Fraction of glucose c
Mole fraction of water c
=
=
=
=
= 5 nB
= 0.01
= 0.989.
in
B
A
= 0.055 nA
Calculation of Molarity :
Density =
1.2 =
Volume of the solution =
Number of moles of glucose =
= 83.33 ml.
= 0.055
3
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83.33 ml contains 0.055 moles of glucose.
1000 ml contains =
Molarity = 0.677 M or 0.67 M of solution.
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6. How many ml of a 0.1M HCl are required to react completely with 1
gm mixture of Na2CO3 and NaHCO3 containing equimolar amounts of two ?
•
Solution:
The mixture of Na2 CO3 and NaHCO3 react with HCl according to the following equation.
One mole of Na2Co3 and one mole of NaHCO3 requires 3 moles of HCl for complete neutralisation.
Molar mass of Na2CO3 = 106 amu [ 46 + 12 + 48]
Molar mas of NaHCO3 = 84 amu [23 + 1 + 12 + 48]
Total mass of the mixture = 190 amu
190 gms of the mixture reacts with 109.5 gms or HCl
1 gm of the mixture reacts with
= 0.57631 gms of HCl
1 gms of the mixture reacts with 0.57631 gms. Of HCl completely
according to chemical equation.
Molarity =
0.1 =
V ml = 157.8 ml.
7. Calculate the percentage composition in terms of mass of a solution
obtained by mixing 300 gms of a 25% and 400 gms of a 40% solution by
mass.
•
Solution:
Two solutions labeled as A and B are 300 gms of 25% and 400 gms of a 40% solution by mass.
Composition of A =
Solution B
in
Solution A
= 75 gms
Composition of B =
= 160 gms.
Both the solutions are mixed and the composition of the mixture = 235 gms.
The % of A in the mixture =
= 31.9
32%.
Similarly % of B in the mixture =
= 68.08%
The % composition of the mixture = 32% and 68.08% respectively.
4
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8. Calculation of molarity of solution.
•
Solution:
Mass of water =200g
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Mass of ethylene glycol =222.6g
Total mass of solution =422.6g
Density of solution =1.072g ml-1
Volume of solution =
394.2 ml of solution contain 222.6 / 62 mol of ethylene glycol.
1000 ml of solution contain
9. A sample of drinking water was found to be severely contaminated
with CHCl3 which is supposed to be carcinogen.The level of contamination
was 15 ppm (by mass).
(i) Express this in percent by mass (ii) Determine the molality of
chloroform in the water sample.
•
Solution:
(i) 15 ppm means that 15 g (of CHCl3) is present in 106 g (million grams) of water.
Percentage of solution
106 g of water contains = 15 g of CHCl3
100 g of water contains =
= 15 × 10-4.
(ii) Molality of Solution :
No. of moles =
106 g of water contains =
in
Molecular mass of CHCl3 = (12 + 1 + 106.5) = 119.5
of CHCl3
1000 g of water contains =
= 1.25 × 10-4 m
i.e., Molality of the chloroform solution = 1.25 × 10-4 m.
5
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10. What role does the molecular interaction play in solution of alcohol
and water ?
•
Solution:
The molecular interaction that exists between alcohol and water molecule is H-bond. The alcohol
water mixture shows positive deviation. The interaction between ethanol-water molecules is weaker
than the interaction between ethanol – ethanol and water – water molecules.
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11. Why do gases always tend to be less soluble in liquids as the
temperature is raised ?
•
Solution:
The solubility of a gas decreases with increase in temperature. This is because on heating the
solution of a gas, some gas is usually expelled out of the solution. The dissolution of a gas in a liquid
is an exothermic process i.e., it is accompanied by evolution of heat i.e.,
Applying Le Chatelier’s principle, it is evident that increase of temperature would shift the equilibrium
in the backward direction, thus the solubility would decrease.
12. State Henry’s Law and mention some of its important applications.
•
Solution:
Henry’s Law states the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
It is also true that the solubility of a gas in a liquid solution is a function of the partial pressure of the
gas. Since mole fraction of the gas is proportional to the partial pressure of the gas. Partial pressure
of gas in solution = KH x c where c is the mole fraction and KH is the Henry’s Law constant.
Some applications of Henry’s Law can be stated below.
(i) To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high
pressure.
(ii) To minimize the painful effects accompanying the decompression of deep-sea divers, oxygen
diluted with less soluble He gas is used as breathing gas.
(iii) In lungs, where oxygen is present in air with high partial pressure, haemoglobin combines with
oxygen to form oxy-haemoglobin. In tissues where partial pressure of O2 is low, Oxy-haemoglobin
releases O2 for utilization cellular activities.
in
13. The partial pressure of ethane over a saturated solution containing
6.56 X 10-2 gm of ethane is 1 bar. If the solution contains 5.00 X 10-2g of
ethane, then what shall be the partial pressure of the gas?
•
Solution:
Applying Henry’s law, m = KH p
6.56
10-2 g =KH
KH =6.56
1 bar
10-2 bar-1.
6
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Now,5.0
10-2 g =6.56
10-2
p
P =0.762 bar
[mass of gas is taken to be proportional to mole fraction here]
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14. What is meant by positive and negative deviations from Raoult’s Law
and how is the sign of
related to positive and negative deviations from
Raoults Law ?
•
Solution:
According to Raoult’s Law there exists two types of solutions which show positive deviation and
negative deviation.
Positive Deviation: In this type of deviation A – B interactions i.e., solute – solvent interactions are
weaker
than solvent – solvent and solute – solute interactions. The escaping tendency of the components are
mole,
as the result
is > zero. Eg: ethanol water mixture. Ethanol cyclo hexane mixture.
Negative deviation: In this type of deviation A – B interactions are stronger than A – A or B – B
interactions.
If the components of the solution mixture are capable of forming H-bonds between them the
escaping tendency becomes less
as HCl, H2O mixture.
is < zero. Eg: chloroform acetone mixture, mineral acid such
Vapour pressure of solution
in
7
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15. An aqueous solution of 2% non-volatile solute exerts a pressure of
1.004 bar at the boiling point of the solvent. What is the molecular mass of
the solute?
•
Solution:
A 2% non-volative solute refers to the concentration of the solution with respect to mass percentage.
I.e., 2 gms dissolved in 100 gms of solution. The v.p of the solvent is 1.004 bar
in
MB =
WB of the solute = 2 gms
MA, mol mass of the solvent = 18 gms
WA, Wt of the solvent. = 98 g
P0A = 1 atm =1.013 bar
.
8
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16. Heptane and octane form ideal solution at 373K. The vapour
pressures of the two liquid components are 105.2 K Pa and 46.8 K Pa
respectively. What will be the vapour pressure in bar of a mixture of 25.0
gm of heptane and 35.0 of octane ?
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Solution:
According to Raoults Law.
Partial vapour pressure of the component = V.P of pure component X mole fraction
Number of moles of heptane =
= 0.25
Number of moles of octane =
= 0.307
Total number of moles = 0.25 + 0.307 = 0.557
Mole fraction of heptane =
= 0.448
Mole fraction of octane =
= 0.551
Partial pressure of heptane = 105.2 × 0.448 = 47.1296 K Pa
Partial pressure of octane = 46.8 × 0.551 = 25.7868 K Pa
Vapour pressure of the mixture = 47.1296 + 25.7868 = 72.916 K Pa
17. The vapour pressure of water is 12.3 K Pa at 300 K. Calculate vapour
pressure of 1 molal solution of a solute in it.
•
Solution:
According to Raoults Law, the relative lowering of vapour pressure is directly proportional to the mole
fraction of the solute
=
Vapour pressure of pure water at 300 K = 12.3 K Pa
Concentration of the solute = 1m
=
in
= 0.018
12.3 - PA = 0.018 × 12.3
12.3 - PA = 0.2214
Vapour Pressure of the solution containing 1 molar of the solute PA = 12.08 K Pa.
18. Calculate the mass of a non-volatile solute (molecular mass 40)
which should be dissolved in 114 gms of Octane to reduce its vapour
pressure to 80%.
9
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•
Solution:
pA= 80% i.e.,
=
Molar mass of Octane = 114 gms
Weight of the solvent i.e., Octane = 144 gms
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=
0.2 =
0.2 =
0.20 =
WB = 8 gms
The wt of non-volatile solute = 8 gms.
19. A solution containing 30 gms of a non-volatile solute exactly in 90
gms of water has a vapour pressure of 2.8 K Pa at 298 K. Further 18 gms of
water is then added to solution. The new vapour pressure becomes 2.9 K
Pa at 298 K. Calculate
(i) Molar mass of the solute
(ii) Vapour pressure of water at 298 K.
•
Solution:
(i) To calculate vapour pressure of water: -
=
=
in
=
- (i)
- (ii)
=
To cancel MB eg (i) is multiplied by 5 and eq (ii) by 6 we get
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=
=
We get
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=
- (iii)
=
- (iv)
Subtracting eq (iv) from (iii) we get
-1 +
=0
=1
(ii) Vapour pressure of pure solvent at 298 K,
To calculate molar mass
= 3.4 K Pa
=
=
=
Molar mass of the solute MB = 34 g mol-1
20. 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 273.15 K.
•
Solution:
The depression in freezing point of a solution is related to its molality by the expression.
in
∆Tf = Kfm
Kf for water = 1.86
For 5% solution of cane sugar the freezing point is 271 K.
The freezing point of pure water = 273.15 K
∆Tf = 273.15 K - 271 K = 2.15 K.
2.15 =
--- (i)
For II solution of 5% glucose
273.15 - T =
--- (ii)
Let us consider 273.15 - T = x
Dividing equation (i) by (ii)
11
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= 0.526
x X 0.526 = 2.15
x=
= 4.085
x = 273.15 - T
4.85 = 273.15 - T
T = 273.15 - 4.085
T = 269.07 K
The freezing point of 5% solution of glucose = 269.07 K.
21. Two elements A and B form compounds having molecular formula
AB2 and AB4 when dissolved in 20 g of C6H6. If 1 g of AB2 lowers the
freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K, (The molar
depression constant for benzene is 5.1 K kg mol-1.) calculate atomic mass
of A and B.
•
Solution:
For compound with molecular formula AB2.
The depression in freezing point = 2.3 K
Wt of the solute AB2 = 1 gm
Wt of the solvent = 20 g.
=
2.3 =
A + 2B =
=
= 110.86
III rly for compound AB4
Wt. Of the solute AB4 = 1
Wt. Of the solvent = 20
Depression in Freezing point = 1.3 K
A + 4B =
=
= 196.15
A + 2B = 110.86 - (i)
A + 4B = 196.15 - (ii)
Subtracting eq. (ii) from (i)
2B = 85.29
in
1.3 =
B=
= 42.64 gms
From eq. (i)
A + 2B = 110.85
A + 85.29 = 110.85
A = 110.85 – 85.29
12
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= 25.56 gms
The atomic masses of A and B are 25.56 gms and 42.64 gms respectively.
22. At 300 K 36g of glucose present per litre in its solution has an
osmotic pressure of 4.98 bar. If the osmotic pressure of solution is 1.52
bar at the same temperature. What would be its concentration.
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•
Solution:
From the osmotic pressure expression p =
Where p is the osmotic pressure
RT
is the concentration.
R = 0.0821 Litre atm K-1 mol-1 - universal gas constant.
4.98 =
- (i)
For the solution at the same temperature.
1.52 = x ´ 0.0821 ´ 300 - (ii)
Dividing eq. (i) by eq. (ii) we get
=
x=
= 0.061 M
concentration of the solution = 0.061 M.
23. Suggest the most important type of intermolecular attractive
interaction in the following pairs:
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) Methanol and acetone
(v) Acetonitrile (CH3CN) and acetone (C3H6O).
•
Solution:
(i) The intermolecular attractive forces between the A-A and B-B are nearly equal to those between
A-B.
(iii) A-B interactions are more than those between A-A and B-B.
in
(ii) A-B interactions are weaker than those between A-A and B-B.
(iv) dipole-dipole
(v) dipole-dipole
24. Based on solute – solvent interactions, arrange the following in
order of increasing solubility in n-octane and explain. Cyclohexane, KCl,
CH3OH, CH3CN.
13
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•
Solution:
KCl > CH3OH > CH3CN > Cyclohexane.
25. Amongst the following compounds, identify which are insoluble,
partially soluble and highly soluble in water?
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(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol chloroform
(vi) pentanol.
•
Solution:
Toluene and chloroform – insoluble in water.
Phenol and pentanol - partially soluble in water.
Formic acid and ethylene glycol – soluble in water.
26. If the density of some lake water is 1.25gmL-1 and contains 92g of
Na+ ions per kg of water, Calculate the molality of Na+ ions the lake.
•
Solution:
23g of Na+ ion corresponds to one mole
Therefore, 92g of Na+ ions will correspond to 92/23 = 4 mol
27. If the solubility product of CuS is 6 x 10
molarity of CuS in the aqueous solution.
, Calculate the maximum
Solution:
Suppose the solubility of CuS is x mol-1
This would give x mol-1 of Cu2+ ions and x mol L-1 of S2- ions on dissociation.
[Cu2+] = x mol-1
[S2-] = x mol-1
Ksp = [Cu2+][S2-] = (x) x (x) = x2
Ksp = 6 x 10 -16
Therefore, x = 6 x 10 -16
in
•
–16
x=
= 2.45 x 10-8
Maximum molarity = 2.45 x 10-8M
14
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28. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile
(CH3CN) when 6.5g of C9H8O4 is dissolved in 450g of CH3CN.
Solution:
Mass of aspirin = 6.5g
Mass of acetonitrile = 450g
Total mass = 450 + 6.5 = 456.5g
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29. Nalorphene (C19H21NO3), similar to morphine, is used to combat
withdrawal symptoms in narcotic users. Dose of nalorphene generally
given is 1.5 mg. Calculate the mass of 1.5 -103 m aqueous solution
required for the above dose.
Solution:
•
Molar mass of C19H21NO3 = 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g mol-1
o
o
o
o
o
o
1.5 x 10-3 m solution contains 1.5 x 10-3 x 312 g in one kg of water
1.5 x 10-3 x 312 =0.468g
Mass of solution =1000 + 0.468
=1000.468g
For 0.468g or 468mg of nalorphene the amount of solution required is 1000.468g
For 1.5 mg of nalorphene
in
Molality (m) =
1.5 =
Mass of solute = 1.5 x 311 = 466.5g
30. Calculate the amount of benzoic acid (C6H5COOH) required for
preparing 250mL of 0.15M solution in methanol.
15
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Solution:
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Number of moles of benzoic acid = Molarity x Volume in litres
= 0.15 x 0.25
= 0.0375 mol
Molecular mass of benzoic acid = 122 g mol-1
1 mol of benzoic acid corresponds to 122g of it
Therefore, 0.0375 mol of benzoic acid will correspond to 122 x 0.0375 = 4.575g
31. The depression in freezing point of water observed for the same
amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases
in the order given above. Explain briefly.
Solution:
•
o
Fluorine is more electronegative than chlorine. Hence trifluroacetic acid is stronger than trichloro
acetic acid which is stronger than acetic acid.
o
Extent of dissociation is in the order
o
Acetic acid<trichloroacetic acid<trifluoroacetic acid
o
∴€No of ions produced and depression in freezing point are also in the same order.
32. Calculate the depression in freezing point of water when 10g of
CH3CH2CHClCOOH is added to 250g of water. Ka = 1.4 x 10-3, Kf = 1.86
K.kg.mol-1.
•
Solution:
Molecular mass of CH3CH2CHClCOOH = 122.5g.mol-1
33. 19.5g 0f CH2FCOOH is dissolved in 500g of water. The depression in
the freezing point of water observed is 1.0° C. Calculate the Van’t Hoff
factor and dissociation constant of flouroacetic acid.
in
•
Solution:
Molecular mass of CH2FCOOH = 78g.mol-1
Number of moles of CH2FCOOH =
= 0.25mol
Given that, Observed ∆Tf = 1K
We have, ∆Tf =
=
= 0.93 K
16
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i=
o
=
= 1.0753.
Calculation of dissociation constant
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34. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the
vapour pressure of water at 293 K when 25g of glucose is dissolved in
450g of water.
•
Solution:
Vapour pressure of pure water
= 17.535mmHg
Vapour pressure of water in solution
=?
in
-
= 0.0056 x
- 0.0982 =
= 0.0056 x 17.535 = 0.0982
= 17.535 - 0.0982 = 17.437 mm Hg.
17
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35. Henry’s law constant for the molality of methane in benzene at 298
K is 4.27 x 105 mm Hg. Calculate the solubility of methane in benzene at
298 K under 760 mm Hg.
•
Solution:
P = KH.x
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= 178 x 10-5.
36. 100g of liquid A (molar mass 140 g mol-1) was dissolved in 1000g
liquid B (molar mass 180 g mol-1). The vapour pressure of pure liquid B
was found to be 5oo torr. Calculate the vapour pressure of pure liquid A
and its vapour pressure in the solution if the total vapour pressure of the
solution is 475 torr.
•
Solution:
Given that, Molecular mass of A, MA = 140 gmol-1
mass of A, WA = 100g
Molecular mass of B, MB = 180 gmol-1
mass of B, WB = 1000g
Total pressure = 475 torr
nA = 100/140 = 0.7143
nB = 1000/180 = 5.56
nA + nB = 6.2743
XA =
=
= 0.1138
in
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37. Benzene and toluene form ideal solution over the entire range of
composition. The vapour pressure of pure benzene and naphthalene at 300
K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole
fraction of benzene in vapour phase if 80g of benzene is mixed with 100g
of toluene.
Solution:
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•
38. The air is a mixture of a number of gases. The major components are
oxygen and nitrogen with approximate proportion of 20% is to 79% by
volume at 298 K. The water is in equilibrium with air at a pressure of 10
atm. At 298K are 3.39 x 107 mm and 6.51 x 107 mm respectively. Calculate
the composition of these gases in water.
•
X(Nitrogen) =
KHM2 = 6.51 x 107 mm =
in
Solution:
We have, p = KH.x
KH for oxygen = 3.39 x 107 mm
KH for nitrogen = 6.51 x 107 mm
=
= 8.56 x 104
XM2 =
=
= 1.17 x 10-4
We know that 1<H2O contains 55.55 moles of water. Suppose nN2 = number of moles of nitrogen in
19
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solution, then
XN2 =
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1.17 x 10-4 =
=
Thus n = 55.55 x 1.17 x 10-4 = 6.5 x10-3 and
millimoles Nitrogen = 6.5 x10-3x 103= 6.5m mol
39. Determine the amount of CaCl2 (i=2.47) dissolved in 2.5 litre of
water such that its osmotic pressure is 0.75 atm at 27°C.
•
Solution:
∏ = iCRT = i x n/V x RT
n = ∏ x V = 0.75 x 2.5 = 0.030 mol
i RT
2.47 x 0.083 x 300
Molecular mass of CaCl2 = 111 gmol-1
1 mol of calcium chloride corresponds to 111g
0.03 mol of calcium chloride will correspond to 0.03 x 111 = 3.33 g
40. Determine the osmotic pressure of a solution prepared by dissolving
25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely
dissociated.
•
Solution:
Molecular mass of potassium sulphate = 174.26 gmol-1
174.26g of potassium sulphate corresponds to 1 mole
25 x 10-3g of potassium sulphate will correspond to 25 x10-3
174.26
= 1.435 x 10-4mol
∏ = iCRT = i n/V RT =
= 5.26 x 10-3 atm
in
20
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