Page 1
UNIT 2
GRAM ATOM
Used to specify the amounts of chemical elements. It is defined as the mass in grams of an
element which is equal numerically to its atomic weight.
)1(weightatomic
gramsin.wtelementanofatomsgram −−−−−−−=
Similarly, the mass in kilograms of a given element that is numerically equal to its atomic
weight is called a kilogram-atom.
Similarly, kilogram atoms of element
)2(weightatomic
kgin.wtelementanofatomskg −−−−−−−=
For chemical compounds, a mole is defined as the amount of substance equal to its molecular
weight / formula weight.
GRAM MOLE
Used to specify the amount of chemical compounds.It is defined as the mass in grams of
substance that is equal numerically to its molecular weight.
)3(weightmolecular
gin.wt compound of gmoles −−−−−−−=
Gram mole of a substance is the mass in grams of the substance that is numerically equal to its
molecular weight.
Similarly,
)4(weightmolecular
kgin.wt compound of kgmoles −−−−−−−=
Mole is defined as the amount of substance equal to its molecular weight.
(1) Calculate the kilogram atoms of carbon which weighs 36 kg
Solution: 36 kg carbon
Atomic weight of carbon = 12
Page 2
312
36
carbonofweightatomic
kgin.wtcarbon of katom ===
(2) Calculate the kilograms of ‘Na’ of which the amount is specified as 3 katom.
Solution: 3 katom Na
Atomic weight of Na = 23
Naofweightatomic
Naofkgin.wt Na of katom =
∴ Kg of Na = katom of Na x Atomic weight of Na = 3 x 23 = 69
(3) How many kilograms of ethane are there in 210 kmol?
Solution: Basis: 210 kmol ethane.
Atomic weights: C=12, H=1, Chemical formula of ethane = C2H6
Molecular weight of ethane = 2x12+1x6 = 30
6H2Cofweight.mol
6H2Cofkgin.wt 6H2C of kmol =
∴ kg of ethane (C2H6) = kmol of ethane x Molecular weight of ethane = 210 x 30 =
6300 kg
∴ 210 kmol of ethane = 6300 kg ethane
(4) Convert 88 kg of carbon dioxide into its amount in molar units.
Solution: Basis: 88 kg of carbon dioxide
Molecular formula of carbon dioxide = CO2
Atomic weights: C=12, O=16
Molecular weight of CO2 = (1x12) + (2x16) = 44
244
88
2COofweight.mol
2OCofkgin.wt2CO of kmol ===
88 g of CO2 = 2 kmol CO2
Page 3
(5) Find the moles of oxygen present in 500 grams
Solution: Basis: 500 g of oxygen
Molecular weight of O2 = 2 x 16 = 32
625.1532
500 2O of kmoles ==
(6) Convert 499 g of CuSO4.5H2O into moles.
Solution: Basis: 499 g of CuSO4.5H2O
Atomic weights: Cu=63.5, S=32, O= 16 and H = 1
Molecular weight of CuSO4.5H2O
= (1x63.5)+(1x32)+(4x16)+5(2x1+1x16) = 249.5
Moles of CuSO4.5H2O = 499 = 2 mol
249.5
25.249
499 O2.5H4CuSO of kmoles ==
The relationship of a compound and its constituents is given for some compounds as follows:
Each mole of NaOH contains one atom of Na
1 mol of NaOH≡ 1 atom of Na ≡ 1 atom of H
Each mole of NaOH contains 1 atom of Na. The sign ≡ refers to ‘equivalent to’ and not ‘equal
to’.
Similarly for H2SO4 and ‘S’
1 mol of H2SO4 = 1 atom of S (atom is written for gram-atom).
1 mol of H2SO4 = 1 atom of S (atom is written for gram-atom)
1 kmol of H2SO4 = 1 katom of S (katom is written for kilogram-atom)
i.e., each mole of H2SO4 contains 1 atom of S.
For CuSO4.5H2O and CuSO4
1 mol of CuSO4.5H2O ≡ 1 mol CuSO4
1 kmol of CuSO4 . 5H2O ≡ 1 kmol of CuSO4
(7) How many kmoles of H2SO4 will contain 64 kg S
Page 4
Solution: Basis: 64 kg of S
Atomic weight of S = 32
232
64
katom
Sofkg S of atoms ===
Each moles of H2SO4 contains one atom of S.
1 kmol of H2SO4≡ 1 katom of S
kmol22x1
1 4SO2H of moles ==
(8) Find kmoles of K2CO3 that will contain 117 kg of K?
Solution: Basis: 117 kg of K
Atomic weight of K = 39
katom339
117K of atoms ==
Each mole of K2CO3 contains 2 atom of K
2 atom of K ≡ 1 mole of K2CO3
2 katom of K ≡ a kmol of K2CO3
kmol5.13x2
1 3CO2K of moles ==
(9) How many kilograms of carbon are present in 64 kg of methane?
Solution: Basis: 64 kg of methane
Atomic weight of C = 12
Molecular weight of CH4 = 16
1 katom of carbon ≡ 1 kmol of CH4
∴ 12 kg of carbon ≡ 16 kg of CH4
i.e., in 16 kg of CH4, 12 kg of carbon are present.
So amount of carbon present in 64 kg of methane = 4864x16
12=
(10) Find equivalent moles of Na2SO4 in 644 kg of Na2SO4.10H2O crystals
Solution: Basis: 644 kg of Na2SO4.10H2O crystals
Molecular weight of Na2SO4 = 2x23+1x32+4x16 = 142
Molecular weight of Na2SO4.10H2O
= (2x23)+1x32+4x16+10 (2x1+1x16) = 322
mol2322
644 O2.10H4SO2Na of moles ==
Page 5
The relationship betweenNa2SO4 and Na2SO4.10H2O is :
1 mol Na2SO4.10H2O ≡ 1 mol Na2SO4
∴ 2 mol of Na2SO4.10H2O ≡ ?
Moles of Na2SO4 = 2 mol
In the formula of Na2SO4.10H2O, the moles of Na2SO4 are equal (one in each) and hence, the
equivalent moles of Na2SO4 in crystals are 2 mol.
(11) Find nitrogen (N) content of 100 kg urea sample containing 96.43% pure urea.
Solution: Basis: 100 kg of urea sample
It contains 96.43 kg of urea
Molecular weight of urea [NH2CONH2] = 60
1 kmol of NH2CONH2≡ 2 katom of N
60 kg of NH2CONH2≡ 28 kg of N
∴96.43 kg of NH2CONH2≡ ?
Nitrogen content of 100 kg sample
= kg4543.96x60
28=
(12) Find kilograms of C2H6 that contain 4 katom of carbon.
Solution: Basis: 4 katom carbon
Molecular weight of C2H6 = 30
Relationship between C2H6 and ‘C’ is
2 katom of C ≡ 1 kmol of C2H6
∴ 4 katom of C ≡ ?
Moles of C2H6 = 1 x 4 = 2 kmol
Weight of C2H6 = kmol of C2H6 x Molecular weight of C2H6 = 2 x 30 = 60 kg
(13) How many grams of carbon are present in 264 g of CO2?
Solution: Basis: 264 g of CO2
Atomic weight of C = 12,
Molecular weight of CO2 = 44
1 mol of CO2≡ 1 atom of C
44 g of CO2≡ 12 g of C
∴264 g of CO2≡ ?
Amount of carbon present in 264 g of CO2 = g72264x44
12 =
Page 6
EQUIVALENT WEIGHT
It is defined as the ratio of the atomic weight or molecular weight to its valence.The valence of
an element or a compound depends on the numbers of hydroxyl ions (OH-) donated or the
hydrogen ions (H+) accepted for each atomic weight or molecular weight. Equivalent weight =
Molecular weight
Valence
valence
weightmolecularweight equivalent =
NORMALITY
It is designated by the symbol ‘N’. It is defined as the number of gram equivalents of solute
dissolved in one litre of solution.
Normality (N) = gram-equivalents of solute= gram of solute_____________________
Volume of solution in Litre eq wt. of solute x Volume of solution in Litre
Linsolutionofvolumexsoluteof.eqwt
soluteofg
Linsolutionofvolume
soluteoftsgequivalenNormality ==
where Linsolutionofvolume
soluteofg)L/g(ionconcentrat =
By using definition of normality, it is possible to find out the concentration of solute in g/l.
Concentration (g/L) = Normality x Equivalent weight
MOLARITY
It is defined as the number of gram moles of solute dissolved in one litre of solution.
Designated by the symbol ‘M’.
Linsolutionofvolume
soluteofg)M(Molarity =
Page 7
MOLALITY
It is defined as the number of gram-moles (mol) of solute dissolved in one kilogram of
solvent.
kginsolventofmass
soluteofgmoles)M(Molality =
Ppm and ppb
solutionL1
solutemg1
solutionL610
soluteg1
solutiong610
soluteg1ppm1 ===
since density of solution(very dilute) = 1 g/cc
solutionL910
soluteg1
solutiong910
soluteg1 1ppb ==
(14) Find the equivalent weights of (1) HCl, (2) NaOH, (3) Na2CO3 and (4) H2SO4
Solution: (1) HCl:
Molecular weight of HCl = (1x1)+(1x35.5) = 36.5
Valence of HCl = 1
5.361
5.36
HClofvalence
HClofwtmol HCl ofweight equivalent ===
(2) NaOH:
Molecular weight of NaOH = (1x23)+(1x16)+(1x1) = 40
Valence of NaOH = 1
401
40 NaOH ofweight equivalent ==
Page 8
1
(3) Na2CO3 :
Molecular weight of Na2CO3=(2x23)+(1x12)+(3x16) = 106
Valence of Na2CO3 = 2
532
106
3CO2Naofvalence
3CO2Naofwtmol 3CO2Na ofweight equivalent ===
(4) H2SO4
Molecular weight of H2SO4 = (2x1) + (1x32) + (4x16) = 98
Valence of H2SO4 = 2
492
98 4SO2H ofweight equivalent ==
(15) Calculate the equivalent weights of the following compounds:
(1) H3PO4, (2) CaCl2, (3) FeCl3, (4) Al2(SO4)3, and (5) KMnO4
Solution: (1) H3PO4
Molecular weight of H3PO4 = (3x1)+(1x31)+(4x16) = 98
Valence of H3PO4 = 3
67.323
98 3CO2Na ofweight equivalent ==
(2) CaCl2
Molecular weight of CaCl2 = (1x40)+(2x35.5) = 111
Valence of CaCl2 = 2
5.552
111
2CaClofvalence
2CaClofwtmol 2CaCl ofweight equivalent ===
(3) FeCl3:
Molecular weight of FeCl3 = (1x56)+(3x35.5) = 162.5
Valence of FeCl3 = 3
∴ Equivalent weight of FeCl3 = 162.5 = 54.17
3
17.543
5.162 ofweight equivalent
3
33 ===
FeClofvalence
FeClofwtmolFeCl
(4) Al2(SO4)3:
Page 9
Molecular weight of Al2(SO4)3 = (2x27)+(3x32)+(12x16) = 342
Valence of Al2(SO4)3 = 6
576
342
(SO4)Al
(SO4)Al (SO4)Al ofweight equivalent
32
3232 ===
ofvalence
ofwtmol
(5) KMnO4:
Molecular weight of KMnO4 = (1x39)+(1x55)+(4x16) = 158
Valence of KMnO4 = 5,
6.315
158
KMnO
KMnO (SO4)Al ofweight equivalent
4
432 ===
ofvalence
ofwtmol
(16) 98 grams of sulphuric acid (H2SO4) are dissolved in water to prepare one litre of
solution. Find normality and molarity of the solution.
Solution: Basis: One litre of solution
Amount of H2SO4 dissolved = 98 g
Molecular weight of H2SO4 = 98
492
98
4SO2Hofvalence
4SO2Hofwtmol 4SO2H ofweight equivalent ===
Gram-equivalents (g eq) of H2SO4 = 98 = 2
49
249
98
4SO2Hofwteq
4SO2Hofwt 4SO2H ofsequivalent g ===
21
2
Lin.sol4SO2Hof.vol
4SO2Hofeqg (N)Normality ===
11
1
Lin.sol4SO2Hof.vol
4SO2Hofmoles (M)Molarity ===
(17) Find grams of HCl needed to prepare 1 litre 2N HCl solution.
Solution: Basis: 1 litre 2N HCl solution.
Page 10
g731x5.36x2ginHClofwt
LinsolutionofvolumexHClof.wteq
ginHClofwt2
Lin.solClHof.vol
HClofeqg (N)Normality
==
=
=
(18) The concentration of an aqueous solution of acetic acid is specified as 30% on weight
basis. Find the molarity of the solution.
Solution: Basis: 100 kg of solution.
Amount of acetic acid in it = 30 kg
Amount of water (solvent) in it = 70 kg
Amount of acetic acid = 30 x 103 g
Molecular weight of CH3COOH = 60
m142.770
500
solventofkg
acidaceticofmolesmolality
mol50060
310x30 acidaceticofmoles
===
==
(19) A solution of caustic soda contains 20% NaOH by weight. Taking density of the
solution as 1.196 kg/L find the normality, molarity, and molality of the solution.
Solution: Basis: 100 kg of solution.
The solution contains 20 kg of NaOH and 80 kg water (solvent).
Density of solution = 1.196 kg/L
L62.83196.1
100solutionofvolume ==
mol 500=kmol 0.5 =40
20=solution in NaOH of Moles
M98.583.52
500 =
Linsolution of volume
NaOH ofmoles g= (M)Molarity =
For NaOH as valence = 1
Equivalent weight = Molecular weight
∴ Normality (N) = Molarity (M) = 5.98
Page 11
(mol/kg) m 6.2580
500 =
solvent of kg
NaOH ofmoles g= (m)Molality =
(20) A chemist is interested in preparing 500 ml of 1 normal, 1 molar and 1 molal solution
of H2SO4. Assuming the density of H2SO4 solution to be 1.075 g/cm3, calculate the
quantities of H2SO4 to be taken to prepare these solutions.
Solution: Basis: 500 ml of solution.
Volume of solution = 500 ml = 0.5 L
Lin solution of volumw
4SO2Hof sequivalent g= (N)Normality
∴ gram-equivalents of H2SO4
= Normality x Volume of solution
= 1 x 0.5 = 0.5 g eq
Molecular weight of H2SO4 = 98
492
98 =4SO2 Hweight of equivalent =
Amount of H2SO4 required for 1 normal solution
= 0.5 x 49 = 24.5 g
Linsolutionof.vol
4SO2Hofgmolesmolarity =
Moles of H2SO4 = Molarity x Volume of solution = 1 x 0.5 = 0.5 mol
Amount of H2SO4 required to make 1 molar solution = 0.5 x 98 = 49 g
Let x be the quantity in grams of H2SO4 required for making 1 molal solution.
Density of solution = 1.075 g/cc
Quantity of solution = 500 x 1.075 = 537.5 g
grams of solvent = grams of solution – grams of solute = 537.5 – x
Weight of solvent = (537.5 - x) x 10-3
kg
Molecular weight of H2SO4 = 98
98
x4SO2Hofmoles =
Page 12
kginsolventofwt
soluteofgmoles4SO2HofMolality =
3-10 . x)- (537.5
98/x1=
Solving we get, x = 47.97 g
Amount of H2SO4 required for preparing 1 molal solution = 47.97 g.
(21) 2000 ml solution of strength 0.5 N H2SO4 is to be prepared in laboratory by adding
98% H2SO4 (sp.gr.1.84) to water. Calculate the volume in ml of 98% to be added to get
solution to required strength.
Solution: Basis: 2000 ml of 0.5 N H2SO4 solution.
Lin solution of volume
4SO2Hof sequivalent g= (N)Normality
Volume of solution = 2000 ml = 2 L
gram equivalents of H2SO4 = Normality x Volume of solution in L = 0.5 x 2 = 1 g eq
Amount of H2SO4 required = 1 x 49 = 49 g
g500.98
49= required 4SO2H98% ofamount =
Specific gravity of 98% H2SO4 = 1.84
ml72.284.1
50
gravityspecific4SO2H98%ofg
= required 4SO2H98% ofvolume ==
(22) A H2SO4 solution has a molarity of 11.24 and molality of 94. Calculate the density of
solution.
Solution: Basis: 1 litre of solution
Molarity = 11.24 and Molality = 94
Now,
kginsolventofwt
soluteofgmoles4SO2HofMolality =
Page 13
1
4SO2Hofgmoles24.11 =
∴Moles of H2SO4 = 11.24 x 1 = 11.24 mol
Amount of H2SO4 = 11.24 x 98 = 1101.52 g = 1.101 kg
Molality = mol H2SO4/ kg of solvent
1
4SO2Hofgmolessolutionofamount =
kg1195.094
24.11solventofamount ==
Amount of solution = 1.101 + 0.1195
= 1.2205 kg
L/kg2205.11
2205.1solution4SO2Hofdensity ==
(23) 2 litres of NH3 at 303 K (300C) and 20.265 kPa is neutralized by 135 ml of solution of
H2SO4. Find the normality of the acid.
Solution: Basis: 2 litres of NH3
PV = nRT
RT
PV3NHofmolesn ==
Where P = 20.265 kPa, T= 303 K, V = 2 l
R = 8.31451 m3
.kPa/(kmol.K) = 8.31451 kPa/(mol.K)
m0161.0)303.(314.8
)2.(265.203NHofmolesn ===
2 NH3 + H2SO4 (NH4)2SO4
For neutralization of 2 moles of NH3, 1 mole of H2SO4 is required.
310x045.8)0161.0.(2
1required4SO2Hofmoles −
==
Amount of H2SO4 required
= 8.045 x 10-3
x98= 0.788 g
0161.049
788.0required4SO2Hoftsgequivalen ==
Volume of H2SO4 solution = 135 ml = 0.135 L
Page 14
N12.0135.0
0161.04SO2HofN ==
(24) A sample of Na2CO3. H2O weighing 0.62 grams is added to 100 ml of 0.1
H2SO4solution. Will the resulting solution be acidic, basic or neutral? (Get the answer
by numerical method.)
Solution: Basis: 100 ml of 0.1 N H2SO4solution and 0.62 g of Na2CO3. H2O sample.
Normality of H2SO4 solution = 0.1 N
Volume of H2SO4 solution = 100 ml = 0.1 l
Gram equivalents of H2SO4 = Normality x Volume of solution
= 0.1 x 0.1 = 0.01
Amount of H2SO4 in solution = 0.01 x 49 = 0.49g
Molecular weight of Na2CO3 = 124, Molecular weight of H2SO4 = 98
Na2CO3. H2O + H2SO4� Na2SO4 + 2H2O + CO2
1 mol Na2CO3.H2O = 1 mol H2SO4
124 g Na2CO3.H2O = 98 g H2SO4
∴ 0.62 Na2CO3.H2O = ?
∴ Amount of H2SO4 required for 0.63 g Na2CO3.H2O g49.0)62.0(124
98==
For neutralizing, we need 0.49 g H2SO4 as per reaction and we have added solution
containing 0.49 g H2SO4. None of the components is in excess. Hence, resulting
solution is neutral.
(25) Do the following conversions:
a. 294 g/L H2SO4 to normality, b. 5N H3PO4 to g/L, c. 54.75 g/L HCl to molarity, d.
3M K2SO4 to g/L,e. 4.8 mg/ml CaCl2 to normality
Sol: (a) 294 g/l H2SO4 to normality
Basis: 1 litre of solution.
H2SO4 in solution = 294 g
geq649
294required4SO2Hoftsgequivalen ==
N61
6N ==
(b) 5N H3PO4 to g/L
Molecular weight of H3PO4 = 98
Valence of H3PO4 = 3
67.323
98required4PO3Hofweightequivalent ==
Concentration in grams per litre of solution = Normality x Equivalent weight
Concentration of solution = 5 x 32.67 = 163.35 g/L
(c) 54.75 g/L HCl to molarity
Basis: 1 l of solution
Page 15
Amount of HCl in g/l = 54.75 g
mol5.15.36
75.54HClofmoles ==
M5.11
5.1solutionHClmolarity ==
(d) 3M K2SO4 to g/L
Basis: 1 l of solution.
Moles of solution = 3 x 1 = 3 mol
Molecular weight of K2SO4 = 174
∴ Amount of K2SO4 = 3 x 174 = 522 g
L/g5221
522solutionHClmolarity ==
(e) 4.8 mg/ml CaCl2 to normality
Basis: 1 L of solution
4.8 mg/ml CaCl2 = 4.8 g/l CaCl2
CaCl2 in solution = 4.8 g
Molecular weight of CaCl2 5.552
111==
Gram equivalents of CaCl2 = 0865.05.55
48=
Normality = 0865.01
0865.0==
(26) An aqueous solution of K2CO3 is prepared by dissolving 43 kg of K2CO3 in 100 kg of
water at 293 K (200C). Calculate molarity, normality, molality of solution. Density of
solution is 1.3 kg/L.
Solution: Basis: 43 kg of K2CO3 and 100 kg of water
Weight of K2CO3 solution = 43 + 100 = 143 kg
Volume of solution = L.1103.1
143=
Molecular weight of K2CO3 = 138
∴ Equivalent weight of K2CO3 692
138==
∴ Moles of K2CO3 in solution = mol6.311kmol3116.0138
43==
Molarity of solution = M832.2110
6.311=
Page 16
Gram equivalents of K2CO3 = eq19.62369
)1000.(43=
Normality of solution = N665.5110
19.623==
Molality of solution = N3116.0100
6.311=
METHODS OF EXPRESSING THE COMPOSITION OF MIXTURES AND
SOLUTIONS
There are various methods used in expressing the composition of mixtures and
solutions.The methods are being explained by considering the system composed of two
components namely, A and B and is also used for more than two components.
The composition of a mixture or solution are expressed in weight percent, volume percent,
mole percent.
WEIGHT PERCENT
The weight of any component expressed as a percentage of the total weight of the system.
∴ Weight % of A = )1(100xW
AW
systemofweighttotal
Aofweight−−−−−−=
Where, WA = weight of the component A
W= WA + WB = weight of the system …. for a binary system of A and B.
Weight percent of component A present in system is defined as the weight of the
component A expressed as a percentage of the total weight of the system.
VOLUME PERCENT
The pure component volume of any component is expressed as a percentage of the total
volume of the system.
Volume % of A = )2(100xV
AV
systemofvolumetotal
Aofvolumecomponentpure−−−−−=
Where, VA = pure component volume of A, VB = total volume of the system
VA+VB … for a binary system of A and B.
MOLE FRACTION
The ratio of the number of moles of a particular component to the total moles of a system.
∴ For a binary system of A and B :
Mole fraction of A =systemofmolestotal
Aofmoles
Page 17
)3(
BM
BW
AM
AW
AM
AW
Ax −−−−−−−−−−
+
=
Where, XA = mole fraction of A
XB = mole fraction of B.
)4(
BM
BW
AM
AW
MB
BW
Bx −−−−−−−−−−
+
=
From equation (3) and (4), we have
Mole % of A = Mole fraction of A x 100
Adding equations (3) and (4), we get
)5(1
BM
BW
AM
AW
MB
BW
BM
BW
AM
AW
AM
AW
Ax −−−−−−−=
+
+
+
=
∴the sum of the mole fractions of all the components present in a given system is equal
to unity, or
NOTE: The sum of all the mole percentage for a given system totals to 100.
WEIGHT FRACTION
The ratio of the weight of a particular component to the total weight of the system.
∴In a binary system of A and B :
)6(BWAW
AW'Ax −−−−−−−−−−
+=
Page 18
Where xA is the weight fraction of A.
Similarly for B,
)7(BWAW
BW'Bx −−−−−−−−−−
+=
From equations (6) & (7), we get
1BWAW
BW
BWAW
AW'Ax =
++
+=
Also, Weight % of A = Weight fraction of A x 100
∴The sum of the weight fractions of all the components present in a given system is
equal to unity, or
NOTE: The sum of all the weight percentages for a given system totals to 100.
(28) An aqueous solution of sodium chloride is prepared by dissolving 25 kg of sodium
chloride in 100 kg of water. Determine (a) weight % and (b) mole % composition of
solution.
Solution:Basis: 25 kg of sodium chloride and 100 kg of water
Amount of solution = 25 + 100 = 125 kg
Weight % NaCl in solution = %20100x125
25
kginsolutionofweighttotal
kginNaClofweight==
Weight % H2O = 100 – Weight % of NaCl
= 100 – 20 = 80 %
Molecular weight of NaCl = 58.5, Molecular weight of H2O = 18.
∴ Moles of NaCl = kmol427.05.58
25=
Moles of H2O = kmol56.518
10=
Total moles of solution = 0.427 + 5.56 = 5.987 kmol
Mole % NaCl in solution = %13.7100x987.5
427.0
solutionofkmol
NaClkmol==
Page 19
Mole % H2O in solution = 100 – Mole % of NaCl
= 100 – 7.13 = 92.87 %
(29) Sodium chloride weighing 200 kg is mixed with 600 kg potassium chloride. Calculate
the composition of the mixture in (i) weight % and (ii) mole %
Solution: Basis: 200 kg NaCl and 600 kg KCl
Weight of NaCl in the mixture = 200 kg
Weight of KCl in the mixture = 600 kg
Total weight of the mixture = 600 + 200 = 800 kg
∴ Weight % of NaCl in the mixture
= %75100x800
200
kginmixtureofweighttotal
kginNaClofweight==
Weight % KCl = 100 – weight % of NaCl = 100 – 25 = 75%
Molecular weight of NaCl = (1 x 23) + (1 x 35.5) = 58.5
Molecular weight of KCl = 1 x 39 + 1 x 35.5 = 74.5
Moles of NaCl = kmol419.35.58
200=
Moles of KCl= kmol05.85.74
600=
Total moles of the mixture = 3.419 + 8.05 = 11.469 kmol
Mole % NaCl = %81.29100x469.11
419.3
mixtureofkmol
NaClofkmol==
Mole % KCl = 100 – mole % NaCl
100 – 29.81 = 70.19%
(30) A saturated solution of salicylic acid (HOC6H4COOH) in methanol (CH3OH) contains
64 kg salicylic acid per 100 kg methanol at 298 K (250C). Find the composition of
solution in (a) weight % and (b) mole %
Sol: Basis: 100 kg of methanol in the saturated solution
Amount of salicylic acid corresponding to 100 kg methanol in saturated solution
Weight of solution = 100 + 64 = 164 kg
∴ Weight % salicylic acid solution = %02.39100x164
64=
Weight % methanol = 100 – 39.02 = 60.98%
Molecular weight of CH3OH = 32, Molecular weight of HOC6H4COOH = 138
Mole of methanol = 100/32 = 3.125 kmol
Moles of Salicylic acid = 64/138 = 0.464 kmol
Page 20
Total amount of solution = 3.125 + 0.464 = 3.589 kmol
Mole % of methanol = %07.87100x589.3
125.3=
Mole % of salicylic acid = 100 – 87.07 = 12.93%
(31) At 298 K (250C) the solubility of methyl bromide in methanol is 44 kg per 100 kg.
Calculate (i) the weight fraction and (ii) the mole fraction of methanol in the saturated
solution.
Solution: Basis : 100 kg of methanol in the saturated solution
Solution contains 44 kg of methyl bromide
Weight of the saturated solution = 100 + 44 = 144 kg
Weight fraction of methanol in the saturated solution = 694.0144
100=
Molecular weight of CH3OH = 32, Molecular weight of CH3Br = 94.91
Moles of CH3OH in solution = 100/32 = 3.125 kmol
Moles of CH3Br in solution = kmol4636.091.94
44==
∴ Total moles of solution = 3.125 + 0.4636 = 3.5886 kmol
Mole fraction of methanol in saturated solution= 871.05886.3
125.3=
(32) Ethanol and water forms aazeotrope containing 96% ethanol by weight. Find the
composition of azeotrope by mole %.
Solution: Basis: 100 kg of ethanol-water mixture
It contains 96 kg of ethanol and 4 kg of water
Molecular weight of H2O = 18. Molecular weight of C2H5OH = 46
Moles of ethanol = 96/46 = 2.087 kmol
Moles of water = 4/18 = 0.222 kmol
Moles of azeotrope mixture = 2.087 + 0.222 = 2.309 kmol
Mole % ethanol in azeotrope mixture = %38.90100x309.2
087.2==
(33) Nitric acid and water forms maximum boiling azeotrope containing 62.2 % water by
mole. Find the composition of the azeotrope in weight %.
Solution: Basis: 100 kmol of HNO3 + water azeotrope
It contains 62.2 kmol of H2O and 37.88 kmol of HNO3
Molecular weight of HNO3 = 63. Molecular weight of H2O = 18
Amount of HNO3 in azeotrope = 37.8 x 63 = 2381.4 kg
Amount of H2O in azeotrope = 62.2 x 18= 1119.6 kg
Amount (weight) of azeotrope = 2381.4 + 1119.6 = 3501 kg
Page 21
Weight % HNO3 in azeotrope = %02.68100x3501
4.2381==
(34) An aqueous solution contains 15% ethanol by volume. Find the weight % ethanol, if
densities of ethanol and water are 0.79 g/cm3 respectively.
Solution: Basis: 100 cm3 of aqueous solution.
Volume of ethanol in solution = 0.15 x 100 = 15 cm3
Volume of water in solution = 0.85 x 100 = 85 cm3
Amount of ethanol in solution = Volume x Density = 15x 0.79 = 11.85 g
Amount of water in solution = 85 x 1 = 85 g
∴ Total amount of solution = 11.85 + 85 = 96.85 g
∴ Weight % ethanol in solution = %23.12100x85.96
85.11==
(35) The available nitrogen (N) in the urea sample is found to be 45 % by weight.
Calculate the actual urea content in the sample.
Solution: Basis: 100 kg of urea sample.
It contains 45 kg of nitrogen as N
Atomic weight of N = 14. Molecular weight of NH2CONH2 = 60
1 kmol NH2CONH2≡ 2 katom N
∴ 60 kg NH2CONH2 ≡ 28 kg N
∴ Actual urea in sample = kg43.9645x28
60==
Weight % urea content of sample = %43.96100x100
43.96100x
samplekg
ureakg==
(36) The nitrogen content of NH4NO3 sample is given as 34.5 % by weight. Find the actual
ammonium nitrate content of the sample.
Solution: Basis: 100 kg of sample
It contains 34.5 kg of nitrogen as N
Molecular weight of NH4NO3 = 80, Atomic weight of N = 14
1 kmol NH4NO3≡ 2 katom N
∴ 80 kg NH4NO3≡ 28 kg N
∴ Amount of NH4NO3 in sample = kg57.985.34x28
80=
% NH4NO3 content of sample = %57.98100x100
57.98100x
100
57.98==
∴ Actual NH4NO3 content of sample is 98.57 %
Page 22
(37) Calculate the available nitrogen content of solution having 30 % urea
(NH2CONH2), 20 % ammonium sulphate and 20 % ammonium nitrate.
Solution: Basis: 100 kg of solution.
It contains 30 kg urea, 20 kg of ammonium sulphate and 20 kg ammonium nitrate.
Molecular weight of urea (NH2CONH2) = 60
1 kmol NH2CONH2 ≡ 2 katom N
60 kg NH2CONH2≡ 28 kg N
∴ Nitrogen available from urea = kg1430x60
28=
Molecular weight of (NH4)2SO4 = 132
1 kmol (NH4)2SO4 ≡ 2 katom N
132 kg (NH4)2SO4≡ 28 kg N (on weight basis)
∴ Nitrogen available from (NH4)2SO4 = kg24.420x132
28=
Molecular weight of NH4NO3 = 80
1 kmol NH4NO3≡ 2 katom N
80 kg NH4NO3≡ 28 kg of N (on weight basis)
∴ Nitrogen available from NH4NO3 = kg720x80
28=
∴ Total nitrogen available from the solution= 14 + 4.24 + 7 = 25.24 kg
∴ Nitrogen available from the solution = %24.2530x100
24.25=
(38) Spent acid from fertilizer plant has the following composition by weight: H2SO4 =
20%, NH4SO4 = 45%, H2O = 30% and organic compounds = 5%. Find the total
acid content of the spent acid in terms of H2SO4 after adding the acid content,
chemically bound in ammonium hydrogen sulphate.
Solution: Basis: 100 kg of spent acid
It contains 20 kg of H2SO4 and 45 kg of NH4SO4
1 kmol NH4SO4 ≡ 1 kmol H2SO4
115 kg NH4SO4 ≡ 98 kg H2SO4
∴ H2SO4 chemically bound in NH4SO4 = kg35.3845x115
98=
H2SO4 from the spent acid = free H2SO4 + H2SO4 chemically bound in NH4SO4
= 20 + 38.35 = 58.35 kg
Total acid content of the spent acid as weight % H2SO4
Page 23
%35.58100x100
35.58=
(39) What will be % Na2O content of lye containing 73% (by weight) caustic soda ?
Solution: Basis: 100 kg of lye
It contains 73 kg of caustic soda (NaOH)
2 NaOH = Na2O = Na2O + H2O
Molecular weight of NaOH = 40, Molecular weight of Na2O = 62
2 kmol NaOH ≡ 1 kmol Na2O
80 kg NaOH ≡ 62 kg Na2O
∴ Amount of Na2O in the lye = kg575.5673x80
62=
Weight % Na2O in the lye = %575.56100x100
56.575=
(40) An aqueous solution of soda ash contains 20 % soda ash by weight. Express the
composition as weight % Na2O.
Solution: Basis: 100 kg of an aqueous solution of soda ash
Amount of soda ash in the solution = 0.2x100= 20 kg
Na2CO3 = Na2O + CO2
Molecular weight of Na2CO3 = 106, Molecular weight of Na2O = 62
1 kmol Na2CO3≡ 1 kmol Na2O
106 kg Na2CO3≡ 62 kg of Na2O
Weight of Na2O in the solution = kg7.11100x106
62=
Weight % Na2O in the solution = %7.11100x100
11.7=
(41) A sample of caustic soda flakers contains 74.6 % Na2O by weight. Determine the
purity of the flakes.
Solution: Basis: 100 kg of caustic soda flakes
It contains 74.6 kg of Na2O
2 NaOH = Na2O + H2O
2 kmol of NaOH ≡ 1 kmol of Na2O
80 kg of NaOH ≡ 62 kg of Na2O
Amount of NaOH in the flakes =80 x74.6=96.26 kg
62
% purity of the flakes = kg NaOH x 100 kg flakes = %26.96100x100
96.26=
(42) The strength of a phosphoric acid sample is found to be 35% P2O5 by weight.
Determine the actual concentrations of H3PO4 (by weight) in the acid.
Page 24
Solution: Basis: 100 kg of phosphoric acid sample
It contains 35 kg of P2O5.
2H3PO4 = P2O5 + 3H2O
Molecular weight of H3PO4 = 96, Molecular weight of P2O5 = 142
2 kmol H3PO4≡ 1 kmol P2O5
196 kg of H3PO4≡ 142 kg of P2O5
∴Amount of H3PO4 in sample = kg31.4835x142
196=
Weight % of H3PO4 in phosphoric acid sample= %31.48100x100
48.31=
(43) Caustic soda flakes received from a manufacturer are found to contain 60 ppm silica
(SiO2), Convert this impurity into weight %.
Sol: Basis: 106 kg of caustic soda flakes
Silica in flakes is given as 60 ppm
∴ Amount of silica in flakes = kg60610x610
60=
Weight % of silica in flakes = 006.0100x610
60=
(44) A sample of water contains 2000 ppm solids. Express the concentration of solids in the
sample in weight percent.
Solution: Basis: 106
kg of sample of water.
Solids in water = 2000 ppm
Amount of solids in water = kg2000610x610
2000=
Weight % of solids = %2.0100x610
2000=
GASES
The composition are expressed in terms of volume percent, as the volume can be measured
easily. The density can be calculated if the parameters such as temperature and pressure are
known, which in turn gives mass of the gas. The relationship among mass, temperature and
volume should be a known one, if it that needed to dealing with substances existing in the
gaseous state.
Page 25
IDEAL GAS LAW
BOYLE’S LAW
Given mass of an ideal gas, the product of the pressure and volume is constant at a constant
temperature i.e., P x V = Constant --------- (A)
Where P is the absolute pressure and V is the volume occupied by the gas.
CHARLE’S LAW :
Given mass of an ideal gas, the ratio of the volume to temperature is constant at a given
pressure, i.e.,
)B(ttanconsT
V−−−−−−−−−=
Where T is the absolute temperature
By combining the above two laws, it can be formulated as an ideal gas laws as
)C(ttanconsT
PxV−−−−−−−−−=
The constant of the above equation is designated by the symbol R and is known as
Universal gas constant
Therefore, PV = RT ---- (D)
When V is the volume in cubic meters of n kmol of gas, then the ideal gas is written as
PV = nRT ------ (E)
The ideal gas law states three facts :
(i) volume of gas is directly proportional to numberof moles.
(ii) volume is directly proportional to the absolutetemperature
(iii) volume is inversely proportional to the pressure.
When P (absolute pressure) is in kPa, V is in kmol and T is in K, then the units of R will be
m3.
kPa/(kmol.K).
When P is in Mpa, V is in m3, T is in K and n is in kmol, then numerical value of R is
0.008314 with the units m3.
Mpa/(kmol.K).
When P is in kPa, V is in m3, T is in K and n is in kmol, then numerical value of R is
8.31451 with units m3.
kPa/(kmol.K)
When the mass of a gas is not known, and if we know the volume occupied at specified
temperature and pressure and conditions are changed and we know two of the three
variables in final state, then the third one can be calculated by means of proportionality
indicated by the gas law.
Let V1, T1, and P1 be the volume, temperature and pressure of n kgmol of gas at conditions-
1
Let V2, T2, and P2 be the volume, temperature and pressure of n kgmol of gas at conditions
-2
then, P1V1 = nRT1 ---- (F)
and P2V2 = nRT2 ---- (G)
Combining the above two (F & G) equations, we get
Page 26
)H(2T
2V2P
1T
1V1P−−−−−−−−−−−−−=
In the ideal gas law given by equation (5), V is called the molar volume. At 273.15 K*
(00C) and 101.325 kPa (1 atm), the volume occupied by 1 kmol of a gas is 22.4136 m
3 i.e.,
V = 22.4136 m3/kmol (or 22.4136 l/mol). These conditions are said to be normal
temperature and pressure (NTP).
In USA, 101.325 kPa (1 atm) and 288.7 K (15.60C) are considered to be standard
temperature and pressure conditions (STP).
If two of three variables are known(temperature, pressure and volume) at one specified
condition and the third one can be calculated, thus are calculated with the help of equation
(8) where the other situation may be taken as NTP i.e., P2= 101.325 kPa; V2 = 22.4 m3 per
kmol and T2 = 273 K.
GASEOUS MIXTURE
The composition of component gases are expressed in terms of volume percent.
The molecules of the each component gas are distributed throughout the entire volume of
the container in a closed vessel.The total pressure exerted by the entire mixture is equal to
the sum of the pressure exerted by each component gas molecules.
PARTIAL PRESSURE
The pressure exerted by that component gas, if that alone is present in the same volume
and at the same temperature .
PURE COMPONENT VOLUME
The volume occupied by that component gas, if that alone is present at the same pressure
and temperature as the gas mixture.
DALTON’S LAW
It states that the total pressure exerted by a gaseous mixture is equal to the sum of the
partial pressures of the component gases present in gas mixture
P = pA+pB+pC ---------------------(I)
AMAGAT’S LAW
The total volume occupied by the gaseous mixture is equal to the sum of the pure
component volumes of component gases.
Mathematically, V = VA + VB+VC + ………… (J)
Where V is the total volume and VA, VB, and VC etc., are the pure component volumes of
component gases A,B,C etc., respectively.
RELATIONSHIP BETWEEN PARTIAL PRESSURE, MOLE FRACTION OF
COMPONENT GAS TO TOTAL PRESSURE
The gas mixture consisting of component gases A,B,C are considered.
Page 27
Let V be the total volume of the gas mixture and P be the total pressure exerted by the gas
mixture.
VA,VB,VC etc are the pure component volumes of A,B,C etc., respectively.
PA,PB,PC are the partial pressures of component gases A,B,C etc., respectively.
The ideal gas law for the component gas A is
pAV = nART
)K(V
RTAnAp −−−−−−−−−−−−=
Where R = gas constant
T = temperature of gas mixture
Similarly, for component B:
)L(V
RTBnBp −−−−−−−−−−−−=
Similarly, for component C :
)M(V
RTCnCp −−−−−−−−−−−−=
Adding Equations (J), (K), and (L) we get
)N(V
RT........)C
nBnA
n(
Cp
Bp
Ap −−−−−−−−−−−−
+++=++
According to the Equation (1)
P = pA+pB+pC+ ------- (O)
Dividing the equation (J) by Equation (M) we get
)P(CnBnAn
An
CpBpAp
Ap−−−−−−−−−−−
++=
++
nA+nB+nC + ------- = n = total moles
∴pA = xA P
Page 28
The mixture of ideal gases (i.e., one which follows an ideal gas law):
The partial pressure of a component gas mixture is equal to the product of the total pressure
and the mole fraction of that component.
Multiplying both sides of Equation (O) by 100,
)Q(100xCnBnAn
An100x
CpBpAp
Ap−−−−−−−−−−−
++=
++
i.e., pressure % of A = mole % of A --- (Q)
This result can be proved for other components of the gas mixture
∴ For gases behaving ideally,
Pressure % = Mole % ------ (R)
When an ideal gas law is applicable, it is written for any component gas A as :
PVA = nA RT ------ (S)
Similarly, for component B :
PVB = nB RT ------ (T)
For component C :
PVC = nCRT ------ (U)
Let VA,VB,VC etc., are the pure component volumes of components A,B,C …. etc.,
V = Total volume of the gas mixture
Adding Equations (S), (T), and (U), we get
P (VA+VB+VC + ------) = (nA+nB+nC+-----) RT – (V)
Dividing Equation (S) by Equation (V) gives
)W(CnBnAn
An
CVBVAV
AV−−−−−−−−−−−
++=
++
According to Amagat’s law,
V = VA+VB+VC + ------- ----- (X)
W/X gives :
AxV
AV=
i.e., Volume fraction of A = mole fraction of A ---- (Y)
Where, CnBnAn
AnAx
++=
∴ Volume % of A = Mole % of A --- (Z)
Page 29
This result can be proved for other components of the gas mixture.Combining
results obtained, i.e., equations (Q) and (Z) for gases behaving ideally/for an ideal gas
mixture, we have: Pressure % = Mole % = Volume %
AVERAGE MOLECULAR WEIGHT OF GAS MIXTURE
Required for calculating the weight of the gas mixture.It is calculated by assuming the
quantity of the gas mixture as one mole.
The weight of the one mole of the gas mixture represents its average molecular weight.
The composition of the gas mixture is specified in terms of volume % (on a volume basis).
Assuming the applicability of ideal gas, thus the volume analysis treated same as the mole
analysis. The mole analysis (i.e., volume % = mole %) of each component is calculated
which is needed for calculating the average molecular weight of the gas mixture.
The gas mixture consisting of components A, B, C etc., is considered.
Let MA, MB, MC etc., and XA,XB,XC, etc., are the molecular weights and mole fractions of
components gases respectively.
Let Mavg be the average molecular weight of the gas mixture. Then,
Mavg = MA.xA+MB.xB+MC.xC +…….. (1)
Average molecular weight of the gas mixture is the sum of the product of molecular weight
and mole fraction of all the individual component gases present in the gas
mixture.
In general, Mavg =
Where, Mi = molecular weight of ith
component gas, Xi = mole fraction of ith
component
gas
DENSITY OF GAS MIXTURE
The density of a gas mixture at a given temperature and pressure is very easily calculated
by using the ideal gas law.For calculating the density, the known average molecular weight
of the gas mixture plays a major role.
The ideal gas equation for a gas mixture is PV = nRT ----- (1)
)2(V
n
RT
P−−−−−−−−=
Where
n= kmol of gas mixture
V = volume of gas mixture, m3
R = universal gas constant 8.31451 m3.
kPa/(kmol.K),
T = temperature K,
P = pressure in kPa,
moles of gas mixture = )3(avgM
mixturegasofkg−−−−−−−−
∴ From equation (1) and (2) can be written as
Page 30
RT
P
V.avgM
mixturegasofwt=
RT
.avgPM
V
mixturegasofwt=
ρmix = Density of gas mixture = Volume
mixturegasofwt
RT
avgPMmix =ρ
When R is expressed in m3.
kPa/(kmol.K), P in kPa and T in K then the density will
have the units of kg/m3
Specific gravity of a gas
= density of the gas =molecular weight of the gas
density of air at the same T and P molecular weight of air
(45) Calculate the volume occupied by 20 kg of chlorine gas at a pressure of 100 kPa and
298 K (250C).
Solution: Basis: 20 kg Cl2 gas.
Moles of Cl2 gas = 20 = 0.2817 kmol
71
XA is the PV = nRT
∴ V = nRT
P
Where, n = 0.2817 kmol, T = 298 K, P = 100 kPa,
R = 8.31451 m3.
kPa/(kmol.K)
Volume, V = 0.2817 x 8.31451 x 298 = 6.98 m3
100
(46) 15 kg of carbon dioxide is compressed at a temperature of 303 K (300C) to a volume
of 0.5 m3. Calculate the pressure required for given duty. Assume ideal gas law is
applicable.
Solution: Basis: 15 kg of carbon dioxide gas
Molecular weight CO2 = 44
∴ Moles of CO2= 15 = 0.341 kmol
44
PV = nRT ∴ P = nRT
V
Where, n = 0.341kmol, T = 303 K,V = 0.5m3,
R = 8.31451 m3.
kPa/(kmol.K)
Page 31
Pressure, P = 0.341 x 8.31451 x 303 =1718.16 kPa
0.5
(47) 5 kg of oxygen contained in a closed container of volume 1 m3 is heated without
exceeding a pressure of 709.28 kPa. Calculate the maximum temperature of gas
attained.
Solution: Basis: 5 kg oxygen. Molecular weight of O2 = 32,
∴ Moles of O2= 5 = 0.5162 kmol
32
PV = nRT
∴ T = PV
nR
Where, V = 1 m3, P = 709.28 kPa, n=0.5162 kmol, R = 8.31451 m3.
kPa/(kmol.K)
Maximum temperature of gas, T = 709.28 x 1 = 546.13 K
0.5162 x 8.31451
∴ Temperature of gas attained is 546.13 K (273.130C)
(48) Calculate the weight of sulphur dioxide in a vessel having 2 m3 volume, the
pressure and temperature being 97.33 kPa and 393 K (1200C).
Solution: Basis: 2 m3 volume of SO2 gas
PV = nRT
∴ n = PV
RT
Where, P = 97.33 kPa, V = 2 m3, T = 393 K,
R = 8.31451 m3.
kPa/(kmol.K)
∴Moles of SO2 gas = 97.33x 2 = 0.0596 kmol
8.31451 x 393
Weight of SO2gas = 0.0596 x 64 = 3.8144 kg
(49) A certain quantity of gas contained in a closed vessel of volume 1 m3 at a
temperature of 298 K (250C) and pressure of 131.7 kPa is to be heated such that the
pressure should exceed 303.98 kPa. Calculate the temperature of gas attained.
Solution: Basis: 1 m3 volume of gas at 298 K
P1V1= P2V2
T1 T2
Where, P1 = 131.7 kPa, P2 = 303.98 kPa, T1 = 298K,
V1=V2 = 1 m3(as vessel being closed), T2 = ?
∴ 131.7 x 1 = 303.98 x 1
298 T2
∴ T2 = 687.82 K
∴ Temperature of gas attained = 687.82 K (414.820C)
(50) A gas contained in a closed vessel at a pressure of 121.59 kPa g and 299 K
(260C) is heated to a temperature of 1273 K (1000
0C). Find the pressure to which a
closed vessel should be designed.
Solution: Basis: A gas at 299 K in closed vessel
P1V1= P2V2
T1 T2
Page 32
As vessel being closed, V1 = V2
∴ P1= P2
T1 T2
∴ P2= P1T2
T1
Where, P1 = 121.59 kPa g, T1 = 299 K, T2 = 1273 K
Absolute pressure = Gauge pressure + Atmospheric pressure
∴ P1 = 121.59 + 101.325 = 222.915 kPa
P2 = 222.915 x 1273 = 949.07 kPa
∴ Pressure to which vessel should be designed= 949.07 kPa = 0.95 MPa.
(51) A sample of gas having volume of 0.5 m3 is compressed in such a manner so that
pressure is increased by 60%. The operation is done for a fixed mass of a gas at
constant temperature. Calculate the final volume of gas.
Solution: Basis: 0.5 m3 of gas sample.
Let initial pressure be P1 kPa.
It is given that pressure of gas increases by 60%
∴ Final pressure = P2 = 1.6 P1, Final volume = V2= ?
Temperature and mass are constant, therefore
P1V1 = P2V2
V2 = P1V1 =P1 x 0.5 = 0.3125 m3
P2 1.6 P1
∴ Final volume of gas = 0.3125 m3
(52) A sample of gas having volume of 1 m3 is compressed in such a manner so that its
pressure is increased by 85%. The operation is done for a fixed mass of gas at
constant temperature. Calculate the final volume of gas.
Solution: Basis: 1 m3 of gas
Initial volume = 1m3, Initial pressure = P1, kPa.
It is given that pressure increases by 85%.
Final pressure = P2 = 1.85 P1, kPa;
Final volume of gas = P1, kPa.
For given mass at constant temperature, we have, P1V1 = P2V2
V2 = P1V1 = P1 x 1 = 0.54 m3
P2 1.85 P1
∴ Final volume of gas = 0.54 m3.
(53) A certain sample of gas at a pressure of 202.65 kPa pressure is expanded so that the
volume is increased by 50%. The operation is done for a fixed mass of gas at constant
temperature. Calculate the final pressure of gas.
Solution: Basis: a gas at initial pressure of 202.65 kPa.
Initial volume = V1 m3=?,
Initial pressure = P1 = 202.65 kPa.
It is given that volume increases by 50%.
∴ Final volume = V2 = 1.5 V1 m
3.
For a given mass at constant temperature, we have, P1V1 = P2V2
Page 33
P2 = P1V1 =202.65 x V1= 135.1 kPa.
V2 1.5 V1
∴ Final pressure of gas = 135.1 kPa.
(54) A sample of gas having volume of 1 m3 is compressed to half of its original volume.
The operation is carried for a fixed mass of gas at constant temperature. Calculate
the percent increase in pressure.
Solution: Basis: 1 m3 of gas.
Initial volume = V1 = 1m3,
Final volume = V2 = 0.5 V1 = 0.5 m3.
Initial pressure = P1, kPa, Final pressure = P2, kPa.
For given mass of gas at constant temperature, we have,P1V1 = P2V2
P2= V1= 1.0
P1 V2 1.5
∴ P2 = 2 P1
% increase in pressure = P2 – P1 x 100
P1
= 2 P1 – P1x 100
P1
∴ % increase in pressure = 100
(55) A cylinder contains 15 kg of liquid propane. What volume in m3 will propane
occupy if it is released and brought to NTP conditions?
Solution:Basis: 15 kg of propane
Molecular weight of propane = 44
∴ Moles of propane = 15 = 0.3409 kmol
44
PV = nRT
P= 101.325 kPa, R = 8.31451 m3.
kPa/(kmol.K) , T = 273K, V = volume in m
3 at NTP=?
NTP conditions ⇒ P = 101.325 kPa, T = 273 K (00C)
V = 0.3409 x 8.31451 x 273 = 7.637 m3
101.325
∴Volume of propane gas at NTP = 7.637 m3.
(56) A certain quantity of a gas contained in a closed vessel of volume 1 m3 at a
temperature of 298 K (250C) and pressure of 121.59 kPa is to be heated such that
pressure should not exceed 405.3 kPa. Calculate temperature of gas attained.
Solution:Basis: 1 m3 of gas at temperature of 298 K and pressure of 121.59 kPa.
P1V1= P2V2
T1 T2
∴ T2 = P2V2x T1
P1V1
Where, P1 = initial pressure = 121.59 kPa.
T1 = initial pressure = 298 K, P2 = final pressure =405.3 kPa, T2 = final temperature of
gas = ?
As vessel is closed, V1= V2 = 1 m3.
Page 34
∴ T2 = 405.3 x 1 x 298 = 993.33 K
121.59 x 1
∴∴∴∴Temperature of gas attained = 993.33 K (720.330C)
(57) A gas mixture contains 0.274 kmol of HCl, 0.337 kmol of N2 and 0.089 kmol of
O2. calculate (a) Average molecular weight of gas and (b) Volume occupied by this
mixture at 405.3 kPa and 303 K (300C)
Solution:Basis: A gas mixture containing 0.274 kmol HCl, 0.337 kmol N2 and 0.089
kmol O2.
Total moles of gas mixture= 0.274 +0.337+0.089 = 0.7 kmol
Mole fraction of HCl (XHCl) = 0.274 = 0.391
0.7
Mole fraction of N2 (XN2)=0.337 = 0.481
0.7
Mole fraction of O2 (XO2) = 0.080 = 0.127
0.7
Molecular weight of HCl = MHCl = 36.5, Molecular weight of N2 = MN2 = 28,
Molecular weight of O2 = MO2 =32
Mavg = ∑Mi Xi
= MHCl. XHCl + MN2.xN2 +MO2.xO2 = 36.5 x 0.391+28x0.481+32x0.127
∴ Mavg = 31.80
PV = RT
PV = nRT
V = nRT
P
Where, V = volume in m3, n= moles of gas mixture = 0.7 kmol, P= 405.3 kPa, T = 363
K, R = 8.31451 m3.
kPa/(kmol.K)
V = 0.7 x 8.31451 x 303 = 4.35 m3
405.3
(58) In case of gas mixture cited in previous example, calculate the partial pressure of each
component gas at 405.3 kPa and 303 K (300C).
Solution:Basis: Given gas mixture
Total pressure = 405.3 kPa
Mole fraction of HCl = xHCl = 0.391
Mole fraction of N2 = xN2 = 0.481
Mole fraction of O2 = xO2 = 0.127
Partial pressure of HCl = pHCl = xHCl.P = 0.391 x 405.3 = 158.5 kPa
Partial pressure of N2 = pN2 = xN2.P= 0.481 x 405.3 = 194.95 kPa
Partial pressure of HCl = pO2 = xO2.P= 0.127 x 405.3 = 51.47 kPa
(59) A mixture of H2 and O2 contains 11.1% H2 by weight. Calculate (a) Average molecular
weight of gas mixture and (b) Partial pressure of O2 and H2 at 100 kPa and 303K(300C).
Sol: Basis: 100 kg of gas mixture.
It contains 11.1 kg of H2 and 88.9 kg of O2.
Molecular weight of O2= 32, Molecular weight of H2 = 2
Page 35
Moles of H2 = 11.1 = 5.55 kmol
2
and Moles of O2 = 88.9 = 2.78 kmol
32
Amount of gas mixture = 5.55 + 2.78 = 8.33 kmol Mole fraction of H2 = xH2
=5.55 = 0.67
8.33
Mole fraction of O2 = xO2 = 2.78 = 0.33
8.33
Mavg = Average molecular weight of gas mixture= M.H2.xH2+M.O2.xO2
= 2 x 0.67 + 32 x 0.33 = 11.9
P = Total pressure = 100 kPa
Partial pressure of H2 = pH2 = xH2.P= 0.67 x100 = 67 kPa (502.54 torr)
Partial pressure of O2 = pO2 = xO2.P = 0.33 x 100 = 33 kPa (247.52 torr)
(60) A mixture of nitrogen and carbon dioxide at 298 K (250C) and 101.325 kPa ha an
average molecular weight of 31. What is the partial pressure of nitrogen?
Solution:Basis: Average molecular weight of 31 of gas mixture.
Let xN2 and xCO2 be the mole fractions of N2 and CO2 respectively.
Molecular weight of N2 = 28, Molecular weight of CO2 = 44.
Mavg = ∑Mixi
Mavg = MN2.xN2 + MCO2.xCO2
31 = 28 xN2 + 44xCO2 ------- (1)
∑xi = 1
xN2 + xCO2 = 1 ------ (2)
∴ xCO2 = 1-xN2 ------ (3)
Put the value of xN2 from equation (2) in equation (1) and solve for xN2
31 = 28xN2 + 44 (1-xN2)
16 xN2 = 13
∴ xN2 = 0.8125
∴ xCO2= 1-0.8125 = 0.1875
Partial pressure of N2 = xN2.P = 0.8125 x 101.325 = 82.33 kPa(617.52 torr)
(61) A mixture of CH4 and C2H6 has the average molecular weight of 22.4. Find mole %
CH4 and C2H6 in the mixture.
Solution: Basis: Average molecular weight of 22.4 of gas mixture.
Let xCH4 and xC2H6 be the mole fraction of CH4 and C2H6 respectively.
Mavg = ∑Mixi
Mavg= M.CH4.xCH4+M.C2H6.xC2H6
22.4 = 16 xCH4+30xC2H6
∑xi = 1
xCH4+xC2H6 = 1
xC2H6 = 1-xCH4
Page 36
Put the value of xC2H6 from equation (3) into equation (1) and solve for xCH4.
22.4 = 16 xCH4 + 30 (1-xCH4)
xCH4 = 0.543
xC2H6 = 1- 0.543 = 0.457
Mole % of CH4 = Mole fraction of CH4 x 100= 0.543 x 100 = 54.30
Mole % of C2H6 = 0.457 x 100 = 45.70
(62) Assuming air to contain 79% N2 and 21 % O2 by volume, calculate the density of air at
NTP.
Solution: Basis: Air containing 21% O2 and 79% N2 by volume
For ideal gases, mole % = volume %
∴ Mole % N2 = 79, Mole % of O2 = 21
Mole fraction of N2 (xN2) = Mole % N2= 79 = 0.79
100 100
Mole fraction of O2 (xO2)=Mole % O2= 21 = 0.21
100 100
Mavg = MN2.xN2 + MO2.xO2 = 28 x 0.79 + 82 x 0.21 = 28.84
Density of air, ρ = PMavg
RT
Where, Mavg = 28.84, T=273 K, P = 101.325 kPa, R = 8.31451 m3.
kPa/(kmol.K).
ρ = 101.325 x 28.84 = 1.2874 kg/m3
8.31451 x 273
Density of air = 1.2874 kg/m3.
(63) A mixture of CH4 and C2H6 has density 1.0 kg/m3 at 273 K (0
0C) and 101.325 kPa
pressure. Calculate the mole % and weight % of CH4 and C2H6 in the mixture.
Solution: Basis: 1 kg/m3 density of gas mixture at 273 K and 101.325 kPa.
Density of gas mixture, ρ = PMavg
RT
∴ Mavg= ρ. RT
P
Where, ρ = 1 kg/m3, T = 273 K, P = 101.325 kPa, R = 8.31451 m3.
kPa/(kmol.K) .
∴ Mavg= 1 x 8.31451 x 273 = 22.4
101.325
Let xCH4 and xC2H6 be the mole fractions of CH4 and C2H6 respectively.
Mavg = ∑Mixi = MCH4.xCH4 + MC2H6.xC2H6
∴ 22.4 = 16xCH4 + 30 xC2H6 ---- (1)
∑xi = 1
xCH4 + xC2H6 = 1 ---- (2)
∴ xC2H6 = 1 – xCH4 -------- (3)
Put the value of xC2H6 from equation (3) into equation (1) and solve for xCH4
22.4 = 16xCH4 + 30 (1-xCH4)
∴ xCH4 = 0.543
xC2H6 = 1-0.543 = 0.457
Page 37
Mole % of CH4 = xCH4 x 100 = 0.543 x 100 = 54.3
Mole % of C2H6 = xC2H6 x 100 = 0.457 x 100 = 45.7
Weight of CH4 in 1 kmol mixture = 0.543 x 16 = 8.69 kg
Weight of C2H6 in 1 kmol mixture = 0.457 x 30=13.71 kg
Weight of gas mixture = 22.4 kg
Weight of CH4 in mixture = 8.69 x 100 = 38.8
22.4
Weight of C2H6 in mixture = 100 – 38.8 = 61.2
(64) A natural gas has the following composition by volume: CH4 = 82%, C2H6 = 12% and
N2 = 6%. Calculate the density of gas at 288 K (150C) and 101.325 kPa and
composition In weight percent.
Solution: Basis: 100 kmol of gas
It contains 82 kmol of CH4, 12 kmol of C2H6 and 6 kmol of N2.
Mole fraction of CH4 = xCH4= 82= 0.82
100
Mole fraction of C2H6 = xC2H6= 12= 0.12
100
Mole fraction of N2 = xN2 = 6 = 0.06
100
Molecular weight of CH4 = 16, Molecular weight of C2H6 = 30, Molecular weight of N2
= 28
Mavg = Average molecular weight of gas
Mavg = 16 x 0.82 + 30 x 0.12+28 x 0.06 = 18.4
Density of gas = ρ= PMavg
RT
Where, P = 101.325 kPa, Mavg = 18.4, R = 8.31451 m3.
kPa/(kmol.K) T= 288 K
Density, ρ= 101.325 x 18.4 = 0.78 kg/m3
8.31451 x 288
CH4 in gas = 82 x 16 = 1312 kg
C2H6 in gas = 12x30 = 360 kg
N2 in gas = 6x28 = 168 kg
Weight % CH= = kg of CH4x 100
kg of gas
Composition by Weight:
Component Quantity in kg Weight %
CH4 1312 71.30
C2H6 360 19.57
Page 38
N2 168 9.13
Total 1840 100
(65) Calculate the density of air containing 21% O2, 79% N2 by volume at 503 K (2300C)
and 1519.875 kPa.
Solution: Basis: Air containing 21% O2 and 79% N2 by volume.
For ideal gas, Mole % = Volume %
Mole % O2 = 21, Mole % N2 = 79.
Mole fraction of O2 = xO2 =21 = 0.21
100
Mole fraction of N2 = xN2= 79= 0.79
100
Molecular weight of O2 = 32, Molecular weight of N2 = 28
Mavg = MO2.xO2 + MN2.xN2 = 32 x 0.21 + 28x 0.79 = 28.84
Density of air, ρ = Pmavg
RT
Where, P = 1519.875 kPa, Mavg = 28.84,R = 8.31451 m3.
kPa/(kmol.K), T = 503 K
ρ = 1519.875 x 28.84 = 10.481 kg/m3
8.31451 x 503
(66) A gas mixture has the following composition by volume: SO2 = 8.5%, O2=10% and N2
= 81.5% . Find (a) the density of gas mixture at a temperature of 473 K (2000C) and
202.65 kPa g and (b) composition by weight.
Solution: Basis: 100 kmol of gas mixture
For ideal gas, Mole % = Volume %
It contains 8.5 kmol of SO2, 10 kmol of O2 and 81.5 kmol of N2.
Mole fraction of SO2 = xSO2= 8.5= 0.085
100
Mole fraction of O2 = xO2= 10= 0.10
100
Mole fraction of N2 = xN2 = 81.5 =0 .815
100
Molecular weight of SO2 = MSO2 = 64, Molecular weight of O2 = MO2 = 32,
Molecular weight of N2 = MN2 = 28.
Mavg = MSO2.xSO2+MO2.xO2+MN2.xN2 = 64 x 0.085 + 32 x 0.1 + 28 x 0.815 = 31.46
Absolute pressure = Gauge pressure + Atmospheric pressure
∴ P = 202.65 + 101.325 = 303.975 kPa
Density of gas mixture ,ρ = PMavg
RT
Where, P = 303.975, Mavg = 31.46, R = 8.31451 m3.
kPa/(kmol.K), T = 473 K
ρ = 303.975 x 31.46 = 2.43 kg/m3
8.31451 x 473
Page 39
SO2 in gas mixture = 8.5 x 64 = 544 kg
O2 in gas mixture = 10 x32 = 320 kg
N2 in gas mixture = 81.5 x 28 = 2282 kg
Amount of gas mixture = 3146 kg
Weight % SO2 in gas mixture =kg of SO2x 100
kg of gas mixture
Composition by Weight:
Component
Quantity in kg Weight %
SO2 544 17.30
O2 320 10.17
N2 2282 72.53
Total 3146 100
(67) In one case 28.6 litres of NO2 at 80 kPa and 298 K (250C) is
allowed to stand until the equilibrium is reached. At equilibrium, the pressure is found
to be 66.662 kPa. Calculate the partial pressure of N2O4 in the final mixture.
Solution: Basis: 26.6 l of NO2 at 80 kPa and 298 K.
Volume of NO2 = 26.6 lit = 0.0266 m3
P1V1 = n1RT
Initial moles, n1 = P1V1
RT1
Where, P1 = 80kPa, T1 = 298 K, R= 8.31451 m3.
kPa/(kmol.K), V1 = 0.0266 m
3
n1 = 80 x 0.0266 = 8.6 x 10-4 kmol = 0.86 mol
8.31451 x 298
2 NO2 = N2O4
Let x be the mol of N2O4 in final gas mixture
NO2 reacted = 2x mol
NO2 unreacted = 0.86 – 2x mol
Final moles = n2 = 0.86-2x + x = 0.86 - x mol
For initial conditions = P1V1=n1RT1
For final conditions = P2V2 = n1RT2
Page 40
But here, V1 = V2 and T1 = T2
∴ P1= n1
P2 n2
80 = 0.86 =
66.62 0.86 – x
Solving we get,
Final moles = 0.86 – x = 0.1434 mol
Mole fraction of N2O4 in final gas mixture, xN2O4
= 0.1434 = 0.20
0.7166
Partial pressure of N2O4 = xN2O4 .P = 0.20 x 66.662 = 13.33 kPa (99.98 torr)
(68) A closed vessel contain a mixture of 40% NO2 and 60% N2O4 at a temperature of 311
K (380C) and a pressure of 531.96 kPa. When the temperature is increased to 333 K
(600C), some of N2O4 dissociates to NO2 and a pressure rises to 679.95 kPa. Calculate
the composition of gases at 600C by weight.
Sol: Basis: 100 kg of gas mixture at 311 K.
NO2 in gas mixture = 40 kg, N2O4 in gas mixture = 60 kg
∴ Moles of NO2= 40 = 0.87 kmol
46
Mole of N2O4 = 60 = 0.652 kmol
92
Initial moles, n1= 0.87 + 0.652 = 1.522 kmol
N2O4 = 2 NO2
Let x be the kmol of N2O4 dissociated at 333 K
NO2 formed = 2x kmol
N2O= at 333 K = 0.652 – x kmol
NO2 at 333 K = 0.87 + 2x kmol
∴ Total moles at 333 K = (0.652-x) + (0.87+2x)
= 1.522 +x kmol
P1V1 = n1RT1
P2V2 = n2RT2
But V1 = V2 for it being closed vessel.
Taking ratio of equations (1) and (2), we get,
P1 = n1.T1
P2 n2 T2
P1 = 531.96 kPa, P2 = 679.95 kPa, n1= 1.522,
n2= 1.522+x, T1= 311 K, T2 = 333 K.
∴531.96 = 1.522 x 311
679.95 1.522 + x 333
Solving we get, x = 0.295 kmol
NO2 at 333 K = 0.87 + 2x = 0.87 +2 x 0.295 = 1.46 kmol
N2O4 at 333 K = 0.652 – x = 0.652 – 0.295 = 0.357 kmol
Amount of NO2 at 333 K = 1.46 x 46 = 67.16 kg
Page 41
Amount of N2O4 at 333 K = 0.357 x 92 = 32.84 kg
Composition by Weight:
Component Quantity in kg Weight %
NO2 67.16 67.16
N2O4 52.84 32.84
Total 100 100
(69) A volume of moist air 30 m3 at a total pressure of 101.328 kPa and a temperature of
303K (300C) contain water vapour in such that proportions that its partial pressure is
2.933 kPa. Without total pressure being changed, the temperature is reduced to 288 K
(150C) an some of water vapour is removed by condensation. After cooling, it is found
that the partial pressure of water vapour is 1.693 kPa. Calculate (a) volume of air at 288
K (150C) and (b) weight of water condensed.
Solution: Basis: 30 m3 of moist air at 303 K
Ideal gas law is : PV = nRT
n = PV
RT
Where, n= moles of moist air, P = 101.325 kPa, R= 8.31451 m3.
kPa/(kmol.K),
V = 30 m3, T = 303 K
Moles of air, n = 101.325 x 30 = 1.2066 kmol
8.31451 x 303
Let n1 be the kmol of air and n2 be the kmol of moisture/water vapour
P1 = Partial pressure of air at 303 K
= 101.325 – 2.933 = 98.392 kPa.
P2 = Partial pressure of moisture at 303 K = 2.933 kPa
For ideal gas, Pressure % = Mole %
Pressure fraction = Mole fraction
For air, 98.392 = n2
101.325 n1
∴n1= 98.392x 1.2066 = 1.172 kmol at 303 K
101.325
For water vapour /moisture,
2.933 = n1
101.325 n
n2 = 2.933 x 1.2066 = 0.035 kmol at 303 K
101.325
At 288 K, partial pressure of water vapour = 1.633 kPa
Let n3 be the moles of water vapour at 288 K
Mole of moist air = 1.172 + n3
Page 42
1.693 = n3
101.325 1.172 + n3
n3 = 0.02 kmol
Moles of moist air at 288 K = 1.172 + 0.02 = 1.192 kmol
Let n’ = 1.192 kmol
PV = n’RT
Where, V = volume of air at 288 K, n’ = 1.192 kmol, R= 8.31451 m3.
kPa/(kmol.K), T =
288 K, P = 101.325 kPa
V = n’RT= 1.192 x 8.31451 x 288 = 28.27 m3
P 101.325
Moles of condensed = n1 – n2 = 0.035 – 0.02 = 0.015 kmol
Amount of water condensed = 0.015 x 18 = 0.27 kg
(70) In manufacture of hydrochloric acid, gas containing 20% HCl and 80% air to volume
enters an absorption tower at a temperature of 323 K (500C) and pressure of 99.325 kPa.
98 percent of HCl is absorbed in water and remaining gas leaves the tower at a
temperature of 293 K (200C) and a pressure of 97.992 kPa. Calculate (a) the weight of
HCl absorbed/ removed per m3 of gas entering the system and (b) the volume of gas
leaving per m3 of gas entering the system.
Solution: Basis: 100 kmol of gas entering the absorption tower.
PV = nRT
V = nRT
P
Where, V = volume of gas entering, m3; n = 100 kmol, R= 8.31451 m
3.kPa/(kmol
.K), P
= 99.325 kPa; T=323 K
V = 100 x 8.31451 x 323 = 2704 m3 .
99.325
Moles of HCl in gas entering = 20 kmol
Moles of air in gas entering = 80 kmol
Moles of HCl absorbed/removed = 0.98 x 20 = 19.6 kmol
Moles of HCl unabsorbed and appearing in gas leaving = 20 – 19.6 = 0.4 kmol
HCl absorbed based on 100 kmol gas entering
= 19.6 x 36.5 = 715.4 kg
Volume of 100 kmol gas entering = 2704 m3
∴ Amount of HCl absorbed per 1 m3 of gas entering
= 715.4 x 1 = 0.2646 kg
2704
Moles of gas entering = 80 + 0.4 = 80.4 kmol
PV = nRT
∴ V = nRT
P
Page 43
Where, V = volume of gas leaving, m3; n = 80.4 kmol, R= 8.31451 m3.
kPa/(kmol.K), P
= 97.992 kPa, T= 293 K
V = 80.4 x 8.31451 x 293 = 1999 m3
97.992
Volume of gas leaving based on 100 kmol or 2704 m3 of gas entering system = 1999 m
3
∴ Volume of gas leaving per 1 m3 of gas entering the system
= 1999 x 1 = 0.74 m3
2704
(71) Equal masses of CO and N2 are mixed together in a container at 300 K (270 C). The
total pressure was found to be 405.3 kPa. Find the partial pressure of CO gas.
Solution: Basis: Let m1 be the mass of CO and m1 be the mass of N2 (as equal mass of
CO and N2).
Mole of CO = m1
28
Moles of N2= m1
28
Total moles of gas mixture = 2 m1
28
Mole fraction of CO (xCO)= m1/28
2m1/28
Partial pressure of CO = xCO. P = 0.5x405.3 = 202.65 kPa
(72) The analysis of the gas sample is given below (on volume basis): CH4 = 66%, CO2 =
30%, NH3 = 4%. Find (a) the average molecular weight of the gas and (b) density of the
gas at 202.65 kPa g pressure and 303 K (300C).
Solution: Basis: Gas sample containing CH4, CO2 and NH3
For ideal gases, Volume % = Mole %
Mole % of CH4 = 66, Mole % of CO2 = 32, Mole % of NH3 = 4
Mole fraction of CH4=xCH4= Mole % of CH4=66=0 .66
100
Mole fraction of CO2 = xCO2 =30 = 0.3
100
Mole fraction of NH3 = xNH3= 4= 0.04
100
Mavg = Average molecular weight of gas
M.CH4 .xCH4 +M.CO2.xCO2 + M.NH3.xNH3= 16 x 0.66 + 44 x 0.30 + 0.04 x 17 = 24.22
ρ = Density of gas = PMavg
RT
Absolute pressure = Gauge pressure + Atmospheric pressure
P = 202.65 + 101.325 = 303.975 kPa, T = 303 K, Mavg = 24.44,
R= 8.31451 m3.
kPa/(kmol.K).
ρ = 303.975 x 24.44 = 2.95 kg/m3
8.31451 x 303
Page 44
(73) By electrolysing a mixed brine, a gaseous mixture is obtained at the cathode having the
following composition by weight: Cl2 = 67%, Br== 28% and O2 = 5%. Calculate (a)
composition of gas by volume, (b) average molecular weight and (c) density of gas
mixture at 298 K (250C) and 101.325 kPa. (Atomic weights : Cl = 35.5, Br = 80, O=16)
Solution: Basis: 100 kg gas mixture
It contains 67 kg of Cl2, 28 kg of Br2, and 5 kg of O2
Moles of O2= 5= 0.1562 kmol
32
Composition of Gas Mixture by Volume :
Component Quantity in kmol Volume % (mole % )
Cl2 0.9437 74.02
Br2 0.175 13.73
O2 0.1562 12.25
Total 1.2749 100
Mole fraction of Cl2 = xCl2 = 74.02 = 0.7402
100
Mole fraction of Br2 = xBr2 = 13.73 = 0.1373
100
Mole fraction of O2 = xO2= 12.25 = 0.1225
100
Mavg = Average molecular weight of gas mixture
= M.Cl2 .xCl2 + M.Br2.xBr2 + M.O2.xO2= 71 x 0.7402 + 160 x 0.1373 + 32 x 0.1225 = 78.44
Also, Mavg = kg of gas mixture
kmol of gas mixture
= 100 = 78.44
1.2749
ρ = Density of gas mixture = PMavg
RT
Where, P = 101.325 kPa, Mavg = 78.44, R= 8.31451 m3.
kPa/(kmol.K), T = 298 K
∴
Page 45
ρ= 101.325 x 78.44 = 3.208 kg/m3
8.31451 x 298
(74) An inert gas (molecular weight 28) is admitted at the rate of 1 m3/min at 202.65 kPa and
303 K (300C) to a pipe line in which natural gas is flowing. The analysis of this gas at a
very long distance shown 2.9% by volume inert gas. Calculate the flow rate of natural
gas through the pipe line per min at 101.325 kPa and 303 K (300C).
Sol: Basis: 1 m3/ min inert gas
Molal flow rate of inert gas is
n’ = PV’
RT
Where, P = 202.65 kPa, V’ = 1m3/min, T = 303 K, R= 8.31451 m
3.kPa/(kmol
.K)
n’ = 202.65 x 1 = 0.0804 kmol/min
8.31451 x 303
Let x be the kmol/min of natural gas flowing through pipe line
Mole% inert gas = Volume % inert gas = 2.9
∴ 2.9 = 0.0804 . 100
0.0804 + x
Solving we get, x = 2.692 kmol/min
Volumetric flow rate of natural gas = xRT
P
Where, x = 2.692 kmol/min, R= 8.31451 m3.
kPa/(kmol.K), T = 303 K,
P = 101.325 kPa
Volumetric flow rate of natural gas
= 2.692 x 8.31451 x 303 = 66.93 m3/min
101.325
(75) A producer gas has the following composition by volume . CO= 21%, CO2 = 5%, O2 =
3% and balance being N2. Calculate the volume of gas in m3 at 298 K (25
0C) and 99.325
kPa per kg of carbon present.
Solution: Basis: 100 kmol of producer gas
It contains 21 kmol of CO, 5 kmol of CO2, 3 kmol of O2 and 71 kmol of N2.
1 kmol of CO ≡ 1 katom C
∴ Carbon from CO = 1 x 21 = 21 katom
1
Carbon in producer gas = 21 + 5 = 26 katom = 26x 12 = 312 kg
PV = nRT
∴ V = nRT
P
Where, n = 100 kmol, V = Volume in m3, T = 298 K,P = 99.325 kPa,
R= 8.31451 m3.
kPa/(kmol.K).
V = 100 x 8.31451 x 298 = 2494.56 m3
99.325
Volume of producer gas corresponding to 312 kg carbon in it = 2492.56 m3.
∴ Volume of producer gas per kg of carbon present
Page 46
= 2494.56 x1 = 8.0 m3.
312
(76) Calculate the number if cubic meters of acetylene gas at temperature of 313 K (400C)
and a pressure of 100 kPa that may be produced from 5 kg of calcium carbide.
Solution: Basis: 5 kg of calcium carbide
CaC2 + 2H2O � C2H2 + Ca(OH)2
1 kmol of CaC2≡ 1 kmol of C2H2
64 kg CaC2≡ 1 kmol of C2H2
5 kg CaC2= ?
Moles of C2H2 produced = 1x 5 = 0.0781 kmol
64
PV = nRT
V = nRT
P
Where, n = 0.0781 kmol, T = 313 K, P = 100 kPa, R= 8.31451 m3.
kPa/(kmol.K).
Volume of acetylene gas produced is
V = 0.0781 x 8.31451 x 313 = 2.03 m3.
100
(77) Nitrogen is to be marketed in cylinder having volume of 0.08 m3 each containing 3.5 kg
of nitrogen. Calculate the pressure for which cylinders must be designed if they are
subjected to a maximum temperature of 323 K (500C).
Solution: Basis: 3.5 kg nitrogen
Molecular weight of N2 = 28
∴ Moles of N2 = 3.5 = 0.125 kmol
28
According to Ideal gas law –
PV = nRT
Where, P = pressure in kPa, V = 0.08 m3, n = 0.0125 kmol, T = 323 K,
R= 8.31451 m3.
kPa/(kmol.K).
∴ P V = nRT
P =0.125 x 8.31451 x 323 = 4196.23 kPa.
0.08
The pressure to which cylinders must be designed
= 4196.23 kPa = 4.2 MPa.
(78) It is desired to market oxygen in small cylinders having volumes of 0.015 m3 and each
containing 0.5 kg of oxygen. If the cylinders may be subjected to a maximum
temperature of 323 K (500C), calculate the pressure for which they must be designed
assuming applicability of ideal gas law.
Solution: Basis: 0.5 kg oxygen
∴ Moles of O2 = 0.5 = 0.0156 kmol
32
According to Ideal gas law –
Page 47
PV = nRT
P = nRT
V
Where, n= 0.0156 kmol, R= 8.31451 m3.
kPa/(kmol.K), V = 0.015 m3, T = 323 K
P = 0.0156 x 8.31451 x 323 = 2793 kPa
0.015
∴ Pressure for which cylinders must be designed
= 2793 kPa = 2.8 MPa.
(79) A gaseous mixture has the following composition by volume: CO2 = 8%, CO = 14%,
H2O = 5%, CH4 = 1% and N2 = 66%. Calculate (i) Average molecular weight of gas
mixture and (ii) Density of gas mixture at 303 K (300C) and 101.325 kPa.
Solution: Basis: A gas mixture containing CO2, CO, O2, H2O, CH4 and N2 at 303 K
and 101.325 kPa.
For ideal gases, Volume % = Mole %
Mole fraction of CO2, xCO2=Mole% of CO2=8= 0.08
100
Mole fraction of CO, xCO =14 =0.14
100
Mole fraction of O2, xO2= 6=0.06
100
Mole fraction of H2O, xH2O = 5 = 0.05
100
Mole fraction of CH4, xcH4 = 1 = 0.01
100
Mole fraction of N2, xN2 = 66 = 0.66
100
Molecular weight of CO2 = 44, Molecular weight of CO= 28, Molecular weight of O2 =
32, Molecular weight of H2O = 18, Molecular weight of CH4 = 16, Molecular weight of
N2 = 28
Average molecular weight of gas mixture is
Mavg = ∑Mixi
= 44 x 0.08 + 28 x 0.14 + 32 x 0.06 + 18 x 0.05 + 16 x 0.01 + 28 x 0.66 = 28.90
Density of gas mixture is
ρ = PMavg
RT
Where, P = 101.325 kPa, Mavg = 28.90, R= 8.31451 m3.
kPa/(kmol.K), T = 303 K
ρ = 101.325 x 28.90 = 1.162 kg/m3.
8.31451 x 303
(80) The gas acetylene is produced according to the following reaction:
CaC2 + 2H2O � C2H2 + Ca(OH)2. Calculate the number of hours of service that can be
derived from 1 kg of calcium carbide in an acetylene lamp burning 0.10 m3 of gas per
hour at temperature of 298 K (250C) and pressure of 99.325 kPa.
Solution: Basis: 1 kg calcium carbide
CaC2 + 2H2O � C2H2 + Ca(OH)2
Page 48
Molecular weight of CaC2 = 64
∴ Moles of CaC2= 1= 0.01562 kmol
64
From reaction, 1 kmol CaC2 = 1 kmol C2H2
Moles of C2H2 produced =1x 0.01562 = 0.01562 kmol
1
Now, PV = nRT
V = nRT
P
Where, n = 0.01562 kmol, T = 298 K, P = 99.325 kPa, R= 8.31451 m3.
kPa/(kmol.K).
Volume of C2H2 gas produced
= 0.01562 x 8.31451 x 298 = 0.3896 m3
99.325
Burning rate of acetylene gas = 0.10 m3/h
No. of hours of service = Volume of acetylene gas
Burning rate of acetylene
= 0.3896 = 3.896 h = 3.9 h
0.1
(81) When heated to 373 K (1000C) and 95.992 kPa pressure, 17.2 grams of N2O4 gas
occupies a volume of 11.45 litres. Assuming that the ideal gas law applies, calculate the
percentage dissociation of N2O4 to NO2.
Solution: Basis: 17.2 g of N2O4.
Molecular weight of N2O4 = 92
∴ Initial moles of N2O4 = 17.2 = 0.187 mol
92
N2O4 = 2NO2
Let x be mol of N2O4 dissociated as per the reaction.
∴ NO2 formed = 2x mol
N2O4 undissociated = (0.187 - x) mol
Total moles of gas after dissociation
= (0.187 – x) + 2x
= (0.187 + x) mol
Now, PV = nRT
Where, P = 95.992 kPa, n = (0.187+x) x 10-3
kmol, V = 11.45 x 10-3
m3, T = 373 K,
R= 8.31451 m3.
kPa/(kmol.K)
95.992 x 11.45x10-3
= (0.187+x) x 10-3
x 8.31451 x 373
Solving we get, x = 0.1674 mol
% dissociation of N2O4 =
moles of N2O4 dissociated x 100
initial moles of N2O4
= 0.1674 = 89.52
0.187
Page 49
(82) A 40 ml sample of a mixture of H2 and O2 was placed in a vessel at 291 K (180C) and
101.325 kPa. A spark was applied so that the formation of water was complete. The
remaining pure gas was H2, What was the initial mole % H2 in the mixture ?
Solution: Basis: 40 ml of sample of mixture of H2 and O2
H2 + ½ O2� H2O
PV = nRT
n = PV
RT
Where, P = 101.325 kPa, T = 291 K,R= 8.31451 m3.
kPa/(kmol.K),
V = 40 ml = 40x 10-6
m3
n = 101.325 x 40x10-6
8.31451 x 291
= 1.675 x 10-6
kmol = 1.675 x 10-3
mol
(H2O + O2) gas mixture = 1.675 x 10-3
kmol
H2 gas remained = 10 ml
Moles of H2 remained = 10x10-6
x 101.325
8.31451 x 291
= 4.188 x 10-7
kmol = 4.188 x 10-4
mol
Let x be the initial moles of H2 and O2 respectively
∴ x + y = 1.675 x 10-3
O2 reacted = y mol
H2 reacted = 1 y = 2y mol
0.5
Material balance of H2:
x moles of H2 = 2y + 4.188 x 10-4
x = 2y + 4.188 x 10-4
From equations (1) and (2), we get,
3y + 4.188 x 10-4 = 1.675 x
and x = 1.256x 10-3
mol
Initial mole % of H2 = 1.256 x 10-3
x 100 = 75
1.675 x 10-3
Initial mole % of O2 = 100 – 75 = 25
(83) The combustion of 4.73 kg of sample of coal yielded 5.30 m3 of carbon dioxide gas
measured at NTP. Find the carbon content of the sample.
Solution: Basis: 4.73 kg of a coal sample
C + O2 = CO2
At NTP, the volume of CO2 gas = 5.30 m3
At NTP, the volume of 1 kmol of CO2 gas is 22.4 m3.
∴ Amount of CO2 yielded =1 x 5.3= 0.237 kmol
22.4
= 0.273 x 44 = 10.428 kg
From reaction, 1 katom C ≡ 1 kmol CO2
12 kg C ≡ 44 kg CO2
Page 50
Carbon in the sample = 12 x 10.248 = 2.844 kg
44
Carbon content of sample in weight %
= 2.844 x 100 = 60.13
33.73
(84) At a temperature of 299 K (260C),ethanol exerts a vapour pressure of 8 kPa. Calculate
the composition of a saturated mixture of air and ethanol vapour at a temperature of 299 K
(260C) and a pressure of 100 kPa in terms of (i) volume, (ii) weight, (iii) kg of vapour per
m3 of mixture and (iv) kg of vapour per kg of vapour free air.
Solution: Basis: 1 m3 of gas mixture
Pure component volume of ethanol vapour in 1 m3 of mixture.
= 1 x 8 = 0.08 m3
1
Composition by Volume:
Ethanol vapour 0.08 m3 8.0%
Air 0.92 m3 92%
Total 1.0 m3 100
Amount of ethanol vapour present in 1 kmol of mixture
= 0.08 x 1 = 0.08 kmol = 0.08 x 46 = 3.68 kg
Amount of air in mixture
= 0.92 x 1 = 0.92 kmol = 0.92 x 28.84 = 26.53 kg
∴ Amount of mixture = 3.68 + 26.53 = 30.21 kg
Composition by Weight:
Ethanol vapour 0.08 m3 8.0%
Page 51
Air 0.92 m3 92%
Total 1.0 m3 100
Now, PV = nRT
V = nRT
P
Volume of 1 kmol of mixture = nRT
P
Where, n = 1 kmol, R= 8.31451 m3.
kPa/(kmol.K), T = 299 K, P = 100 kPa.
∴ V = 1 x 8.31451 x 299 = 24.86 m3
100
Weight of ethanol vapour present per m3 of mixture = 3.68= 0.148 kg
24.86
Kg of ethanol vapour per kg of vapour free air
3.68 = 0.1387
26.53
(85) The vapour pressure of ether (molecular weight 74) is 58.928 kPa at 293 K (250C). If 3
g of compound A are introduced and dissolved in 50 g of ether at this temperature, the
vapour pressure falls to 56.795 kPa. Calculate the molecular weight of A. Assume that the
solution of A in ether is very dilute.
Solution: Basis: 3 g of a compound A in 50 g of ether at 293 K
V.P. of ether in solution = V.P. of pure ether x Mole fraction of ether in solution
56.795 = 58.928.xi
∴ x1 = 0.9638
x1 + x2 = 1
x2 = 1-x1 = 1 – 0.9638 = 0.0362
Mole fraction of A in solution = 0.0362
Mole fraction of ether in solution = 0.9638
Mole fraction of ether in solution = 50 = 0.6757 mol
74
Total moles of solution = 0.6757 = 0.7011 mol
0.9638
Mole of solution = 0.0362 x 0.7011 = 0.0254 mol
Mol of A = gram of A =
Molecular weight of A
∴ Molecular weight of A = 3 = 118
0.254
Page 52
(86) In the manufacture of formaldehyde; O2, methanol and steam are mixed in proportion
1.5: 2:1.33 (by weight ) at 283 K (100C). The total pressure is 70.928 kPa g. Calculate the
partial pressure of each of the components present in this mixture.
Solution: Basis: 100 kg of solution
In gas mixture, the proportion of O2:HCHO: Steam (H2O) is 1.5:2:1.33 by weight.
∴ O2 in the mixture = 1.5 x 100 = 31.06 kg
4.83
= 31.06 = 0.97 kmol
32
Methanol in mixture = 2 x 100 = 41.4 kg
4.83
= 41.4 = 1.3 kmol
32
Steam in mixture = 1.33 x 100 = 27.54 kg
4.83
= 27.54 = 1.53 kmol
18
Moles of gas mixture = 0.97 + 1.3 + 1.53 = 3.8 kmol
Mole fraction of O2 = xO2 = 0.97 = 0.256
3.8
Mole fraction of methanol = xCH3OH =1.3 = 0.342
3.8
Mole fraction of Steam = xH2O = 1.53 = 0.403
3.8
Total pressure = 70.928 kPa g
P = 70.928 + 101.325
∴ P = 172.253 kPa
Partial pressure of O2= xO2.P = 0.256 x 172.253 = 44.10 kPa
Partial pressure of methanol = 0.342 x 172.253= 58.91 kPa
Partial pressure of steam = 0.403x 172.253= 69.42 kPa.
(87) Ammonia under a pressure of 1519.875 kPa and 298 K (250C) is heated to 620 K
(3470C) in a closed vessel in the presence of catalyst. Under these conditions, NH3 is
partially decomposed. The vessel is such that the volume remain effectively constant,
whereas the pressure increases to 5066.25 kPa. Calculate the % of NH3 decomposed.
Solution: Basis: 100 kmol NH3 initially present at
298 K
Ideal gas law for initial and final conditions is
P1V1= n1RT1
P2V2 = n2RT2
∴ P1V1 =n1. T1
P2V2 n2 T2
But V1 = V2 as vessel being closed
Page 53
∴ P1 =n1. T1
P2 n2 T2
Where, n1 = initial moles = 100 kmol, P1 = 1519.875 kPa, T1 = 298 K, n2 = fional
moles of gas mixture = ?, P2 = 5066.250 kPa, T2 = 620 K
∴ n2 = n1x T1 x P1
T2 P2
= 100 x298 x5066.250= 1690.21 kmol
620 1519.875
Total moles after decomposition = 160.21 kmol
Decomposition of NH3 takes place as :
2 NH3 = N2 + 3 H2
Let x be the kmol NH3 decomposed
NH3undecomposed= (100 - x) kmol
N2 produced = 1 .x = 0.5 x kmol
2
H2 produced = 3 . x = 1.5 x kmol
2
Moles of NH3+N2+H2 after decomposition =160.21
100 – x + 0.5 x + 1.5 x = 160.21
x = 60.21 kmol
% decomposition of NH3
= moles of NH3 decomposed x 100
Initial moles NH3
= 60.21 x 100 = 60.21
100
(88) A gas containing 96% ethylene and 4% butenes by volume is passed through a bed of a
activated carbon where 98% of the original butenes are adsorbed and none of the ethylene.
In five hours of continuous operation if quantity of butenes removed is 0.5 kmol, find (i)
Mole % ethylene in gas leaving carbon bed and (ii) Molar flow rate of the feed gas to the
carbon bed.
Solution: Basis: Five hours of operation
Amount of butenes removed = 0.5 kmol
∴ Butenes removed = 0.5 = 0.1 kmol/h
5
Let y be the molar flow rate (kmol/h) of gas to the carbon bed
Butenes in gas fed = 0.04 y kmol/h
Butenes adsorbed = 0.98 x 0.04 y = 0.0392 y kmol/h = 0.0392 y = 0.1
∴ y = 2.551 kmol/h
Molar flow rate of gas to carbon bed = 2.551 kmol/h
Butenes not adsorbed = 0.04 y – 0.1= 0.04 x 2.551 – 0.1 = 0.00204 kmol/h
Ethylene in inlet gas leaving carbon bed = 0.96 x 2.551 = 2.449 kmol/h
Ethylene in gas leaving carbon bed = Ethylene in inlet gas = 2.449 kmol/h
Gas leaving carbon bed = 2.449 + 0.00204 = 2.451 kmol/h
Page 54
Mole% ethylene in gas leaving carbon bed
= 2.449 x 100 = 99.92%
2.451
(89) A sample of gas having volume of 10 l at 101.325 kPa pressure and at temperature of
298 K (250C) is compressed to a high pressure so that its volume reduces by 4.5 l. if the
pressure rises by 0.1 MPa, what will be the rise in temperature?
Solution: : Basis: 10 l of gas at 298 K
PV = nRT
Where, V = 10 l = 10 x 10-3
m3, R= 8.31451 m
3.kPa/(kmol
.K), T = 298 K,
P = 201.325 kPa.
n = PV
RT
= 101.325 x 10 x 10-3
= 4.09 x 10-4
kmol
8.31451 x 298
Final pressure = 101.325 + 100 = 201.325 kPa.
(as pressure increases by 0.1 MPa i.e., 100 kPa) Final volume = 10 – 4.5 = 5.5 l = 5.5 x
10-3
m3
PV = nRT
Where, P=201.325 kPa, V = 5.5 x 10-3
m3
201.325 x 5.5 x 10-3
= 4.09 x 10-4
x 8.31451 x T
∴ T = 325.6 K
∴ Rise in temperature = 325.6 – 298 = 27.6 K (27.6 deg C )
(90) Exhaust gas having 75% N2 and 25% CO2 (by volume) is passed through a absorption
column. 97% CO2 is absorbed in KOH solution. The gas enters the system at a temperature
323 K (500C) and 740 torr and leaves at 303 K (30
0C) and 737 torr. Calculate (i) The
volume of gases leaving per 100 litres entering. (ii) The weight of CO2 absorbed per 100
litres of gas entering
Solution: Basis: 100 mol of gas entering
It contains 75 mol N2 and 25 mol of CO2
Amount of CO2 absorbed = 0.97 x 25=24.25 mol
Amount of CO2 unabsorbed=25-24.25 =0.75 mol
Total gas leaving system = 75 + 0.75 =75.75 mol
PV = nRT
Where, P = 740 torr = 98.66 kPa
∴ V = 100 x 8.31451 x 323 = 2722 l
98.66
∴ Volume of 100 mol of entering gas = 2722 l
Weight of CO2 absorbed per 100 l gas entering
= 24.25 x 100 = 0.89 mol
2722
= 0.89 x 44 = 39.16 g
Page 55
Volume of gas leaving based on 100 mol gas entering is V’ = n’ RT’
P’
= 75.75 x 8.31451 x 303 = 1942 l
98.26
Volume of gas leaving per 100 l gas entering
= 1942x 100 = 71.43 l
2722
(91) Cracked gas form a petroleum refinery contains 45% CH4, 1% C2H6, 25% C2H4, 7%
C3H8, 8% C3H6 and 5% C4H10 by volume. Calculate (i) the average molecular weight of
gas mixture, (ii) the composition by weight, and (iii) the specific gravity of the mixture
taking average molecular weight of air as 28.84
Solution: Basis: 100 kmol of cracked gas.
It contains 45 kmol CH4, 10 kmol C2H6, 25 kmol C2H4, 7 kmol C3H8, and 5 kmol C4H10
Molecular weight data:
CH4= 16, C2H6= 30, C2H4 = 28, C3H6= 42, and C4H10 = 58
Weight of methane = 45 x 16 = 720 kg
In the same way, calculate the weight of each components of cracked gas
Composition of refinery gas:
Average molecular weight of refinery gsa is
Mavg = 2854 = 26.54
100
OR Mavg = M.CH4.xCH4 + M.C2H6.xC2H6 +
M.C2H4.xC2H4 + M.C3H8.xC3H8 + M.C3H6.xC3H6 + M.C4H10.xC4H10
= 16 x 0.45 + 30 x 0.1 + 28 x 0.25 + 44 x 0.07 + 42 x 0.08 + 58 x 0.05 = 26.54
Specific gravity of gas mixture = 26.54 = 0.92
28.84
Page 56
Component Kmol Mol.Wt kg Weight%
CH4 45 16 720 27.13
C2H6 10 30 300 11.30
C2H4 25 28 700 26.37
C3H8 7 44 308 11.61
C3H6 8 42 336 12.66
C4H10 5 58 290 10.93
Total 100 - 2654 100
Page 57
(92) The composition of gas mixture in manufacture of nitric acid at a pressure of 0.709
MPa and 923 K (6500C) is as follows:
N2 = 70.5%, O2 = 18.8%, H2O = 1.2% and NH3 = 9.5%. Calculate the density of gas
mixture using ideal gas law
Sol: Basis: 100 kmol of gas mixture at 923 K
It contains 70.5 kmol of N2, 18.8 kmol O2, 1.2 kmol H2O and 9.5 kmol NH3.
Mole fraction of N2 = xN2 = 70.5 = 0.705
100
Mole fraction of O2 = xO2 = 18.8 = 0.188
100
Mole fraction of H2O = xH2O = 1.2 = 0.012
100
and Mole fraction of NH3 = xNH3 = 9.5 = 0.095
100
Molecular weight of N2 = 28, Molecular weight of O2 = 32, Molecular weight of H2O =
18 and Molecular weight of NH3 = 17
Average molecular weight of gas mixture is
Mavg= ∑Mixi
= 28 x 0.705 + 32 x0.188 + 18 x 0.012 + 17 x 0.095 = 27.59
P = 0.709 MPa g = 810.325 kPa, T = 923 K, R= 8.31451 m3.
kPa/(kmol.K)
Density of gas mixture is given by
ρ = PMavg
RT
= 810.325 x 27.59 = 2.913 kg/m3
8.31451 x 923
(93) The Orsat (dry) analysis of a flue gas from a boiler house is as given below: CO2 =
10%, O2= 7.96%, N2 = 82% and SO2 = 0.04% by volume. The flue gas pressure is 100 kPa
(750 torr) and temperature is 463 K (1900C). SO2 is undesirable from the point of view of
occupational hazards (environmental pollution). Express the concentration of SO2 in ppm
and mg/m3.
Sol: Basis: 100 kmol of a flue gas
It contains 10 kmol of CO2, 7.96 kmol O2, 82 kmol N2, and 0.04 kmol SO2.
Amount of CO2 = 10 x 44 = 440 kg
Amount of O2 = 7.96 x 32 = 254.72 kg
Amount of N2 = 82 x 28 = 2296 kg
Amount of SO2 = 0.04 x 64 = 2.56 kg
Amount of flue gas = 440 + 254.72 + 2296 + 2.56 = 2993.28 kg
Concentration of SO2 in ppm= 2.56 x106
2993.28
= 855.25
Page 58
Volume of flue gas, V = nRT
P
= 100 x 8.31451 x 463 = 3849.62 m3
100
Concentration of SO2 in mg/m3
= 2.56 x 103x10
3= 665
3849.62
Molecular weight of steam (water) = v = V
M
= 0.518 (m3/kmol) = 0.0288 m
3/kg
18 kg/kmol)
(94) 630 grams of oxalic acid dihydrate were dissolved in water. Total volume of solution
being 3000 ml. what will be the normality and molarity of the solution?
Sol: Basis: 630 g of oxalic acid
Oxalic acid = C2H2O4.2H2O
Molecular weight of oxalic acid = 126
Equivalent weight of oxalic acid = 63
Volume of solution = 3000 ml = 3 l
Moles of oxalic acid = 630 = 5 mol
126
Equivalent weight of oxalic acid = 630 = 10g eq
63
Normality = 10 = 3.33 N
3
Molarity = 5 = 1.67 M
3
Reference Books
1) B. I. Bhatt and S. M.Vora, Stoichiometry, McGraw Hill, New Delhi, 4th
ed. 2004.
2) David Mautner Himmelblau, Basic Principles and Calculations in Chemical
Engineering, Prentice Hall, New Delhi, 6th
ed. 2002.
3) O. A. Hougen, K. M. Waston and R. A. Ragatz, Chemical Process Principles Part–I,
Material and Energy Balances, CBS Publishers and Distributors, New Delhi, 2nd
ed.
1995.
4) K. A. Gavhane, Introduction to Process Calculations (Stoichiometry), Nirali Prakashan,
Pune, 15th
ed. 2003.
5) K.V.Narayanan and B.Lakshmikutty, Stoichiometry and Process Calculations, 1st ed.,
Prentice Hall of India, New Delhi, 2006.
