UNIT 2GRAM 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. gram atoms of an element = wt . in grams − − − − − − − (1) atomic weight 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 kg atoms of an element = wt .in kg − − − − − − − (2) atomic weight 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. gmoles of compound = wt .in g − − − − − − − (3) molecular weight Gram mole of a substance is the mass in grams of the substance that is numerically equal to its molecular weight. Similarly, kgmoles of compound = wt .in kg molecular weight − − − − − − − (4) 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 katom of carbon = wt .in kg 36 = =3 atomic weight of carbon 12 (2) Calculate the kilograms of ‘Na’ of which the amount is specified as 3 katom. Solution: 3 katom Na Atomic weight of Na = 23 katom of Na = wt .in kg of Na atomic weight of Na ∴ 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 kmol of C 2 H 6 = wt .in kg of C 2 H 6 mol . weight of C 2 H 6 ∴ 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 kmol of CO 2 = wt .in kg of C O 2 88 = = 2 mol . weight of CO 2 44 88 g of CO2 = 2 kmol CO2 (5) Find the moles of oxygen present in 500 grams Solution: Basis: 500 g of oxygen Molecular weight of O2 = 2 x 16 = 32 kmoles of O 2 = 500 = 15 . 625 32 (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 kmoles of CuSO 4 .5H 2 O = 499 = 2 249 . 5 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 Solution: Basis: 64 kg of S Atomic weight of S = 32 atoms of S = kg of S 64 = = 2 katom 32 Each moles of H2SO4 contains one atom of S. 1 kmol of H2SO4≡ 1 katom of S moles of H 2 SO 4 = 1 x 2 = 2 kmol 1 (8) Find kmoles of K2CO3 that will contain 117 kg of K? Solution: Basis: 117 kg of K Atomic weight of K = 39 atoms of K = 117 = 3 katom 39 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 moles of K 2 CO 3 = 1 x 3 = 1 . 5 kmol 2 (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 = 12 x 64 = 48 16 (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 moles of Na 2 SO 4 .10H 2 O = 644 = 2 mol 322 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 = 28 x 96 . 43 = 45 kg 60 (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 = 12 x 264 = 72 g 44 Designated by the symbol ‘M’. It is defined as the number of gram equivalents of solute dissolved in one litre of solution.EQUIVALENT WEIGHT It is defined as the ratio of the atomic weight or molecular weight to its valence. Normality (N) = gram-equivalents of solute= gram of solute_____________________ Volume of solution in Litre eq wt. 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. it is possible to find out the concentration of solute in g/l. Molarity ( M ) = g of solute volume of solution in L . of solute x volume of solution in L where concentrat ion (g / L ) = g of solute volume of solution in L By using definition of normality.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 equivalent weight = molecular weight valence NORMALITY It is designated by the symbol ‘N’. of solute x Volume of solution in Litre Normality = gequivalen ts of solute g of solute = volume of solution in L eqwt . MOLALITY It is defined as the number of gram-moles (mol) of solute dissolved in one kilogram of solvent.5 Valence of HCl = 1 equivalent weight of HCl = mol wt of HCl 36 .5 valence of HCl 1 (2) NaOH: Molecular weight of NaOH = (1x23)+(1x16)+(1x1) = 40 Valence of NaOH = 1 equivalent weight of NaOH = 40 = 40 1 .5) = 36.5 = = 36 . (2) NaOH. (3) Na2CO3 and (4) H2SO4 Solution: (1) HCl: Molecular weight of HCl = (1x1)+(1x35. Molality ( M ) = gmoles of solute mass of solvent in kg Ppm and ppb 1ppm = 1g solute 1 g solute 1 mg solute = = 10 6 g solution 10 6 L solution 1 L solution since density of solution(very dilute) = 1 g/cc 1ppb = 1g solute 1 g solute = 10 9 g solution 10 9 L solution (14) Find the equivalent weights of (1) HCl. 5 valence of CaCl 2 2 (3) FeCl3: Molecular weight of FeCl3 = (1x56)+(3x35.5) = 111 Valence of CaCl2 = 2 equivalent weight of CaCl 2 = mol wt of CaCl 2 111 = = 55 .17 3 equivalent weight of FeCl 3 = (4) Al2(SO4)3: mol wt of FeCl 3 162 .5 = = 54 .5) = 162. (3) FeCl3. and (5) KMnO4 Solution: (1) H3PO4 Molecular weight of H3PO4 = (3x1)+(1x31)+(4x16) = 98 Valence of H3PO4 = 3 equivalent weight of Na 2 CO 3 = (2) CaCl2 98 = 32 .67 3 Molecular weight of CaCl2 = (1x40)+(2x35.5 = 54.1 (3) Na2CO3 : Molecular weight of Na2CO3=(2x23)+(1x12)+(3x16) = 106 Valence of Na2CO3 = 2 equivalent weight of Na 2 CO 3 = (4) H2SO4 mol wt of Na 2 CO 3 106 = = 53 valence of Na 2 CO 3 2 Molecular weight of H2SO4 = (2x1) + (1x32) + (4x16) = 98 Valence of H2SO4 = 2 equivalent weight of H 2 SO 4 = 98 = 49 2 (15) Calculate the equivalent weights of the following compounds: (1) H3PO4. (4) Al2(SO4)3. (2) CaCl2.17 valence of FeCl 3 3 .5 Valence of FeCl3 = 3 ∴ Equivalent weight of FeCl3 = 162. of H 2SO 4 sol.of H 2 SO 4 sol. Solution: Basis: 1 litre 2N HCl solution. in L 1 Molarity (M) = (17) Find grams of HCl needed to prepare 1 litre 2N HCl solution.Molecular weight of Al2(SO4)3 = (2x27)+(3x32)+(12x16) = 342 Valence of Al2(SO4)3 = 6 equivalent weight of Al 2 (SO4) 3 = mol wt of Al 2 (SO4) 3 342 = = 57 valence of Al 2 (SO4) 3 6 (5) KMnO4: Molecular weight of KMnO4 = (1x39)+(1x55)+(4x16) = 158 Valence of KMnO4 = 5. . in L 1 moles of H 2 SO 4 1 = =1 vol. Find normality and molarity of the solution. equivalent weight of Al 2 (SO4) 3 = (16) mol wt of KMnO 4 158 = = 31. Solution: Basis: One litre of solution Amount of H2SO4 dissolved = 98 g Molecular weight of H2SO4 = 98 equivalent weight of H 2SO 4 = mol wt of H 2SO 4 98 = = 49 valence of H 2SO 4 2 Gram-equivalents (g eq) of H2SO4 = 98 = 2 49 g equivalent s of H 2 SO 4 = wt of H 2 SO 4 98 = =2 eq wt of H 2 SO 4 49 Normality (N) = g eq of H 2 SO 4 2 = =2 vol.6 valence of KMnO 4 5 98 grams of sulphuric acid (H2SO4) are dissolved in water to prepare one litre of solution. 98 . in L wt of HCl in g 2 = eq wt .Normality g eq of HCl vol . Solution: Basis: 100 kg of solution.196 kg/L find the normality.98 M volume of solution in L 83. The solution contains 20 kg of NaOH and 80 kg water (solvent).5 kmol = 500 mol 40 Molarity(M) = g moles of NaOH 500 = = 5. 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 moles molality of 30 x 10 3 = 500 60 moles of acetic acid 500 = = kg of solvent 70 acetic acid = mol = 7 . and molality of the solution. 5 x 1 = 73 g (N) = in L (18) The concentration of an aqueous solution of acetic acid is specified as 30% on weight basis.of H Cl sol . molarity. Taking density of the solution as 1.62L 1. Find the molarity of the solution. of HCl x volume of solution wt of HCl in g = 2 x 36 . Density of solution = 1.196 Molesof NaOHin solution= 20 = 0.196 kg/L 100 volume of solution = = 83.52 For NaOH as valence = 1 Equivalent weight = Molecular weight ∴ Normality (N) = Molarity (M) = 5. Solution: Basis: 100 kg of solution. 142 m (19) A solution of caustic soda contains 20% NaOH by weight. calculate the quantities of H2SO4 to be taken to prepare these solutions.5 = 0.Molality(m) = g molesof NaOH 500 = = 6.5 g eq Molecular weight of H2SO4 = 98 equivalent weight of H2SO4 = 98 = 49 2 Amount of H2SO4 required for 1 normal solution = 0.5 x 49 = 24.5 g molarity = gmoles of H 2 SO 4 vol. 1 molar and 1 molal solution of H2SO4. Density of solution = 1.075 = 537. of solution in L Moles of H2SO4 = Molarity x Volume of solution = 1 x 0.25 m (mol/kg) kg of solvent 80 (20) A chemist is interested in preparing 500 ml of 1 normal.x) x 10-3 kg Molecular weight of H2SO4 = 98 moles of H 2SO4 = x 98 . Volume of solution = 500 ml = 0.075 g/cm3.5 x 98 = 49 g Let x be the quantity in grams of H2SO4 required for making 1 molal solution.5 – x Weight of solvent = (537. Solution: Basis: 500 ml of solution. Assuming the density of H2SO4 solution to be 1.5 L Normality (N) = g equivalent s of H 2 SO 4 volumw of solution in L ∴ gram-equivalents of H2SO4 = Normality x Volume of solution = 1 x 0.5 g grams of solvent = grams of solution – grams of solute = 537.5 = 0.5 mol Amount of H2SO4 required to make 1 molar solution = 0.5 .075 g/cc Quantity of solution = 500 x 1. gr. Calculate the volume in ml of 98% to be added to get solution to required strength. (21) 2000 ml solution of strength 0. 72 ml specific gravity 1 . x = 47. Solution: Basis: 2000 ml of 0.24 and molality of 94.84 volume of 98% H 2 SO 4 required = g of 98% H 2 SO 4 50 = = 2 . Molalityof H 2SO 4 = gmoles of solute wt of solvent in kg .97 g.84 (22) A H2SO4 solution has a molarity of 11.5 N H2SO4 is to be prepared in laboratory by adding 98% H2SO4 (sp.5 N H2SO4 solution. Calculate the density of solution. Normality (N) = g equivalent s of H 2 SO 4 volume of solution in L Volume of solution = 2000 ml = 2 L gram equivalents of H2SO4 = Normality x Volume of solution in L = 0. Solution: Basis: 1 litre of solution Molarity = 11.5 x 2 = 1 g eq Amount of H2SO4 required = 1 x 49 = 49 g amount of 98% H 2 SO 4 required = 49 = 50 g 0.84) to water.24 and Molality = 94 Now.1.97 g Amount of H2SO4 required for preparing 1 molal solution = 47.x) .10-3 Solving we get.Molalityof H 2SO 4 = gmoles of solute wt of solvent in kg 1= x / 98 (537.5.98 Specific gravity of 98% H2SO4 = 1. 24 x 98 = 1101. Solution: Basis: 2 litres of NH3 PV = nRT PV n = moles of NH 3 = RT Where P = 20.265 kPa.24 = 0.045 x 10-3x98= 0.24 = amount of solution = gmoles of H 2SO 4 1 11.0161 m 8.788 gequivalents of H 2SO 4 required = = 0.(2) = 0. V = 2 l R = 8.265 kPa is neutralized by 135 ml of solution of H2SO4.1195 = 1.24 mol Amount of H2SO4 = 11. Find the normality of the acid.2205 = 1.045x10 − 3 2 Amount of H2SO4 required = 8.788 g 0.31451 m3 .2205 kg amount of solvent = 1.101 kg Molality = mol H2SO4/ kg of solvent 11.1195 kg 94 Amount of solution = 1.kPa/(kmol.52 g = 1.(303) 2 NH3 + H2SO4 (NH4)2SO4 For neutralization of 2 moles of NH3. 1 mole of H2SO4 is required.gmoles of H 2SO 4 1 ∴Moles of H2SO4 = 11.K) density of H 2SO 4 solution = (23) n = moles of NH 3 = 20. T= 303 K.2205 kg / L 1 2 litres of NH3 at 303 K (300C) and 20.314.24 x 1 = 11.265.31451 kPa/(mol.0161) = 8.0161 49 Volume of H2SO4 solution = 135 ml = 0.K) = 8. 1 moles of H 2SO 4 required = .101 + 0.135 L .(0. Hence. resulting solution is neutral. None of the components is in excess.1 l Gram equivalents of H2SO4 = Normality x Volume of solution = 0. c.0161 = 0.) Solution: Basis: 100 ml of 0. basic or neutral? (Get the answer by numerical method.75 g/L HCl to molarity Basis: 1 l of solution equivalent weight of H 3 PO 4 required = . H2O sample. 3M K2SO4 to g/L.67 3 Concentration in grams per litre of solution = Normality x Equivalent weight Concentration of solution = 5 x 32.63 g Na2CO3.62 Na2CO3.N of H 2SO 4 = 0.49 g 124 For neutralizing.49 g H2SO4.135 (24) A sample of Na2CO3.1 x 0.62) = 0. H2SO4 in solution = 294 g 294 gequivalen ts of H 2 SO 4 required = = 6 geq 49 N= 6 = 6N 1 (25) (b) 5N H3PO4 to g/L Molecular weight of H3PO4 = 98 Valence of H3PO4 = 3 98 = 32.1 H2SO4solution.49 g H2SO4 as per reaction and we have added solution containing 0.49g Molecular weight of Na2CO3 = 124.1 = 0.1 N Volume of H2SO4 solution = 100 ml = 0.H2O = (0.1 N H2SO4solution and 0.12 N 0. we need 0. 294 g/L H2SO4 to normality. H2O weighing 0. d.35 g/L (c) 54.75 g/L HCl to molarity.e. Do the following conversions: a.01 Amount of H2SO4 in solution = 0.62 grams is added to 100 ml of 0.67 = 163. b.8 mg/ml CaCl2 to normality Sol: (a) 294 g/l H2SO4 to normality Basis: 1 litre of solution.H2O = ? 98 ∴ Amount of H2SO4 required for 0.01 x 49 = 0.H2O = 1 mol H2SO4 124 g Na2CO3. Normality of H2SO4 solution = 0. 4.H2O = 98 g H2SO4 ∴ 0. Molecular weight of H2SO4 = 98 Na2CO3. Will the resulting solution be acidic. H2O + H2SO4 Na2SO4 + 2H2O + CO2 1 mol Na2CO3.62 g of Na2CO3. 5N H3PO4 to g/L. 54. 6 Molarity of solution = = 2 . Moles of solution = 3 x 1 = 3 mol Molecular weight of K2SO4 = 174 ∴ Amount of K2SO4 = 3 x 174 = 522 g molarity HCl solution = 522 = 522 g / L 1 (e) 4. 832 M 110 . 3116 kmol = 311 . normality. Solution: Basis: 43 kg of K2CO3 and 100 kg of water Weight of K2CO3 solution = 43 + 100 = 143 kg 143 Volume of solution = = 110 . 0865 1 (26) An aqueous solution of K2CO3 is prepared by dissolving 43 kg of K2CO3 in 100 kg of water at 293 K (200C). 6 mol 138 311 .3 Molecular weight of K2CO3 = 138 ∴ Equivalent weight of K2CO3 = 138 = 69 2 ∴ Moles of K2CO3 in solution = 43 = 0 .75 g moles of HCl = 54. 5 0 .8 g/l CaCl2 CaCl2 in solution = 4.5 1 .5 mol 36. 0865 55 .8 mg/ml CaCl2 = 4. 0865 Normality = = = 0 . Calculate molarity.5 M 1 molarity HCl solution = (d) 3M K2SO4 to g/L Basis: 1 l of solution.75 = 1.3 kg/L. L 1 .8 mg/ml CaCl2 to normality Basis: 1 L of solution 4.5 = 1 . 5 2 48 Gram equivalents of CaCl2 = = 0 . Density of solution is 1.8 g 111 Molecular weight of CaCl2 = = 55 .Amount of HCl in g/l = 54. molality of solution. (1000 ) = 623 . VOLUME PERCENT The pure component volume of any component is expressed as a percentage of the total volume of the system. WEIGHT PERCENT The weight of any component expressed as a percentage of the total weight of the system. The composition of a mixture or solution are expressed in weight percent.Gram equivalents of K2CO3 43 . 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. VB = total volume of the system VA+VB … for a binary system of A and B. 19 eq 69 623 . volume percent. VA = pure component volume of A. A and B and is also used for more than two components. mole percent. 6 Molality of solution = = 0 . WA = weight of the component A W= WA + WB = weight of the system …. 19 Normality of solution = = = 5 . V pure component volume of A Volume % of A = = A x 100 − − − − − ( 2 ) total volume of system V Where. WA weight of A ∴ Weight % of A = = x 100 − − − − − − (1) total weight of system W Where. 665 N 110 311 . 3116 N 100 = 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. 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. ∴ For a binary system of A and B : moles of A Mole fraction of A = total moles of system . WEIGHT FRACTION The ratio of the weight of a particular component to the total weight of the system. XA = mole fraction of A XB = mole fraction of B. we get WA M A WA WB + MA MB WB MB WA WB + MA MB xA = + = 1 − − − − − − − (5) ∴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. ∴In a binary system of A and B : x 'A = WA − − − − − − − − − −(6) WA + WB .xA = WA M A WA WB + MA MB − − − − − − − − − − (3) Where. we have Mole % of A = Mole fraction of A x 100 Adding equations (3) and (4). xB = WB MB WA WB + MA MB − − − − − − − − − −(4) From equation (3) and (4). 427 + 5. or NOTE: The sum of all the weight percentages for a given system totals to 100. 56 kmol 18 Total moles of solution = 0. Molecular weight of H2O = 18. 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.987 kmol kmol NaCl 0 . 5 10 Moles of H2O = = 5 . we get WA WB x 'A = + =1 WA + WB WA + WB Also. 25 ∴ Moles of NaCl = = 0 . WB x 'B = − − − − − − − − − −( 7) WA + WB From equations (6) & (7). Similarly for B. 427 Mole % NaCl in solution = = x 100 = 7 . 13 % kmol of solution 5 .56 = 5. (28) An aqueous solution of sodium chloride is prepared by dissolving 25 kg of sodium chloride in 100 kg of water. Solution:Basis: 25 kg of sodium chloride and 100 kg of water Amount of solution = 25 + 100 = 125 kg weight of NaCl in kg 25 Weight % NaCl in solution = = x 100 = 20 % total weight of solution in kg 125 Weight % H2O = 100 – Weight % of NaCl = 100 – 20 = 80 % Molecular weight of NaCl = 58.Where xA is the weight fraction of A. 427 kmol 58 . Determine (a) weight % and (b) mole % composition of solution.5. 987 . 5 Moles of KCl= 600 = 8 . 5 Total moles of the mixture = 3.81 = 70.464 kmol .13 = 92. 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 weight of NaCl in kg 200 = = x 100 = 75 % total weight of mixture in kg 800 Weight % KCl = 100 – weight % of NaCl = 100 – 25 = 75% Molecular weight of NaCl = (1 x 23) + (1 x 35. 05 kmol 74 . 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 64 ∴ Weight % salicylic acid solution = x 100 = 39 .419 + 8. 02 % 164 Weight % methanol = 100 – 39.05 Mole % NaCl = = 11.125 kmol Moles of Salicylic acid = 64/138 = 0.02 = 60.5 200 Moles of NaCl = = 3 .5 Molecular weight of KCl = 1 x 39 + 1 x 35.469 kmol kmol of NaCl 3 . Molecular weight of HOC6H4COOH = 138 Mole of methanol = 100/32 = 3.5 = 74. 419 kmol 58 .5) = 58.Mole % H2O in solution = 100 – Mole % of NaCl = 100 – 7.87 % (29) Sodium chloride weighing 200 kg is mixed with 600 kg potassium chloride.98% Molecular weight of CH3OH = 32.19% (30) A saturated solution of salicylic acid (HOC6H4COOH) in methanol (CH3OH) contains 64 kg salicylic acid per 100 kg methanol at 298 K (250C). 469 Mole % KCl = 100 – mole % NaCl 100 – 29. 419 = x 100 = 29 . 81 % kmol of mixture 11 . 5886 (32) Ethanol and water forms aazeotrope containing 96% ethanol by weight. 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 100 Weight fraction of methanol in the saturated solution = = 0 .125 + 0. Molecular weight of H2O = 18 Amount of HNO3 in azeotrope = 37.07 = 12. 694 144 Molecular weight of CH3OH = 32. 125 Mole fraction of methanol in saturated solution= = 0 .125 kmol 44 Moles of CH3Br in solution = = = 0 . 4636 kmol 94 . 087 Mole % ethanol in azeotrope mixture = = x 100 = 90 .2 x 18= 1119.125 + 0.5886 kmol 3 . Molecular weight of CH3Br = 94. 91 ∴ Total moles of solution = 3. Solution: Basis: 100 kg of ethanol-water mixture It contains 96 kg of ethanol and 4 kg of water Molecular weight of H2O = 18.6 = 3501 kg .8 x 63 = 2381.222 = 2.6 kg Amount (weight) of azeotrope = 2381.589 kmol 3 .464 = 3.2 % water by mole. 38 % 2 .4 kg Amount of H2O in azeotrope = 62.91 Moles of CH3OH in solution = 100/32 = 3.Total amount of solution = 3.4 + 1119.222 kmol Moles of azeotrope mixture = 2. 125 Mole % of methanol = x 100 = 87 . Find the composition of azeotrope by mole %. Find the composition of the azeotrope in weight %. 589 Mole % of salicylic acid = 100 – 87. Solution: Basis: 100 kmol of HNO3 + water azeotrope It contains 62.4636 = 3. 07 % 3 .087 kmol Moles of water = 4/18 = 0.2 kmol of H2O and 37.309 kmol 2 .087 + 0. Calculate (i) the weight fraction and (ii) the mole fraction of methanol in the saturated solution. 309 (33) Nitric acid and water forms maximum boiling azeotrope containing 62. Molecular weight of C2H5OH = 46 Moles of ethanol = 96/46 = 2.88 kmol of HNO3 Molecular weight of HNO3 = 63.93% (31) At 298 K (250C) the solubility of methyl bromide in methanol is 44 kg per 100 kg. 871 3 . Find the weight % ethanol.5 kg of nitrogen as N Molecular weight of NH4NO3 = 80. It contains 45 kg of nitrogen as N Atomic weight of N = 14. 43 Weight % urea content of sample = x 100 = x 100 = 96 . Calculate the actual urea content in the sample.85 + 85 = 96.Weight % HNO3 in azeotrope = = 2381 .85 x 100 = 85 cm3 Amount of ethanol in solution = Volume x Density = 15x 0. 57 kg 28 % NH4NO3 content of sample = 98 . 57 98 .79 = 11. 23 % 96 . 85 Weight % ethanol in solution = = ∴ x 100 = 12 .15 x 100 = 15 cm3 Volume of water in solution = 0. Volume of ethanol in solution = 0.79 g/cm3 respectively.57 % . 5 = 98 . Find the actual ammonium nitrate content of the sample.85 g Amount of water in solution = 85 x 1 = 85 g ∴ Total amount of solution = 11. 85 (35) The available nitrogen (N) in the urea sample is found to be 45 % by weight. 4 x 100 = 68 . Atomic weight of N = 14 1 kmol NH4NO3≡ 2 katom N ∴ 80 kg NH4NO3≡ 28 kg N 80 ∴ Amount of NH4NO3 in sample = x 34 .5 % by weight.85 g 11 . Molecular weight of NH2CONH2 = 60 1 kmol NH2CONH2≡ 2 katom N ∴ 60 kg NH2CONH2 ≡ 28 kg N 60 ∴ Actual urea in sample = = x 45 = 96 . Solution: Basis: 100 kg of urea sample. 43 % kg sample 100 (36) The nitrogen content of NH4NO3 sample is given as 34. 57 % 100 100 ∴ Actual NH4NO3 content of sample is 98. 57 x 100 = x 100 = 98 . Solution: Basis: 100 kg of sample It contains 34. Solution: Basis: 100 cm3 of aqueous solution. if densities of ethanol and water are 0. 02 % 3501 (34) An aqueous solution contains 15% ethanol by volume. 43 kg 28 kg urea 96 . H2O = 30% and organic compounds = 5%. NH4SO4 = 45%. 24 kg 132 Molecular weight of NH4NO3 = 80 1 kmol NH4NO3≡ 2 katom N 80 kg NH4NO3≡ 28 kg of N (on weight basis) 28 ∴ Nitrogen available from NH4NO3 = x 20 = 7 kg 80 ∴ Total nitrogen available from the solution= 14 + 4.35 kg Total acid content of the spent acid as weight % H2SO4 . 24 ∴ Nitrogen available from the solution = x 30 = 25 . It contains 30 kg urea. Solution: Basis: 100 kg of solution.(37) Calculate the available nitrogen content of solution having 30 % urea (NH2CONH2).24 + 7 = 25. Molecular weight of urea (NH2CONH2) = 60 1 kmol NH2CONH2 ≡ 2 katom N 60 kg NH2CONH2≡ 28 kg N 28 ∴ Nitrogen available from urea = x 30 = 14 kg 60 Molecular weight of (NH4)2SO4 = 132 1 kmol (NH4)2SO4 ≡ 2 katom N 132 kg (NH4)2SO4≡ 28 kg N (on weight basis) 28 ∴ Nitrogen available from (NH4)2SO4 = x 20 = 4 . 24 % 100 (38) Spent acid from fertilizer plant has the following composition by weight: H2SO4 = 20%.35 = 58. Find the total acid content of the spent acid in terms of H2SO4 after adding the acid content. chemically bound in ammonium hydrogen sulphate. 35 kg 115 H2SO4 from the spent acid = free H2SO4 + H2SO4 chemically bound in NH4SO4 = 20 + 38. 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 98 ∴ H2SO4 chemically bound in NH4SO4 = x 45 = 38 . 20 kg of ammonium sulphate and 20 kg ammonium nitrate.24 kg 25 . 20 % ammonium sulphate and 20 % ammonium nitrate. 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. . Molecular weight of Na2O = 62 2 kmol NaOH ≡ 1 kmol Na2O 80 kg NaOH ≡ 62 kg Na2O 62 ∴ Amount of Na2O in the lye = x 73 = 56 .575 Weight % Na2O in the lye = x 100 = 56 . 575 % 100 (40) An aqueous solution of soda ash contains 20 % soda ash by weight. Express the composition as weight % Na2O. Solution: Basis: 100 kg of caustic soda flakes It contains 74. Determine the purity of the flakes. 7 % 100 (41) A sample of caustic soda flakers contains 74.6=96.7 x 100 = 11 . 7 kg 106 11.26 kg 62 96. 575 kg 80 56.6 % Na2O by weight. Determine the actual concentrations of H3PO4 (by weight) in the acid.2x100= 20 kg Na2CO3 = Na2O + CO2 Molecular weight of Na2CO3 = 106.58 . 35 x 100 = 58 .26 % purity of the flakes = kg NaOH x 100 kg flakes = x 100 = 96 . Molecular weight of Na2O = 62 1 kmol Na2CO3≡ 1 kmol Na2O 106 kg Na2CO3≡ 62 kg of Na2O 62 Weight of Na2O in the solution = x 100 = 11 . 35 % 100 (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. Solution: Basis: 100 kg of an aqueous solution of soda ash Amount of soda ash in the solution = 0. 26 % 100 Weight % Na2O in the solution = (42) The strength of a phosphoric acid sample is found to be 35% P2O5 by weight. temperature and volume should be a known one. Solution: Basis: 106 kg of sample of water. Express the concentration of solids in the sample in weight percent. Molecular weight of P2O5 = 142 2 kmol H3PO4≡ 1 kmol P2O5 196 kg of H3PO4≡ 142 kg of P2O5 196 ∴Amount of H3PO4 in sample = x 35 = 48 . if it that needed to dealing with substances existing in the gaseous state. 31 % 100 Caustic soda flakes received from a manufacturer are found to contain 60 ppm silica (SiO2). as the volume can be measured easily. The relationship among mass. which in turn gives mass of the gas. The density can be calculated if the parameters such as temperature and pressure are known.31 x 100 = 48 .Solution: Basis: 100 kg of phosphoric acid sample It contains 35 kg of P2O5. . 2 % 10 6 GASES The composition are expressed in terms of volume percent. Convert this impurity into weight %. Solids in water = 2000 ppm 2000 Amount of solids in water = x 10 6 = 2000 kg 6 10 Weight % of solids = 2000 x 100 = 0 . 2H3PO4 = P2O5 + 3H2O Molecular weight of H3PO4 = 96. Sol: Basis: 106 kg of caustic soda flakes Silica in flakes is given as 60 ppm 60 ∴ Amount of silica in flakes = x 10 6 = 60 kg 6 10 Weight % of H3PO4 in phosphoric acid sample= (43) Weight % of silica in flakes = 60 x 100 = 0 . 31 kg 142 48. 006 6 10 (44) A sample of water contains 2000 ppm solids. kPa/(kmol. i. T is in K and n is in kmol.e. the ratio of the volume to temperature is constant at a given pressure. V is in m3. P1V1 = nRT1 ---. and P2 be the volume. then numerical value of R is 0.(E) The ideal gas law states three facts : (i) volume of gas is directly proportional to numberof moles.008314 with the units m3. P x V = Constant --------. V is in m3.(A) Where P is the absolute pressure and V is the volume occupied by the gas. V = cons tan t − − − − − − − − − (B) T Where T is the absolute temperature By combining the above two laws. temperature and pressure of n kgmol of gas at conditions -2 then. Let V1. and P1 be the volume.K). then the ideal gas is written as PV = nRT -----.IDEAL GAS LAW BOYLE’S LAW Given mass of an ideal gas.. 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. T2.31451 with units m3. then the third one can be calculated by means of proportionality indicated by the gas law. When P is in Mpa. the product of the pressure and volume is constant at a constant temperature i. T is in K and n is in kmol. then the units of R will be m3. CHARLE’S LAW : Given mass of an ideal gas. T1.K). temperature and pressure of n kgmol of gas at conditions1 Let V2. PV = RT ---. V is in kmol and T is in K.(G) Combining the above two (F & G) equations. (ii) volume is directly proportional to the absolutetemperature (iii) volume is inversely proportional to the pressure..Mpa/(kmol. When P (absolute pressure) is in kPa. then numerical value of R is 8. we get .(D) When V is the volume in cubic meters of n kmol of gas.K) When the mass of a gas is not known. When P is in kPa.e.kPa/(kmol. it can be formulated as an ideal gas laws as PxV = cons tan t − − − − − − − − − ( C ) T The constant of the above equation is designated by the symbol R and is known as Universal gas constant Therefore.(F) and P2V2 = nRT2 ---. V = 22.P1 V1 P 2 V 2 = − − − − − − − − − − − − − (H ) T1 T2 In the ideal gas law given by equation (5). At 273. V is called the molar volume. The molecules of the each component gas are distributed throughout the entire volume of the container in a closed vessel.15 K* (00C) and 101.325 kPa. if that alone is present in the same volume and at the same temperature . 101.4136 l/mol). 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. RELATIONSHIP BETWEEN PARTIAL PRESSURE. These conditions are said to be normal temperature and pressure (NTP)...325 kPa (1 atm) and 288.C etc. If two of three variables are known(temperature.7 K (15. PURE COMPONENT VOLUME The volume occupied by that component gas.B. respectively.4136 m3/kmol (or 22. if that alone is present at the same pressure and temperature as the gas mixture. are the pure component volumes of component gases A..325 kPa (1 atm).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. V2 = 22.4136 m3 i.B. In USA.e. and VC etc.e.. GASEOUS MIXTURE The composition of component gases are expressed in terms of volume percent.60C) are considered to be standard temperature and pressure conditions (STP). MOLE FRACTION OF COMPONENT GAS TO TOTAL PRESSURE The gas mixture consisting of component gases A. . pressure and volume) at one specified condition and the third one can be calculated.C are considered. Mathematically. P2= 101. thus are calculated with the help of equation (8) where the other situation may be taken as NTP i.4 m3 per kmol and T2 = 273 K. V = VA + VB+VC + ………… (J) Where V is the total volume and VA. the volume occupied by 1 kmol of a gas is 22. VB. .. VA.PB. The ideal gas law for the component gas A is pAV = nART pA = n A RT − − − − − − − − − − − −(K ) V Where R = gas constant T = temperature of gas mixture Similarly. (K).C etc. respectively.B.. and (L) we get (n pA +pB +pC = +n + .= n = total moles ∴pA = x A P ..) RT A + nB C − − − − − − − − − − − −(N ) V According to the Equation (1) P = pA+pB+pC+ ------..(O) Dividing the equation (J) by Equation (M) we get pA nA = − − − − − − − − − − − (P ) pA + pB + pC nA + nB + nC nA+nB+nC + ------.B.. respectively.PC are the partial pressures of component gases A.VC etc are the pure component volumes of A.. for component B: pB = n B RT − − − − − − − − − − − − (L ) V Similarly.VB. for component C : pC = n C RT − − − − − − − − − − − −(M ) V Adding Equations (J)..Let V be the total volume of the gas mixture and P be the total pressure exerted by the gas mixture..C etc. PA. (R) When an ideal gas law is applicable.(S) Similarly. 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.e.(T) For component C : PVC = nCRT -----.(Y) Where. Volume fraction of A = mole fraction of A ---. Multiplying both sides of Equation (O) by 100.The mixture of ideal gases (i. pA nA x 100 = x 100 − − − − − − − − − − − ( Q ) pA + pB + pC nA + nB + nC i. Pressure % = Mole % -----..C …. x A = nA nA + nB + nC ∴ Volume % of A = Mole % of A --. pressure % of A = mole % of A --.(U) Let VA. V = VA+VB+VC + ------W/X gives : ----.(X) VA = xA V i. (T)...(Z) .VC etc.e. are the pure component volumes of components A. etc.B. and (U). we get P (VA+VB+VC + ------) = (nA+nB+nC+-----) RT – (V) Dividing Equation (S) by Equation (V) gives VA nA = − − − − − − − − − − − (W ) VA + V B + V C nA + nB + nC According to Amagat’s law. V = Total volume of the gas mixture Adding Equations (S). for component B : PVB = nB RT -----.VB. it is written for any component gas A as : PVA = nA RT -----.e.(Q) This result can be proved for other components of the gas mixture ∴ For gases behaving ideally... B. MB.xC +……. Let MA....It is calculated by assuming the quantity of the gas mixture as one mole. moles of gas mixture = kg of gas mixture − − − − − − − − (3) M avg ∴ From equation (1) and (2) can be written as . etc.e. The weight of the one mole of the gas mixture represents its average molecular weight. 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. m3 R = universal gas constant 8.Combining results obtained.XC..XB... The ideal gas equation for a gas mixture is PV = nRT ----. Then.e.For calculating the density.xA+MB. is considered. thus the volume analysis treated same as the mole analysis.This result can be proved for other components of the gas mixture. MC etc. are the molecular weights and mole fractions of components gases respectively. The mole analysis (i. Mavg = MA.31451 m3.K). (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. volume % = mole %) of each component is calculated which is needed for calculating the average molecular weight of the gas mixture. The composition of the gas mixture is specified in terms of volume % (on a volume basis). T = temperature K. Assuming the applicability of ideal gas. we have: Pressure % = Mole % = Volume % AVERAGE MOLECULAR WEIGHT OF GAS MIXTURE Required for calculating the weight of the gas mixture. Mavg = Where. In general. i. Mi = molecular weight of ith component gas. the known average molecular weight of the gas mixture plays a major role. Let Mavg be the average molecular weight of the gas mixture. equations (Q) and (Z) for gases behaving ideally/for an ideal gas mixture.kPa/(kmol. The gas mixture consisting of components A. and XA. P = pressure in kPa. C etc.xB+MC.(1) P n = − − − − − − − − ( 2) RT V Where n= kmol of gas mixture V = volume of gas mixture. n = 0.5 m3.341kmol.kPa/(kmol. Calculate the pressure required for given duty.5m3. Moles of Cl2 gas = 20 = 0. Assume ideal gas law is applicable. n = 0. R = 8.kPa/(kmol.31451 x 298 = 6.kPa/(kmol. V RT wt of gas mixture PM avg .K).2817 kmol 71 XA is the PV = nRT ∴ V = nRT P Where.K) Volume.98 m3 100 (46) 15 kg of carbon dioxide is compressed at a temperature of 303 K (300C) to a volume of 0. V = 0. Solution: Basis: 15 kg of carbon dioxide gas Molecular weight CO2 = 44 ∴ Moles of CO2= 15 = 0. R = 8. Solution: Basis: 20 kg Cl2 gas.K) .V = 0.31451 m3.2817 kmol. 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). = V RT ρmix = Density of gas mixture = wt of gas mixture Volume ρ mix = PM avg RT When R is expressed in m3. P = 100 kPa.wt of gas mixture P = M avg .341 kmol 44 PV = nRT ∴ P = nRT V Where. T = 298 K.2817 x 8.31451 m3. T = 303 K. Solution: Basis: 2 m3 volume of SO2 gas PV = nRT ∴ n = PV RT Where.31451 m3.0596 x 64 = 3. V = 2 m3.98 kPa. the pressure and temperature being 97.5 (47) 5 kg of oxygen contained in a closed container of volume 1 m3 is heated without exceeding a pressure of 709.5162 kmol 32 PV = nRT ∴ T = PV nR Where.16 kPa 0.59 kPa g and 299 K (260C) is heated to a temperature of 1273 K (10000C). P = 97.7 kPa. P = 0.31451 m3.820C) (50) A gas contained in a closed vessel at a pressure of 121.K) ∴Moles of SO2 gas = 97. ∴ Moles of O2= 5 = 0.82 K (414. Solution: Basis: A gas at 299 K in closed vessel P1V1= P2V2 T1 T2 .28 x 1 = 546.K) Maximum temperature of gas.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 303.31451 ∴ Temperature of gas attained is 546.kPa/(kmol. P2 = 303. P1 = 131. V = 1 m3.130C) (48) Calculate the weight of sulphur dioxide in a vessel having 2 m3 volume.Pressure.341 x 8.28 kPa.98 kPa.31451 x 393 Weight of SO2gas = 0.31451 x 303 =1718. Find the pressure to which a closed vessel should be designed.98 x 1 298 T2 ∴ T2 = 687.0596 kmol 8. R = 8. P = 709.82 K ∴ Temperature of gas attained = 687. T2 = ? ∴ 131. T1 = 298K. Calculate the temperature of gas attained. R = 8.5162 x 8.5162 kmol. Molecular weight of O2 = 32. T = 393 K. T = 709. V1=V2 = 1 m3(as vessel being closed).13 K (273.kPa/(kmol.13 K 0. Solution: Basis: 1 m3 volume of gas at 298 K P1V1= P2V2 T1 T2 Where.28 kPa. Solution: Basis: 5 kg oxygen. n=0.7 x 1 = 303. Calculate the maximum temperature of gas attained.33 kPa.33x 2 = 0.33 kPa and 393 K (1200C). 85 P1. Calculate the final volume of gas. It is given that volume increases by 50%. It is given that pressure increases by 85%. Final volume of gas = P1.3125 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%.65 kPa. Solution: Basis: a gas at initial pressure of 202. T2 = 1273 K Absolute pressure = Gauge pressure + Atmospheric pressure ∴ P1 = 121. The operation is done for a fixed mass of gas at constant temperature. It is given that pressure of gas increases by 60% ∴ Final pressure = P2 = 1.As vessel being closed. Calculate the final volume of gas.85 P1 ∴ Final volume of gas = 0. kPa.5 V1 m3. kPa. we have. Final pressure = P2 = 1.95 MPa.5 m3 is compressed in such a manner so that pressure is increased by 60%. we have.65 kPa. The operation is done for a fixed mass of a gas at constant temperature. Solution: Basis: 1 m3 of gas Initial volume = 1m3.5 m3 of gas sample. V1 = V2 ∴ P1= P2 T1 T2 ∴ P2= P1T2 T1 Where. Initial pressure = P1.59 kPa g.54 m3 P2 1.54 m3.6 P1. Calculate the final pressure of gas.5 = 0. Final volume = V2= ? Temperature and mass are constant.07 kPa ∴ Pressure to which vessel should be designed= 949. The operation is done for a fixed mass of gas at constant temperature.07 kPa = 0. Initial volume = V1 m3=?.6 P1 ∴ Final volume of gas = 0.325 = 222. kPa. P1V1 = P2V2 . (51) A sample of gas having volume of 0. Initial pressure = P1 = 202.65 kPa pressure is expanded so that the volume is increased by 50%.915 kPa P2 = 222. For a given mass at constant temperature. P1V1 = P2V2 V2 = P1V1 = P1 x 1 = 0. Let initial pressure be P1 kPa. (53) A certain sample of gas at a pressure of 202. ∴ Final volume = V2 = 1.3125 m3 P2 1. For given mass at constant temperature.59 + 101. Solution: Basis: 0. P1 = 121.915 x 1273 = 949. therefore P1V1 = P2V2 V2 = P1V1 =P1 x 0. T1 = 299 K. kPa/(kmol. V1= V2 = 1 m3.59 kPa. P1 = initial pressure = 121.3 kPa. For given mass of gas at constant temperature. 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 = 0.5 V1 ∴ Final pressure of gas = 135.5 V1 = 0. Initial volume = V1 = 1m3. Solution: Basis: 1 m3 of gas. T = 273 K (00C) V = 0. Calculate temperature of gas attained. we have. P2 = final pressure =405. T1 = initial pressure = 298 K. V = volume in m3 at NTP=? NTP conditions ⇒ P = 101.5 ∴ 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.3409 x 8.P1V1 = P2V2 P2= V1= 1.59 kPa. Calculate the percent increase in pressure.637 m3.31451 m3.325 kPa. ∴ . T2 = final temperature of gas = ? As vessel is closed. V2 1.325 ∴Volume of propane gas at NTP = 7. Final volume = V2 = 0.3 kPa. (54) A sample of gas having volume of 1 m3 is compressed to half of its original volume.K) .59 kPa is to be heated such that pressure should not exceed 405. T = 273K. Initial pressure = P1. Final pressure = P2.1 kPa. (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 121.0 P1 V2 1. kPa. kPa.P2 = P1V1 =202.637 m3 101.5 m3. R = 8.1 kPa. Solution:Basis: 1 m3 of gas at temperature of 298 K and pressure of 121. The operation is carried for a fixed mass of gas at constant temperature.65 x V1= 135. P1V1= P2V2 T1 T2 T2 = P2V2x T1 P 1 V1 Where.325 kPa.31451 x 273 = 7.3409 kmol 44 PV = nRT P= 101. 7 Molecular weight of HCl = MHCl = 36. Molecular weight of N2 = MN2 = 28.481 x 405.127 x 405.274 kmol of HCl.31451 m3. Calculate (a) Average molecular weight of gas mixture and (b) Partial pressure of O2 and H2 at 100 kPa and 303K(300C). It contains 11.274 kmol HCl. Molecular weight of O2= 32.kPa/(kmol.3 = 51.274 = 0.127 ∴ Mavg = 31.391 Mole fraction of N2 = xN2 = 0.T2 = 405.33 K 121.337+0.59 x 1 ∴Temperature of gas attained = 993.391 0. calculate the partial pressure of each component gas at 405.xN2 +MO2.337 kmol N2 and 0.xO2 = 36.089 kmol O2.5 kPa Partial pressure of N2 = pN2 = xN2.330C) (57) A gas mixture contains 0. Molecular weight of H2 = 2 ∴ .95 kPa Partial pressure of HCl = pO2 = xO2.33 K (720. V = volume in m .481+32x0.127 Partial pressure of HCl = pHCl = xHCl.089 = 0. Sol: Basis: 100 kg of gas mixture.35 m3 405. calculate (a) Average molecular weight of gas and (b) Volume occupied by this mixture at 405.7 Mole fraction of N2 (XN2)=0.P= 0.7 kmol.3 x 1 x 298 = 993. n= moles of gas mixture = 0.127 0.391+28x0.337 kmol of N2 and 0.3 kPa Mole fraction of HCl = xHCl = 0.3 = 194.3 kPa and 303 K (300C) Solution:Basis: A gas mixture containing 0.P= 0.K) V = 0.47 kPa (59) A mixture of H2 and O2 contains 11.089 kmol of O2.3 kPa.080 = 0. Solution:Basis: Given gas mixture Total pressure = 405. Total moles of gas mixture= 0. T = 363 K.274 +0. XHCl + MN2.9 kg of O2.7 x 8.1 kg of H2 and 88.481 Mole fraction of O2 = xO2 = 0.31451 x 303 = 4.3 kPa and 303 K (300C).7 Mole fraction of O2 (XO2) = 0.3 (58) In case of gas mixture cited in previous example.P = 0. Molecular weight of O2 = MO2 =32 Mavg = ∑Mi Xi = MHCl.3 = 158.337 = 0.80 PV = RT PV = nRT V = nRT P 3 Where.7 kmol Mole fraction of HCl (XHCl) = 0.391 x 405.5 x 0. P= 405. R = 8. 0.1% H2 by weight.481 0.5. 0. xH2+M.78 = 0.325 kPa ha an average molecular weight of 31.CH4.55 kmol 2 and Moles of O2 = 88.Moles of H2 = 11.33 Mavg = Average molecular weight of gas mixture= M. Mavg = ∑Mixi (60) (61) Mavg = MN2.4.67 x100 = 67 kPa (502.P = 0.P = 0.33 = 11.C2H6.9 = 2.xC2H6 22.(1) ∑xi = 1 xN2 + xCO2 = 1 -----.33 kmol Mole fraction of H2 = xH2 =5.O2.1875 Partial pressure of N2 = xN2.33 x 100 = 33 kPa (247.(2) ∴ xCO2 = 1-xN2 -----.55 = 0.xCO2 31 = 28 xN2 + 44xCO2 ------.4 of gas mixture.33 Mole fraction of O2 = xO2 = 2. Find mole % CH4 and C2H6 in the mixture.1 = 5.54 torr) Partial pressure of O2 = pO2 = xO2.8125 x 101.52 torr) A mixture of nitrogen and carbon dioxide at 298 K (250C) and 101.H2. Solution: Basis: Average molecular weight of 22.325 = 82. Molecular weight of CO2 = 44. Molecular weight of N2 = 28.8125 = 0.(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. Mavg = ∑Mixi Mavg= M.78 = 8.67 + 32 x 0.4 = 16 xCH4+30xC2H6 ∑xi = 1 xCH4+xC2H6 = 1 xC2H6 = 1-xCH4 .8125 ∴ xCO2= 1-0.52 torr) A mixture of CH4 and C2H6 has the average molecular weight of 22. Let xCH4 and xC2H6 be the mole fraction of CH4 and C2H6 respectively.xCH4+M.P= 0.55 + 2.xO2 = 2 x 0.33 kPa(617.67 8.78 kmol 32 Amount of gas mixture = 5. Let xN2 and xCO2 be the mole fractions of N2 and CO2 respectively.33 8. What is the partial pressure of nitrogen? Solution:Basis: Average molecular weight of 31 of gas mixture.9 P = Total pressure = 100 kPa Partial pressure of H2 = pH2 = xH2.xN2 + MCO2. A mixture of CH4 and C2H6 has density 1.543 xC2H6 = 1-0. Solution: Basis: Air containing 21% O2 and 79% N2 by volume For ideal gases.79 + 82 x 0.325 x 28.4 = 16 xCH4 + 30 (1-xCH4) xCH4 = 0. mole % = volume % ∴ Mole % N2 = 79.79 100 100 Mole fraction of O2 (xO2)=Mole % O2= 21 = 0. Mole % of O2 = 21 Mole fraction of N2 (xN2) = Mole % N2= 79 = 0.84 = 1.325 Let xCH4 and xC2H6 be the mole fractions of CH4 and C2H6 respectively.325 kPa.K). ρ = 101.xCH4 + MC2H6.K) .2874 kg/m3 8.xN2 + MO2. ρ = 1 kg/m3. Mavg = ∑Mixi = MCH4. ρ = PMavg RT ∴ Mavg= ρ. ∴ Mavg= 1 x 8.kPa/(kmol. T = 273 K.(62) (63) Put the value of xC2H6 from equation (3) into equation (1) and solve for xCH4. P = 101.543 x 100 = 54. Mavg = 28. ρ = PMavg RT Where.(3) Put the value of xC2H6 from equation (3) into equation (1) and solve for xCH4 22.4 = 16xCH4 + 30 (1-xCH4) ∴ xCH4 = 0. Solution: Basis: 1 kg/m3 density of gas mixture at 273 K and 101.4 = 16xCH4 + 30 xC2H6 ---.84 Density of air.21 100 100 Mavg = MN2. Calculate the mole % and weight % of CH4 and C2H6 in the mixture.543 = 0.kPa/(kmol. P = 101.543 = 0. R = 8.457 Mole % of CH4 = Mole fraction of CH4 x 100= 0. Density of gas mixture.(2) ∴ xC2H6 = 1 – xCH4 -------. T=273 K.84.457 x 100 = 45. calculate the density of air at NTP. RT P Where.4 101.70 Assuming air to contain 79% N2 and 21 % O2 by volume.30 Mole % of C2H6 = 0.(1) ∑xi = 1 xCH4 + xC2H6 = 1 ---.21 = 28.0 kg/m3 at 273 K (00C) and 101.xC2H6 ∴ 22.0.31451 x 273 Density of air = 1.543 xC2H6 = 1.325 kPa.457 .31451 m3.325 kPa pressure.xO2 = 28 x 0.31451 x 273 = 22.325 kPa. 22.2874 kg/m3.31451 m3. R = 8. 325 kPa. Mavg = 18.06 = 18. Molecular weight of C2H6 = 30.325 kPa and composition In weight percent.4 Weight of C2H6 in mixture = 100 – 38.06 100 Molecular weight of CH4 = 16.12 100 Mole fraction of N2 = xN2 = 6 = 0.543 x 16 = 8.4.78 kg/m3 8.69 x 100 = 38.57 (64) .4 Density of gas = ρ= PMavg RT Where.69 kg Weight of C2H6 in 1 kmol mixture = 0.7 Weight of CH4 in 1 kmol mixture = 0.K) T= 288 K Density.30 19.71 kg Weight of gas mixture = 22. Mole fraction of CH4 = xCH4= 82= 0.31451 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 CH4 C 2 H6 Quantity in kg 1312 360 Weight % 71.kPa/(kmol.4 kg Weight of CH4 in mixture = 8. Molecular weight of N2 = 28 Mavg = Average molecular weight of gas Mavg = 16 x 0.82 100 Mole fraction of C2H6 = xC2H6= 12= 0.8 = 61.2 A natural gas has the following composition by volume: CH4 = 82%.543 x 100 = 54.82 + 30 x 0.3 Mole % of C2H6 = xC2H6 x 100 = 0. ρ= 101. Calculate the density of gas at 288 K (150C) and 101. 12 kmol of C2H6 and 6 kmol of N2.457 x 100 = 45. Solution: Basis: 100 kmol of gas It contains 82 kmol of CH4.31451 m3.4 = 0. P = 101.Mole % of CH4 = xCH4 x 100 = 0.457 x 30=13.12+28 x 0.325 x 18. R = 8. C2H6 = 12% and N2 = 6%.8 22. 1 + 28 x 0. 79% N2 by volume at 503 K (2300C) and 1519.815 100 Molecular weight of SO2 = MSO2 = 64.5%.R = 8. Mavg = 28. T = 503 K ρ = 1519.085 + 32 x 0.84 Density of air.79 = 28.46.13 100 (65) (66) Calculate the density of air containing 21% O2.5 =0 .79 100 Molecular weight of O2 = 32. Mole % N2 = 79.975 x 31.10 100 Mole fraction of N2 = xN2 = 81.xN2 = 64 x 0.481 kg/m3 8.31451 x 473 .875 kPa. P = 1519. Mole % = Volume % Mole % O2 = 21.xO2 + MN2.875 kPa.K).815 = 31.ρ = PMavg RT Where. For ideal gas.65 + 101.975. Molecular weight of N2 = 28 Mavg = MO2.K).43 kg/m3 8. Mole % = Volume % It contains 8.5= 0.84. 10 kmol of O2 and 81.31451 m3. Solution: Basis: Air containing 21% O2 and 79% N2 by volume. Mavg = 31. Molecular weight of O2 = MO2 = 32.xO2+MN2. R = 8. ρ = Pmavg RT Where.5 kmol of N2.46 = 2.21 100 Mole fraction of N2 = xN2= 79= 0.085 100 Mole fraction of O2 = xO2= 10= 0.N2 Total 168 1840 9.46 Absolute pressure = Gauge pressure + Atmospheric pressure ∴ P = 202. Molecular weight of N2 = MN2 = 28. Find (a) the density of gas mixture at a temperature of 473 K (2000C) and 202.31451 x 503 A gas mixture has the following composition by volume: SO2 = 8.5% .kPa/(kmol. Mole fraction of O2 = xO2 =21 = 0.325 = 303. T = 473 K ρ = 303. Mavg = MSO2. P = 303.21 + 28x 0.975 kPa Density of gas mixture .875 x 28. Solution: Basis: 100 kmol of gas mixture For ideal gas.31451 m3.kPa/(kmol.xSO2+MO2.5 kmol of SO2. O2=10% and N2 = 81.xN2 = 32 x 0.84 = 10.65 kPa g and (b) composition by weight. Mole fraction of SO2 = xSO2= 8. V1 = 0. Calculate the partial pressure of N2O4 in the final mixture. R= 8. T1 = 298 K. Volume of NO2 = 26.kPa/(kmol.5 x 64 = 544 kg O2 in gas mixture = 10 x32 = 320 kg N2 in gas mixture = 81.6 x 10-4 kmol = 0.31451 m3.17 N2 2282 72. Solution: Basis: 26.86-2x + x = 0.0266 m3 n1 = 80 x 0.30 O2 320 10. n1 = P1V1 RT1 Where.86 mol 8. the pressure is found to be 66.86 .0266 m3 P1V1 = n1RT Initial moles.6 lit = 0. P1 = 80kPa.662 kPa.x mol For initial conditions = P1V1=n1RT1 For final conditions = P2V2 = n1RT2 .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: Quantity in kg Component Weight % SO2 544 17.31451 x 298 2 NO2 = N2O4 Let x be the mol of N2O4 in final gas mixture NO2 reacted = 2x mol NO2 unreacted = 0.K).6 litres of NO2 at 80 kPa and 298 K (250C) is allowed to stand until the equilibrium is reached.0266 = 8.86 – 2x mol Final moles = n2 = 0.53 Total 3146 100 (67) In one case 28. At equilibrium.SO2 in gas mixture = 8.6 l of NO2 at 80 kPa and 298 K. Final moles = 0.46 kmol N2O4 at 333 K = 0.295 kmol NO2 at 333 K = 0.20 x 66.522 +x kmol P1V1 = n1RT1 P2V2 = n2RT2 But V1 = V2 for it being closed vessel. x = 0.T1 P2 n2 T2 P1 = 531.20 0. T1= 311 K.95 kPa.33 kPa (99.95 kPa. Sol: Basis: 100 kg of gas mixture at 311 K.46 x 46 = 67.652 = 1.1434 mol Mole fraction of N2O4 in final gas mixture. Taking ratio of equations (1) and (2). ∴531. n2= 1. Calculate the composition of gases at 600C by weight. n1= 0.96 kPa.87 + 2x = 0.P = 0.522 kmol N2O4 = 2 NO2 Let x be the kmol of N2O4 dissociated at 333 K NO2 formed = 2x kmol N2O= at 333 K = 0.87 kmol 46 Mole of N2O4 = 60 = 0. we get.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 531.86 – x = 0. NO2 in gas mixture = 40 kg.95 1.96 = 1.62 0. N2O4 in gas mixture = 60 kg ∴ Moles of NO2= 40 = 0.But here. n1= 1. T2 = 333 K.662 = 13.522.652 – x kmol NO2 at 333 K = 0.96 kPa.87 + 2x kmol ∴ Total moles at 333 K = (0. V1 = V2 and T1 = T2 ∴ P 1 = n1 n2 P2 80 = 0.295 = 1.16 kg . When the temperature is increased to 333 K (600C).87 + 0.7166 Partial pressure of N2O4 = xN2O4 .522 + x 333 Solving we get.652 – 0.357 kmol Amount of NO2 at 333 K = 1. P2 = 679. P1 = n1. some of N2O4 dissociates to NO2 and a pressure rises to 679.86 = 66.87+2x) = 1.522+x.652 kmol 92 Initial moles.522 x 311 679.652 – x = 0.87 +2 x 0.295 = 0. xN2O4 = 0.86 – x Solving we get.1434 = 0.652-x) + (0. 325 At 288 K.933 kPa.328 kPa and a temperature of 303K (300C) contain water vapour in such that proportions that its partial pressure is 2.693 kPa.035 kmol at 303 K 101.325 x 30 = 1. After cooling.2066 kmol 8.392 = n2 101. partial pressure of water vapour = 1.2066 = 1.31451 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 = 101.633 kPa Let n3 be the moles of water vapour at 288 K Mole of moist air = 1.933 = n1 101. n = 101.325 n n2 = 2.325 – 2. 98.84 kg Composition by Weight: Component NO2 N2 O4 Total Quantity in kg 67.31451 m3. Solution: Basis: 30 m3 of moist air at 303 K Ideal gas law is : PV = nRT n = PV RT Where.Amount of N2O4 at 333 K = 0.933 = 98. R= 8. T = 303 K Moles of air.16 52. the temperature is reduced to 288 K (150C) an some of water vapour is removed by condensation.325 n1 ∴n1= 98. P = 101.K).325 For water vapour /moisture. V = 30 m3.84 100 Weight % 67. P2 = Partial pressure of moisture at 303 K = 2.933 x 1. n= moles of moist air.84 100 (69) A volume of moist air 30 m3 at a total pressure of 101.357 x 92 = 32.172 + n3 .392x 1. Pressure % = Mole % Pressure fraction = Mole fraction For air.933 kPa For ideal gas.kPa/(kmol. Without total pressure being changed. it is found that the partial pressure of water vapour is 1. 2.2066 = 0.325 kPa.172 kmol at 303 K 101.16 32. Calculate (a) volume of air at 288 K (150C) and (b) weight of water condensed.392 kPa. n’ = 1.015 kmol Amount of water condensed = 0. Solution: Basis: 100 kmol of gas entering the absorption tower.192 kmol Let n’ = 1.K).98 x 20 = 19.6 x 36.035 – 0.1.325 kPa.992 kPa.325 Moles of HCl in gas entering = 20 kmol Moles of air in gas entering = 80 kmol Moles of HCl absorbed/removed = 0.192 x 8.172 + 0.6 kmol Moles of HCl unabsorbed and appearing in gas leaving = 20 – 19.27 kg (70) In manufacture of hydrochloric acid. V = volume of air at 288 K.kPa/(kmol. 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.4 = 80.27 m3 P 101. PV = nRT V = nRT P Where. R= 8.4 x 1 = 0.5 = 715.2646 kg 2704 Moles of gas entering = 80 + 0.kPa/(kmol.172 + n3 n3 = 0.02 kmol Moles of moist air at 288 K = 1.4 kmol PV = nRT ∴ V = nRT P .325 kPa V = n’RT= 1.31451 m3. 99. P = 101.325 1. V = volume of gas entering.31451 x 288 = 28.02 = 1.31451 x 323 = 2704 m3 . P = 99.325 Moles of condensed = n1 – n2 = 0. T = 288 K.K). 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. T=323 K V = 100 x 8.6 = 0.4 kg Volume of 100 kmol gas entering = 2704 m3 ∴ Amount of HCl absorbed per 1 m3 of gas entering = 715. n = 100 kmol.02 = 0.192 kmol. gas containing 20% HCl and 80% air to volume enters an absorption tower at a temperature of 323 K (500C) and pressure of 99. R= 8.015 x 18 = 0.4 kmol HCl absorbed based on 100 kmol gas entering = 19. m3.31451 m3.693 = n3 101.325 kPa.192 kmol PV = n’RT Where. Mole % of CO2 = 32.CH4 .K).325 = 303.3 kPa. T = 303 K. CO2 and NH3 For ideal gases.04 100 Mavg = Average molecular weight of gas M.65 kPa (72) The analysis of the gas sample is given below (on volume basis): CH4 = 66%.992 Volume of gas leaving based on 100 kmol or 2704 m3 of gas entering system = 1999 m3 ∴ Volume of gas leaving per 1 m3 of gas entering the system = 1999 x 1 = 0. Mole % of NH3 = 4 Mole fraction of CH4=xCH4= Mole % of CH4=66=0 .xCO2 + M.xNH3= 16 x 0. T= 293 K V = 80. Mavg = 24.31451 x 293 = 1999 m3 97.65 + 101. n = 80.K).CO2.4 x 8.kPa/(kmol. Find (a) the average molecular weight of the gas and (b) density of the gas at 202. 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. m3.95 kg/m3 8. NH3 = 4%.4 kmol. 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).975 x 24.992 kPa.22 ρ = Density of gas = PMavg RT Absolute pressure = Gauge pressure + Atmospheric pressure P = 202. ρ = 303. P = 97.66 100 Mole fraction of CO2 = xCO2 =30 = 0.30 + 0.74 m3 2704 (71) Equal masses of CO and N2 are mixed together in a container at 300 K (270 C).31451 m3.xCH4 +M.Where.975 kPa. The total pressure was found to be 405.31451 x 303 .NH3. R= 8.65 kPa g pressure and 303 K (300C).3 = 202.kPa/(kmol.3 100 Mole fraction of NH3 = xNH3= 4= 0.31451 m3. P = 0.44 = 2. R= 8. Solution: Basis: Gas sample containing CH4.5x405. V = volume of gas leaving.44.04 x 17 = 24. CO2 = 30%.66 + 44 x 0. Volume % = Mole % Mole % of CH4 = 66. T = 298 K ∴ .73 O2 0. (b) average molecular weight and (c) density of gas mixture at 298 K (250C) and 101. and 5 kg of O2 Moles of O2= 5= 0.325 kPa. R= 8.1225 = 78.02 = 0.7402 + 160 x 0. Mavg = kg of gas mixture kmol of gas mixture = 100 = 78.44 Also. Mavg = 78. Br = 80.1373 + 32 x 0.1373 100 Mole fraction of O2 = xO2= 12.7402 100 Mole fraction of Br2 = xBr2 = 13.Br2.9437 74.31451 m3.02 Br2 0.1225 100 Mavg = Average molecular weight of gas mixture = M.25 Total 1.325 kPa.1562 kmol 32 Composition of Gas Mixture by Volume : Quantity in kmol Volume % (mole % ) Component Cl2 0.73 = 0.44.(73) By electrolysing a mixed brine.2749 100 Mole fraction of Cl2 = xCl2 = 74. a gaseous mixture is obtained at the cathode having the following composition by weight: Cl2 = 67%.xBr2 + M.K).Cl2 .O2.kPa/(kmol.2749 ρ = Density of gas mixture = PMavg RT Where.1562 12. O=16) Solution: Basis: 100 kg gas mixture It contains 67 kg of Cl2.175 13. 28 kg of Br2. Calculate (a) composition of gas by volume.xO2= 71 x 0.xCl2 + M. (Atomic weights : Cl = 35. P = 101.5.25 = 0. Br== 28% and O2 = 5%.44 1. 208 kg/m3 8.93 m3/min 101.31451 m3. CO= 21%. Calculate the volume of gas in m3 at 298 K (250C) and 99. Sol: Basis: 1 m3/ min inert gas Molal flow rate of inert gas is n’ = PV’ RT Where. Calculate the flow rate of natural gas through the pipe line per min at 101.44 = 3.56 m3. x = 2.65 kPa and 303 K (300C) to a pipe line in which natural gas is flowing.9 ∴ 2.0804 .325 kPa Volumetric flow rate of natural gas = 2.0804 + x Solving we get. The analysis of this gas at a very long distance shown 2.692 kmol/min. T = 298 K.K).325 kPa.ρ= 101.0804 kmol/min 8. 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.kPa/(kmol. T = 303 K.9 = 0. Solution: Basis: 100 kmol of producer gas It contains 21 kmol of CO.kPa/(kmol.692 x 8.K) n’ = 202. n = 100 kmol.325 kPa and 303 K (300C).31451 x 298 (74) An inert gas (molecular weight 28) is admitted at the rate of 1 m3/min at 202. V’ = 1m3/min. P = 101. O2 = 3% and balance being N2. 3 kmol of O2 and 71 kmol of N2.692 kmol/min Volumetric flow rate of natural gas = xRT P Where. x = 2.325 Volume of producer gas corresponding to 312 kg carbon in it = 2492.31451 x 298 = 2494. R= 8. R= 8.325 (75) A producer gas has the following composition by volume . P = 202. ∴ Volume of producer gas per kg of carbon present .K). 100 0.9% by volume inert gas. R= 8.65 kPa.31451 x 303 = 66.31451 x 303 Let x be the kmol/min of natural gas flowing through pipe line Mole% inert gas = Volume % inert gas = 2. 5 kmol of CO2.65 x 1 = 0.P = 99. V = 100 x 8.31451 m3.31451 m3.56 m3 99.325 kPa per kg of carbon present. V = Volume in m3.kPa/(kmol. CO2 = 5%.325 x 78. T = 303 K. Solution: Basis: 0. n = 0.125 kmol 28 According to Ideal gas law – PV = nRT Where. 0.125 x 8. Calculate the pressure for which cylinders must be designed if they are subjected to a maximum temperature of 323 K (500C). V = 0. calculate the pressure for which they must be designed assuming applicability of ideal gas law.5 kg oxygen ∴ Moles of O2 = 0.5 kg of oxygen.5 kg of nitrogen.08 m3.= 2494. 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. P = pressure in kPa. T = 313 K. n = 0. ∴ P V = nRT (78) P =0.08 m3 each containing 3.0781 kmol.5 = 0. R= 8.5 kg nitrogen Molecular weight of N2 = 28 ∴ Moles of N2 = 3. P = 100 kPa. T = 323 K.0781 x 8.0781 kmol 64 PV = nRT V = nRT P Where.5 = 0.03 m3.23 kPa = 4.kPa/(kmol. R= 8.31451 x 313 = 2.31451 m3.kPa/(kmol. 100 (77) Nitrogen is to be marketed in cylinder having volume of 0. Volume of acetylene gas produced is V = 0. 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. Solution: Basis: 3.2 MPa.23 kPa.0 m3. If the cylinders may be subjected to a maximum temperature of 323 K (500C).0125 kmol.31451 x 323 = 4196.08 The pressure to which cylinders must be designed = 4196.015 m3 and each containing 0.K).56 x1 = 8.31451 m3.K).0156 kmol 32 According to Ideal gas law – . It is desired to market oxygen in small cylinders having volumes of 0. Solution: Basis: A gas mixture containing CO2. n= 0. V = 0.015 m3.31451 m3.31451 x 303 (80) The gas acetylene is produced according to the following reaction: CaC2 + 2H2O C2H2 + Ca(OH)2. xcH4 = 1 = 0. T = 323 K P = 0.01 + 28 x 0.90 = 1.PV = nRT P = nRT V Where.8 MPa. CH4 = 1% and N2 = 66%. Calculate the number of hours of service that can be derived from 1 kg of calcium carbide in an acetylene lamp burning 0.66 = 28.06 100 Mole fraction of H2O. R= 8.31451 x 323 = 2793 kPa 0.kPa/(kmol. Mavg = 28. CO = 14%.kPa/(kmol. H2O = 5%.05 100 Mole fraction of CH4. xH2O = 5 = 0. CH4 and N2 at 303 K and 101. CO. Molecular weight of O2 = 32.325 kPa.325 kPa. Molecular weight of N2 = 28 Average molecular weight of gas mixture is Mavg = ∑Mixi = 44 x 0.66 100 Molecular weight of CO2 = 44.K).0156 x 8. (79) A gaseous mixture has the following composition by volume: CO2 = 8%.10 m3 of gas per hour at temperature of 298 K (250C) and pressure of 99. H2O. Molecular weight of CO= 28.90.06 + 18 x 0.08 + 28 x 0.K).325 kPa.015 ∴ Pressure for which cylinders must be designed = 2793 kPa = 2. Calculate (i) Average molecular weight of gas mixture and (ii) Density of gas mixture at 303 K (300C) and 101. T = 303 K ρ = 101. xO2= 6=0. xCO =14 =0. 8.325 x 28. xCO2=Mole% of CO2=8= 0. Volume % = Mole % Mole fraction of CO2.05 + 16 x 0.325 kPa.01 100 Mole fraction of N2.90 Density of gas mixture is ρ = PMavg RT Where.31451 m3. R= 8. Solution: Basis: 1 kg calcium carbide CaC2 + 2H2O C2H2 + Ca(OH)2 . xN2 = 66 = 0.162 kg/m3.14 100 Mole fraction of O2.0156 kmol.14 + 32 x 0. Molecular weight of H2O = 18. O2. For ideal gases. P = 101.08 100 Mole fraction of CO. Molecular weight of CH4 = 16. 01562 x 8.45 x 10-3 m3.992 x 11.325 Burning rate of acetylene gas = 0. Volume of C2H2 gas produced = 0.Molecular weight of CaC2 = 64 ∴ Moles of CaC2= 1= 0.1674 = 89.K) 95.992 kPa pressure.1674 mol % dissociation of N2O4 = moles of N2O4 dissociated x 100 initial moles of N2O4 = 0. P = 95.187+x) x 10-3 x 8. T = 373 K.01562 kmol. R= 8. V = 11. 17.45 litres.2 = 0. of hours of service = Volume of acetylene gas Burning rate of acetylene = 0. calculate the percentage dissociation of N2O4 to NO2.2 grams of N2O4 gas occupies a volume of 11.2 g of N2O4. x = 0.896 h = 3. n = (0.kPa/(kmol. ∴ NO2 formed = 2x mol N2O4 undissociated = (0. 1 kmol CaC2 = 1 kmol C2H2 Moles of C2H2 produced =1x 0. PV = nRT Where.10 m3/h No.187 + x) mol Now. PV = nRT V = nRT P Where. Assuming that the ideal gas law applies.K). P = 99.3896 m3 99.45x10-3 = (0.31451 m3. n = 0.01562 kmol 64 From reaction.52 0.187 mol 92 N2O4 = 2NO2 Let x be mol of N2O4 dissociated as per the reaction.187 . T = 298 K.325 kPa.01562 kmol 1 Now.187+x) x 10-3 kmol.187 – x) + 2x = (0.992 kPa.31451 x 373 Solving we get.3896 = 3.187 . Solution: Basis: 17.kPa/(kmol.31451 x 298 = 0.9 h 0.31451 m3.x) mol Total moles of gas after dissociation = (0.1 (81) When heated to 373 K (1000C) and 95.01562 = 0. R= 8. Molecular weight of N2O4 = 92 ∴ Initial moles of N2O4 = 17. 237 kmol 22.675 x 10-3 mol (H2O + O2) gas mixture = 1.675 x 10-3 Initial mole % of O2 = 100 – 75 = 25 (83) The combustion of 4. What was the initial mole % H2 in the mixture ? Solution: Basis: 40 ml of sample of mixture of H2 and O2 H2 + ½ O2 H2 O PV = nRT n = PV RT Where. we get.188 x 10-4 x = 2y + 4.30 m3 of carbon dioxide gas measured at NTP.73 kg of a coal sample C + O2 = CO2 At NTP.188 x 10-4 = 1. Find the carbon content of the sample.3= 0. T = 291 K.188 x 10-4 From equations (1) and (2).325 x 40x10-6 8.325 8. Solution: Basis: 4.30 m3 At NTP.4 m3. 1 katom C ≡ 1 kmol CO2 12 kg C ≡ 44 kg CO2 .kPa/(kmol. The remaining pure gas was H2.273 x 44 = 10.(82) A 40 ml sample of a mixture of H2 and O2 was placed in a vessel at 291 K (180C) and 101. P = 101. V = 40 ml = 40x 10-6 m3 n = 101.675 x and x = 1. the volume of 1 kmol of CO2 gas is 22.188 x 10-4 mol Let x be the initial moles of H2 and O2 respectively ∴ x + y = 1.31451 m3.675 x 10-6 kmol = 1.R= 8.256x 10-3 mol Initial mole % of H2 = 1.K).325 kPa. ∴ Amount of CO2 yielded =1 x 5.428 kg From reaction.31451 x 291 = 1.675 x 10-3 O2 reacted = y mol H2 reacted = 1 y = 2y mol 0.73 kg of sample of coal yielded 5.256 x 10-3 x 100 = 75 1. A spark was applied so that the formation of water was complete.675 x 10-3 kmol H2 gas remained = 10 ml Moles of H2 remained = 10x10-6 x 101.4 = 0.325 kPa.188 x 10 kmol = 4. 3y + 4.5 Material balance of H2: x moles of H2 = 2y + 4. the volume of CO2 gas = 5.31451 x 291 -7 = 4. 0% Air 0. 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. Solution: Basis: 1 m3 of gas mixture Pure component volume of ethanol vapour in 1 m3 of mixture.92 m3 92% Total 1.53 = 30.08 kmol = 0.53 kg ∴ Amount of mixture = 3.92 x 1 = 0.92 kmol = 0.08 m3 1 Composition by Volume: Ethanol vapour 0.844 kg 44 Carbon content of sample in weight % = 2.08 m3 8.73 (84) At a temperature of 299 K (260C).92 x 28.21 kg Composition by Weight: Ethanol vapour 0. (ii) weight.248 = 2.08 x 1 = 0.0% .13 33. = 1 x 8 = 0. (iii) kg of vapour per m3 of mixture and (iv) kg of vapour per kg of vapour free air.ethanol exerts a vapour pressure of 8 kPa.08 m3 8.84 = 26.844 x 100 = 60.Carbon in the sample = 12 x 10.68 + 26.0 m3 100 Amount of ethanol vapour present in 1 kmol of mixture = 0.08 x 46 = 3.68 kg Amount of air in mixture = 0. K). n = 1 kmol.xi ∴ x1 = 0.6757 mol 74 Total moles of solution = 0.0362 x 0.795 kPa. T = 299 K.9638 Mole of solution = 0.P.7011 mol 0. Assume that the solution of A in ether is very dilute.P. Calculate the molecular weight of A.6757 = 0.0362 Mole fraction of ether in solution = 0.7011 = 0.31451 m3. ∴ V = 1 x 8. of ether in solution = V. P = 100 kPa.0 m3 100 Now.795 = 58.928. R= 8.0362 Mole fraction of A in solution = 0.9638 Mole fraction of ether in solution = 50 = 0.kPa/(kmol.86 Kg of ethanol vapour per kg of vapour free air 3.1387 26.0254 mol Mol of A = gram of A = Molecular weight of A ∴ Molecular weight of A = 3 = 118 0.9638 = 0.254 .68 = 0.148 kg 24. the vapour pressure falls to 56.31451 x 299 = 24. PV = nRT V = nRT P Volume of 1 kmol of mixture = nRT P Where.68= 0.92 m3 92% Total 1.53 (85) The vapour pressure of ether (molecular weight 74) is 58.86 m3 100 Weight of ethanol vapour present per m3 of mixture = 3.Air 0.928 kPa at 293 K (250C).9638 x1 + x2 = 1 x2 = 1-x1 = 1 – 0. Solution: Basis: 3 g of a compound A in 50 g of ether at 293 K V. of pure ether x Mole fraction of ether in solution 56. If 3 g of compound A are introduced and dissolved in 50 g of ether at this temperature. 928 + 101.928 kPa g.342 3.53 = 3. whereas the pressure increases to 5066. The vessel is such that the volume remain effectively constant. T1 P2V2 n2 T2 But V1 = V2 as vessel being closed .256 3.4 kg 4.42 kPa.8 Mole fraction of Steam = xH2O = 1.(86) In the manufacture of formaldehyde.06 = 0.403 3.P = 0.256 x 172.8 kmol Mole fraction of O2 = xO2 = 0.403x 172.4 = 1. methanol and steam are mixed in proportion 1.253= 58.3 = 0.54 = 1. Under these conditions.253 kPa Partial pressure of O2= xO2.3 + 1.342 x 172.253 = 44.91 kPa Partial pressure of steam = 0.83 = 27.83 = 31. ∴ O2 in the mixture = 1.97 kmol 32 Methanol in mixture = 2 x 100 = 41.8 Mole fraction of methanol = xCH3OH =1.875 kPa and 298 K (250C) is heated to 620 K (3470C) in a closed vessel in the presence of catalyst. 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. Calculate the % of NH3 decomposed.5:2:1.253= 69.83 = 41. NH3 is partially decomposed.33 (by weight ) at 283 K (100C).5: 2:1. Calculate the partial pressure of each of the components present in this mixture.53 kmol 18 Moles of gas mixture = 0. The total pressure is 70. the proportion of O2:HCHO: Steam (H2O) is 1.5 x 100 = 31.928 kPa g P = 70.25 kPa.10 kPa Partial pressure of methanol = 0. Solution: Basis: 100 kg of solution In gas mixture.97 + 1.97 = 0.8 Total pressure = 70.3 kmol 32 Steam in mixture = 1.33 by weight.33 x 100 = 27.53 = 0. (87) Ammonia under a pressure of 1519.325 ∴ P = 172.54 kg 4. O2.06 kg 4. 551 = 2.1 = 0.04 y = 0. T1 P2 n2 T2 Where.0392 y = 0.x) kmol N2 produced = 1 .0392 y kmol/h = 0. find (i) Mole % ethylene in gas leaving carbon bed and (ii) Molar flow rate of the feed gas to the carbon bed.551 kmol/h Butenes not adsorbed = 0.21 kmol 620 1519.875 kPa.5 = 0.5 x + 1.96 x 2.00204 kmol/h Ethylene in inlet gas leaving carbon bed = 0. n2 = fional moles of gas mixture = ?.250= 1690.21 x 100 = 60. n1 = initial moles = 100 kmol.875 Total moles after decomposition = 160.5 kmol ∴ Butenes removed = 0.1 kmol/h 5 Let y be the molar flow rate (kmol/h) of gas to the carbon bed Butenes in gas fed = 0.21 100 – x + 0.449 kmol/h Ethylene in gas leaving carbon bed = Ethylene in inlet gas = 2. Solution: Basis: Five hours of operation Amount of butenes removed = 0.x = 0.98 x 0.451 kmol/h . In five hours of continuous operation if quantity of butenes removed is 0.04 y – 0.21 kmol Decomposition of NH3 takes place as : 2 NH3 = N2 + 3 H2 Let x be the kmol NH3 decomposed NH3undecomposed= (100 .449 + 0.250 kPa.5 x kmol 2 Moles of NH3+N2+H2 after decomposition =160. P1 = 1519.1 ∴ y = 2.00204 = 2.04 y kmol/h Butenes adsorbed = 0.449 kmol/h Gas leaving carbon bed = 2. P2 = 5066.P1 =n1. T2 = 620 K ∴ n2 = n1x T1 x P1 T2 P2 = 100 x298 x5066.21 x = 60.5 x kmol 2 H2 produced = 3 .551 kmol/h Molar flow rate of gas to carbon bed = 2.04 x 2. T1 = 298 K.5 kmol. x = 1.1= 0.21 kmol ∴ % decomposition of NH3 = moles of NH3 decomposed x 100 Initial moles NH3 = 60.5 x = 160.551 – 0.21 100 (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. 325 x 10 x 10-3= 4.75 mol Total gas leaving system = 75 + 0.6 – 298 = 27. R= 8.25 mol Amount of CO2 unabsorbed=25-24.6 K (27.97 x 25=24. 100 kPa) Final volume = 10 – 4. (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. n = PV RT = 101.325 kPa pressure and at temperature of 298 K (250C) is compressed to a high pressure so that its volume reduces by 4.451 (89) A sample of gas having volume of 10 l at 101.09 x 10-4 kmol 8. if the pressure rises by 0.325 kPa.325 + 100 = 201. (as pressure increases by 0.16 g . Calculate (i) The volume of gases leaving per 100 litres entering.66 ∴ Volume of 100 mol of entering gas = 2722 l Weight of CO2 absorbed per 100 l gas entering = 24.6 K ∴ Rise in temperature = 325.5 = 5.31451 x 298 Final pressure = 101. P = 201.25 =0..75 mol PV = nRT Where.5 x 10-3 m3 201.09 x 10-4 x 8.5 x 10-3 m3 PV = nRT Where. P = 740 torr = 98.25 x 100 = 0. T = 298 K.kPa/(kmol.e.325 x 5.89 x 44 = 39.5 x 10-3 = 4.6 deg C ) (90) Exhaust gas having 75% N2 and 25% CO2 (by volume) is passed through a absorption column. V = 5.5 l. The gas enters the system at a temperature 323 K (500C) and 740 torr and leaves at 303 K (300C) and 737 torr.31451 x T ∴ T = 325.1 MPa i.1 MPa. 97% CO2 is absorbed in KOH solution. V = 10 l = 10 x 10-3 m3.31451 m3.5 l = 5.75 =75.31451 x 323 = 2722 l 98.449 x 100 = 99.66 kPa ∴ V = 100 x 8.Mole% ethylene in gas leaving carbon bed = 2.325 kPa.89 mol 2722 = 0.325 kPa.K). P=201. what will be the rise in temperature? Solution: : Basis: 10 l of gas at 298 K PV = nRT Where.92% 2. 7% C3H8.C3H6.xC3H6 + M.84 . 8% C3H6 and 5% C4H10 by volume.C2H6.xCH4 + M.31451 x 303 = 1942 l 98.54 100 OR Mavg = M.84 Solution: Basis: 100 kmol of cracked gas. 10 kmol C2H6.25 + 44 x 0. 1% C2H6.C2H4. (ii) the composition by weight.Volume of gas leaving based on 100 mol gas entering is P’ = 75.CH4. and C4H10 = 58 Weight of methane = 45 x 16 = 720 kg In the same way. C2H6= 30.1 + 28 x 0.xC2H4 + M. 25 kmol C2H4. calculate the weight of each components of cracked gas Composition of refinery gas: Average molecular weight of refinery gsa is Mavg = 2854 = 26.43 l 2722 V’ = n’ RT’ (91) Cracked gas form a petroleum refinery contains 45% CH4. 25% C2H4.26 Volume of gas leaving per 100 l gas entering = 1942x 100 = 71.07 + 42 x 0.xC2H6 + M. C3H6= 42. 7 kmol C3H8. Calculate (i) the average molecular weight of gas mixture.45 + 30 x 0.xC4H10 = 16 x 0.C4H10. and 5 kmol C4H10 Molecular weight data: CH4= 16.54 Specific gravity of gas mixture = 26.08 + 58 x 0.54 = 0.xC3H8 + M. It contains 45 kmol CH4. and (iii) the specific gravity of the mixture taking average molecular weight of air as 28.05 = 26. C2H4 = 28.C3H8.92 28.75 x 8. Component CH4 C 2 H6 C 2 H4 C 3 H8 C 3 H6 C4H10 Total Kmol 45 10 25 7 8 5 100 Mol.93 100 .Wt 16 30 28 44 42 58 - kg 720 300 700 308 336 290 2654 Weight% 27.37 11.13 11.61 12.66 10.30 26. 012 100 and Mole fraction of NH3 = xNH3 = 9.913 kg/m3 8.56 = 2993. 7. Molecular weight of O2 = 32.5 kmol of N2.31451 m3. Molecular weight of H2O = 18 and Molecular weight of NH3 = 17 Average molecular weight of gas mixture is Mavg= ∑Mixi = 28 x 0. Mole fraction of N2 = xN2 = 70.095 = 27.2 kmol H2O and 9. Amount of CO2 = 10 x 44 = 440 kg Amount of O2 = 7.5%.5%.705 100 Mole fraction of O2 = xO2 = 18.96%. N2 = 82% and SO2 = 0.8 kmol O2.56 kg Amount of flue gas = 440 + 254. O2= 7.96 kmol O2.25 .705 + 32 x0.31451 x 923 (93) The Orsat (dry) analysis of a flue gas from a boiler house is as given below: CO2 = 10%.325 x 27.96 x 32 = 254.188 100 Mole fraction of H2O = xH2O = 1. Sol: Basis: 100 kmol of a flue gas It contains 10 kmol of CO2.709 MPa and 923 K (6500C) is as follows: N2 = 70.325 kPa.72 kg Amount of N2 = 82 x 28 = 2296 kg Amount of SO2 = 0.188 + 18 x 0.kPa/(kmol.K) Density of gas mixture is given by ρ= PMavg RT = 810. 18.04 kmol SO2. The flue gas pressure is 100 kPa (750 torr) and temperature is 463 K (1900C). Calculate the density of gas mixture using ideal gas law Sol: Basis: 100 kmol of gas mixture at 923 K It contains 70. R= 8. 82 kmol N2.5 = 0.72 + 2296 + 2.095 100 Molecular weight of N2 = 28. 1. T = 923 K.56 x106 2993.8%.012 + 17 x 0.2% and NH3 = 9.(92) The composition of gas mixture in manufacture of nitric acid at a pressure of 0.04% by volume. O2 = 18.28 = 855.709 MPa g = 810.59 = 2.8 = 0.2 = 0. SO2 is undesirable from the point of view of occupational hazards (environmental pollution). and 0.5 = 0.5 kmol NH3.28 kg Concentration of SO2 in ppm= 2.04 x 64 = 2. Express the concentration of SO2 in ppm and mg/m3.59 P = 0. H2O = 1. Narayanan and B. A. New Delhi.67 M 3 Reference Books 1) B.Lakshmikutty. what will be the normality and molarity of the solution? Sol: Basis: 630 g of oxalic acid Oxalic acid = C2H2O4. Introduction to Process Calculations (Stoichiometry).62 m3 100 Concentration of SO2 in mg/m3 = 2. 4th ed. 2) David Mautner Himmelblau. A. Pune. Ragatz. CBS Publishers and Distributors. Bhatt and S. 15th ed. New Delhi.56 x 103x103= 665 3849. 2003.62 Molecular weight of steam (water) = v = V M 3 = 0. New Delhi. Hougen. 3) O.0288 m3/kg 18 kg/kmol) (94) 630 grams of oxalic acid dihydrate were dissolved in water.V.Vora. Chemical Process Principles Part–I. Prentice Hall.Volume of flue gas. 2002. M.31451 x 463 = 3849. Material and Energy Balances. 2nd ed. M. McGraw Hill.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. 2006. Nirali Prakashan.. Waston and R. 5) K. Basic Principles and Calculations in Chemical Engineering. A. 4) K. V = nRT P = 100 x 8.33 N 3 Molarity = 5 = 1. 1995. 6th ed.518 (m /kmol) = 0. Total volume of solution being 3000 ml. K. 1st ed. Gavhane. Prentice Hall of India. New Delhi. . Stoichiometry. Stoichiometry and Process Calculations. 2004. I.