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  • 1. ADDITIONAL PROBLEMS ON SOLUTIONS SOLUTIONS Some more Additional Problems on Solutions (Solved) 1. What is the molarity of an aqueous solution of ethyl alcohol (CH3 CH2 OH, molar mass = 46.0 g/mol) which contains 50.0 g of alcohol in 500 mL of the solution? Solution. 50 1000 =100g Mass of ethyl alcohol/ litre = 500 Mass in g / litre 100 = =2.18 No. of moles of alcohol/ litre = = Molar mass in g /mol 46 Thus, the solution contains 2.18 moles of ethyl alcohol per litre of the solution. So, the molarity of the solution is 2.18 mol / L. 2. Calculate the normality· and molarity of H2S04 in a solution containing 9.8 g of H2SO4 per dm3 of the solution. Solution: Therefore, Hence, Mass of H2S04 per litre = 9.8 g Molar mass of H2SO4 = 2 x (1) + 32 + (4 x 16) = 2 + 32 + 64 = 98 g mol-1 98 Number of moles of H2S04 per litre of solution = =0.1 98 Molarity of H2S04 in solution= 0.1 moll 1-1 As there are two equivalents per mole of H2 SO4, therefore Normality of H2S04'in solution= 0.1 x 2 = 0.2.equiv L-1 3. 2.82 g of glucose (molar mass180 g mol-1) are dissolved in 30 g of water Calculate. (i) molality of the solution (ii) molefractions of, (a) glucose (b) water. Solution. Mass of glucose, w8= 2:82 g Molar mass of glucose, Mg = 180 g mot-1 Mass of water, ww = 30 g BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 1
  • 2. ADDITIONAL PROBLEMS ON SOLUTIONS Molar mass of water, Mg = 18 g mol-1 (i) From the definition, molality is the number of moles per 1000 g of the solvent. Then, 2.28 Number of moles of glucose m 30 g water = mol = 0.0.157 mol 180 0.0157 1000 =. 0.522' mol kg-1 So, Molality of glucose = 30 (ii) From above, the number of moles of glucose (ng) and. water (nw) in the solution are given by, Mass of glucose 282 n g = Molar mass of glucose Mass of water 180 = mol =· 0 0157 mol 30 = mol = 1.667mol nw  Molar mass of water 18 ng Therefore, Molefraction of glucose = ng + n w 0.0157 = 0.0157+1.667 ng + n w 0.01 = 0.99 1.667 nw Molefraction of water = = = 1.667+0.0157 4. What is the molality of the aqueous solution of methyl alcohol (CH3 OH. molar mass 32.0 g/mol) which contains 64 g of methyl alcohol in 200g of water? Solution: Mass of methyl alcohol in 200 g of water = 64 g 64 1000 Then, =320g 1 Mass of methyl alcohol in 200 320g So, No. of moles of alcohol per 1000 g of water 32g/ mol = 10 mol Thus, I kg of the solution contains 10 mole of methyl1 alcohol. So, the molality of the solution is 10 mol /kg. BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 2
  • 3. ADDITIONAL PROBLEMS ON SOLUTIONS 5. An agueous of solution of a. dibasic acid (molar mass=118 g/mol) containing 17.7 g of the acid per litre of solution has a density, I. 0077 g/mL. Express the concentration of the solution in as many as you can Solution. Mass of solute, Molar mass" of the solute, w M = 17.7 g/L = 118 g mol-1 Mass of solute (g/L) So 'No ·of moles of solute dissolved = w = -1 Molar mass (g mpl ) M mol/L 17.7 Then Mass of the solvent, (water) So, Volume of the solution Density of solution Mass of solvent (water) Then, Maas of the solvent, (water) N o. of moles of solvent Total number of moles in solution = Mol /L = 0.15 mol /L 118 = 1000 mL · = 1.0077 g / mL = Volume x Density = 1000 mL x 1.0077g/mL = 1007.7 g = Mass of solution - Mass of solute = (1007.7- 17.7 g) = 990 g Moles of solute 990 = = =55.0 Molar mass in g / mol 18 = Moles of solute + Moles of solvent = 0.15 +55 = 55.15 Since, the acid is dibasic, hence its Eq. mass = Molar mass/ 2 =59 g equiv -1 From the above results the concentration of solution can be calculated in various units as follows: (i) Molarity: Already calculated above : Molarity = 0.15 mol/L (ii) Molality: From above, 1000g 1 Mass of solute (acid) per 1000 g of solvent =17.7 =0.151mol / kg 990 g 118 Mass in g per 1000 g solvent No. of moles of the solute/ 1000 g of solvent = 1000 1 = 0.151 mol / kg =17.7 990 So, Molar mass (g/ mol) 118 Molality = 0.151 mol/ kg BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 3
  • 4. ADDITIONAL PROBLEMS ON SOLUTIONS n (iii) Mole- fraction (X) = n acid 0.15 n acid+ water = 0.15 = 0.15+55 =0.027 55.15 6. How much sodium chloride be dissolved to make l litre of 0.1 F solution? Solution:. Sodium chloride is an ionic .compound and is represented by the formula NaCl. its gram-formula mass is, 23 g + 35.5 g = 58.5, g Therefore, So, 1 litre of 1 F NaCl solution, one requires = 58.5 g, of NaCl 1 litre of 0.1 F NaCI solution, one. requires = 58.5 x O.1g = 5.85 g Thus, 5.85 g of NaCl should be dissolved to make up 1 litre of solution. 7. How many grams of KCI would be required to prepare 1 litre of 0.1 M solution? Atomic masses are; K= 39 u, Cl-= 35.5 u Solution: Potassium chloride (KCl) is an ionic compound with a molar mass of 39 + 35.5 = 74.5 g mol-1 Therefore, For 1 litre of 1 M KCl solution, one .requires 74.5 g of KCl 74.5 0.1 For 1 litre of 0.1 M KCl solution, one requires = 7.45g 1 Thus, one requires 7.45 g of KCl to prepare 1 litre of its 0.1 M solution 8. 2.46 g of sodium hydroxide (molar mass = 40 g/mol) are dissolved in water and the solution is made to .100 mL in a volumetric flask. Calculate the molarity of the solution. Solution: Mass of sodium hydroxide = 2.4'6 g Molar mass of sodium hydroxide ·= 40 g moll-1 Volume of the · solution = 100 mL Molarity of the solution =? Molarity of a solution is given by the number of moles of solute present in 1000 mL of the solution, i.e., No. of moles of solute Molarity = Volume of the solution in mL x 1000 ml L-1 BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 4
  • 5. ADDITIONAL PROBLEMS ON SOLUTIONS (Mass of the solute I Molar mass of the solute) = x 1000 mol L -1 Volume of the solution in mL 2.46/ 40) 1000 = 2.46 molL-1= 100 -1 10molL- = 0.615molL 40 So, the molality of the solution is 0615 mol L-1 9. Calculate the molality of a 1 M solution of sodium nitrate. The density of the solution is, the solution is 1.25 g cm-3 (NaN03) = (23 + 14 + 48) g/mol = 8S Solution: Molar mass of sodium nitrate, g/mol Mass of 1dm3 (or 1 litre) of the solution = Volume x Density Therefore, = 1000 cm3 x1.25 g/cm3 = 1250 g Therefore, Mass of water containing 8.5 g .of NaN03 = (1250- 85) g = 1165 g = 1.165 kg So, 1mol -1 Molality (m) of the solution. = =0.86 mol kg 1.165kg 10. 0.75.g of sodium bicarbonate (NaHC03) are dissolved in250 ml of a solution. Calculate its, (i) normality, (ii) molarity Solution. Mass of sodium bicarbonate dissolved = 0.75g Volume of solution = 250 mL Molar mass of sodium bicarbonate . = (23+ 1 + 12+~8) g mol-1 = 0.615 L 1 g mot' Since, one molecule of NaHCO3 contains only one cationic charge (on Na+), hence Molar mass Equivalent mass, (E)= 84 = 1 g equiv -1 1 As per definition, (Mass of NaHCO 3 / Molar mass of NaHC0 3 ) Molarity (M) = Volume of solution in mL -1 0.75g/84 mol 0.75/84mol 1000mL = 250/1000L -1 250 mL x 1000 -1 = 0.0357molL BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 5
  • 6. ADDITIONAL PROBLEMS ON SOLUTIONS (Mass of NaHCO 3 / Equiv. mass of NaHC0 3 ) Normality (N) = 1000 ML / L Volume of solution in mL 11. Calculate the molefraction of water in a mixture of 12 g water .108 g acetic acid and 92 g ethyl alcohol. Solution: Following the procedure of the previous problem, one can write Mass of water 12 = =0.67 n(H2O) = Molar mass 18 Mass of ethanol 92 = = 2.00 n(C2H5OH) = Molar mass 46 Mass of acetic acid 108 n(CH3COOH) = = Molar mass =1.80 60 So, Total number of moles in the solution, ntotat = 0.67 + 2.00 + 1.80 = 4.47 Therefore,  X water =  0.67 =   n total =0.15 4.47 n total nC2H5 OH X ethanol = Xacetic acid =  nH2 O 2.00 = =0.45 4.47  nCH3 cooH n total 1.80 = =0.40 4.47 12. What is the molality of ammonia in a solution containing 0.85 g NH3 in a 100 cm3of a liquid of density 0.85 g cm-3? Mass of ammonia w2 = 0.85 g Molar mass of ammonia, M2=17 g mol-1 Mass of liquid (solvent), Solution: w1= 100 cm3 0.85g cm-3=0.85 g BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 6
  • 7. ADDITIONAL PROBLEMS ON SOLUTIONS So, Molality of ammonia n NH w /M -1 3 = 2 2 mol kg = Mass of solvent in kg 85/1000 or, Molality of ammonia = 0.85/17 mol kg-1 = 0.59 mol kg-1 85/1000 13. An aqueous solution of a dibasic acid (C2H2O4.2H2O) contains 1.26 g of the solute per litre of the solution Calculate the normality end molarity of the solution Solution: Mass of the acid W=1.26 g/L = Molar mass M of dibasic acid 2 12 + 2 1 + 4 16 + 2 2 + 16 -1 = 24 + 2 + 64 + 36 = 126 gmol So, Equivalent mass of the acid = Molar mass 2 = 126 2 = 63 g equiv -1 According to the definition, Molarity = and, Normality Mass of solute in g per L Molar mass = = 1.26 g / L 126 g/mol Mass of solute in g per L Equivalent mass = -1 = 0.01mol L 1.26 g / L 63 g equiv -1 -1 = 0.01mol L 14. What volume of 95% sulphuric acid (density=1.85 g/cm3) and what mass of water must to taken to prepare 100 cm3 of 15% solution of sulphuric acid (density) =1.10 g /cm3)? Solution: Volume of the solution Density of the solution Therefore = 100 cm2 =1.10 g /cm3 Mass of 100 cm3 of solution = 100 x 1.10 g =110 g The given solutions is 15%. This means that 100 g of solution contains 15 g of H2 SO4 BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 7
  • 8. ADDITIONAL PROBLEMS ON SOLUTIONS Then, mass H2SO4 in 110 g (=100 cm3) of solution = 15g 100g 110g = 16.5g and Mass of water in 110 g (=100cm3 ) of solution = =(110-16.5) g =93.5 g So, to obtain 100 cm3 of 15 % solution acid, we require Mass of water = 93.5 g Mass of H2SO4 (100% pure) =16.5 g Since, the given sulphuric acid is 95% pure, hence Mass of H2SO4 (95%) required = Density of 95% H2SO4 100 16.5g 95 = 17.37g =1.85 cm-3 15. Sea water contains 6 x 10-3 g of dissolved oxygen in one litre Express the concentration of oxygen in sea water per million (ppm) units Density of seawater =1.03 g /mL. Solution: Mass of 1 littre of sea water = 1000 mL x 1.03 g /mL =1.030 g Mass of dissolved oxygen in sea water per litre = 6 x 10-3 g 6 10-3 So, Concentration of dissolved oxygen = 106 ppm = 5.8 ppm 1030 g 16. Calculate the molality of a salt (molar mass = 138 g mol-1) solution obtained by dissolving 2.5 g in one litre of the solution. Density of solution is 0.85 g cm-3 Solution: Mass of the salt dissolved =2.5 g Volume of the solution =1000 cm3 Density of the solution =0.85 g cm3 So, Mass of 1000 cm3 of solution Therefore Thus =Volume x Density= (1000 x 0.85) g= 850 g Mass of solvent = (850-2.5) = 847.5 g 847.5 g of solvent contains 2.5 g of solute BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 8
  • 9. ADDITIONAL PROBLEMS ON SOLUTIONS Or, 2.5 847.5 g of solvent contains 1 g solvent contains = 1.38 2.5 1 1.38 847.5 1000 g of solvent contains = mol of solute mol 1000 2.5 8475 138 = 0.0214 mol Therefore, molality of the solutions is 0.0214 mol kg-1 17. Calculate the mass of the solute present in the following solutions: (a) l L of N/10 Na2CO3 solution (b) 2 L of N /10 HCI solution (c) 100 mL of 0.5 M H2SO4 (d) 250 mL of N/10 oxalic acid Solution. (a) Preparation of 1 L of N/ 10 solutions: Molar mass of Na2 CO3= (2 23) + 12 + (3 16) = 106g mol -1 We know, that the gram equivalent mass of Na2Co3 is half of the molar mass Hence, Equivalent mass of Na2 CO3= = 106 = 53 g equiv -1 2 Thus 1 L of 1 N Na2CO3 solution should contain 53 g of Na2 CO3 Then, 1 L of N/ 10 Na2CO3 solution would contain 53 10 g of Na2 Co3 = 5.3 g Na2CO3= 5.3 g Na2CO3 (b) Preparation of 2 L of N/ 10 HCI solution: Equivalent mass of HCI = (1+35.5) = 36.5 g equiv-1 So, 1 L of 1 N HCl solution should contain 36.5 g of HCI. Then, BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 9
  • 10. ADDITIONAL PROBLEMS ON SOLUTIONS 2. L of N /10 HCl solution would contain = 36.5 2 10 73 g of HCl = 10 g HCl =7.3 g HCl (c) Preparation of 100 mL of M H2 SO4 Molar mass of H2 SO4 contains 98 g H2 SO4 Or 1000 mL of 1 M H2SO4 contain 98 g H2SO4 Then, 100 mL of 0.5 M H2SO4 would contain = 98 1000 100 0.5 1 g of H2SO4 = 4.9 g of H2SO4 (d)Preparation of 250 mL of N/10 oxalic acids: Oxalic acid crystals correspond to the formula, (COOH)2 2 H2O Molar mass of oxalic acid crystals = (2 12+4 16 +2 1) +2 (2 +1 +16) =(24 +64 + 2 +36) =126 g mol-1 Basicity of oxalic acid =2 So, Equivalent mass of oxalic acid = Then, 126 = 63 g equiv -1 2 1 L of 1 N solution of oxalic acid contains 63 g oxalic acid 1000 mL of 1 N 250 mL of N 10 “ “ “ “ 63 g oxalic acid ” “ “ “ = = 63 1000 63 1000 250 1 10 g oxalic acid = 1.575 g oxalic acid 18. Find the molarity and molality of a 15 % solution of H2SO4 (density of H2SO4 solution =1.10 g/ cm3 Molar mass of H2SO4 98 g mol-1) Solution: 15% Solution of H2SO4 means that 100 g of solution should contain 15 g of H2SO4 So. Mass of H2SO4 dissolved =15 g BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 10
  • 11. ADDITIONAL PROBLEMS ON SOLUTIONS Mass of solution Density of the solution =1.10 g cm-3 Molar mass of the H2SO4 So, =100 g =98 g mol -1 Mass of water containing 15 g of H2SO4 = (100 – 15 )g =85 g No. of moles of H2SO in 15 g of H2SO = and, Volume of 100 g of solution = 15 g 98 g/ mol mass Density = 15 98 mol = 0.153 mol 100g = 1.1g cm -3 = 90.9 cm 3 90.9 dm3 1000 (i) Calculation of molarity No. of moles of H2SO4 Molarity of H2SO4 solution = Volume of solution in dm 0.153 mol = 3 (90.9 / 1000) bm 3 =1.68 mol dm -3 So, thee molarity of H2SO4 solution is 1.68 (ii) Calculation of molality Molarity of H2SO4 solution = No. of moles of H2SO4 Molality = = Mass of water in g 0.153 85 1000mol kg -1 1000 = 1.8 mol kg -1 19. Urea forms an ideal solution in water Determine the vapour pressure of an aqueous solution containing 10 per cent by mass urea at 40oC (Vapour pressure of water of 40oC =55.3 Hg) Solution: Concentration of urea =10% Let Mass of the solution =100g Then Mass of urea =10 g Therefore, Mass of water =(100-10)g = 90g Molar mass of urea (NH2CONH2) =60g mol-1 BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 11
  • 12. ADDITIONAL PROBLEMS ON SOLUTIONS Then, Mole- fraction of urea, Xurea = Xurea = Then, from Raoult’s law, nurea nurea + nwater 0.1667 5.1667 = 10/60 (10/60 + 90/18 = 1.1667 0.1667 + 5 = 0.032 po - p o = Xurea po 55.3 - Ps = 0.032 55.3 So, the vapour of urea solution at 40o C is 53.5 mm Hg. 20. The vapour pressure of water at a creation temperature is 18.15 torr .and that of a solution containing 9.47 g of sugar, at the same temperature , is 18.06 torr Calculate the molar mass of the sugar. PoA=18.15 torr Solution Given Vapour pressure of water (solvent) Vapour pressure of solution, PA=18.06 torr Mass of sugar (solute) W B = 9.47 g Molar of water (solvent) WA=100g Molar mass of water (solvent) MA=18 g /mol Molar mass of sugar(solute), MB = ? From Eq (3.11), MB = WB MA WA PA 9.47 18 = o P A -P A 100  18.03 18.15-18.06 WB MA Form Eq. (3.13) M = B WA o P A o P A -P A  g / mol =342 g/ mol 9.47 18 = 100   18.03   18.15 -18.06   BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 12
  • 13. ADDITIONAL PROBLEMS ON SOLUTIONS 21 The vapour pressure of a 5% aqueous solution of a non – volatice organic substance at 373 K is 745 mm Hg. Calculate the molar mass of the substance. Solution. From the given data, for 100 g solution, Mass of Solute, W2 = 5 g Mass of solvent (water) W 1=(100-5) g= 95 g=0.095 kg Vapour pressure of solution, Ps =745 mm Hg Vapour pressure of pure solution.(water) at373 K, Po s=760 mm Hg M2=? Molar mass of solution, M1 =18 g/ mol Molar mass of solvent (water), P Form Raoult,s law, - Ps n2 W2 / M2 s = = o n1 +n 2 W 1/ M1+ W /M P s 2 2   760- 745  5/M2 = o o 5/M2 = P s 760 95 /18+ 5/M 2   5.278 +5/M 2  The above equation may be rewritten as  760 5.278 +5 / M2 = 15 5/M 2    5.278 M = 2 +1=1.0556M2 +1 5 50.67 =1.0556 M2 + 1 -1 50.67-1 or M = 2 1.0556 1.0556 = 47g mol 22. Calculate the boiling point of a one molar aqueous solution (densitry: 1.03 g mL-1) of sodium chloride. Kb for water = 0.52 K kg mor-1Atomic mass. Na = 23. CI= 35.5. BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 13
  • 14. ADDITIONAL PROBLEMS ON SOLUTIONS So, = 1 molar = 1 mol L-1 Conc. of the solution Density of solution = 1.03 · g mL-1 Molar mass of NaCl = (23 + 35.5) g/mol = 58.5 g/ mol-1 Mass of 1 litre of solution = 1000 x 1.03 =1030 g Therfore, Mass of water containing 1 mole of NaCI =(1030 - 58.5)g = 971.5 g Solution. 1 1000 mol/kg = Thus, Molality, of the solution, m = Then, So, 1.0293 mol/kg ΔTb= I Kb m = (2 0.52 1.0293) K= 1.07 K Boiling point of solution = (373.15 +1.07) K = 374.22 K 971.5 23. An aqueous solution containing 2.4 g of a substance per 100 g solvent shows an elevation of boiling point of 0.21 K what is the molar mass of the substance? Kb for water (solvent) is 0.52 k kg mol-1 Mass of solute. w2 =2.4 g Mass of solvent, Solution: w1=100g =0.1 kg ΔTb=0.21 K Molar mass of the solute=? (M2) 0.52 2.4 Kb w2 M2= we know, w1 Tb = 0.1 0.21 g/mol = 59.4 g/ mol 24. A solution contains 3.5 g of a non-volatile solute in 125 g of water; and it boils at 373.52 K. Calculate the molar mass of the solute. (Kb for water= 0.52 K/m) Solution· Let, w2 = 3:5 g Mass· of the solute, Mass of water, w1 =125 g = 0.125 kg Elevation of boiling point, Δ Tb = (373.52- 373.0) K = 0.52 K. Molar mass of the solute = M (?) We know that, BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 14
  • 15. ADDITIONAL PROBLEMS ON SOLUTIONS Tb = K b n2  Kb w 2 = w1  / M 0.52 kg mol -1 3.5g M 0.125 kg w1 -1 0.52 3.5 K kg mol g 0.52 K= M 0.125 kg -1 0.52 3.5 K kg mol g M= 3.5 -1 -1 g mol = 28 g mol = 0.52 K 0.125 kg 0.125 25. The boiling point of water becomes 100.52 C, if 1.5 g of non-volatile solute is dissolved in 100 mL of it. Calculate the molar mass of the solute. Kb for water = 0.6 k/m. Solution. Mass of solute = 1.5 g Volume of water (solvent) = 100 mL Taking density of water as 1 g/mL, Mass of water (solvent) = 100 mL x 1 g/mL = 100 g = 0.1 kg· ΔTb = (100.52 - 100) C = 0.52 C or Kb w2 w1 Kb w 2 We know, M ΔTb = M= 0.6 1.5 Tb w1 0.52 0.1 g/ mol = 17.3 g/ mol 26. A solution of 3.795 g sulphur in 100 g carbon disulphide (boiling point, 46.30 C) boils at 46.66 C what is the formula of sulphur molecule in the solution? Kb for carbon disulphide is 2.42 K kg mol-1 Solution. Mass of sulphur (solute) = 3.795 g Mass of carbon disulphide (solvent) = 100 g = 0.1 kg Molar mass of sulphur = M ΔTb = (46.66 - 46.30) C = 0.36 C We know, M= Kbw2 w1 Tb 2.42 3.795 = g/mol =255.1g/mol 0.1 0.36 BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 15
  • 16. ADDITIONAL PROBLEMS ON SOLUTIONS Atomic mass of sulphur = 32 g/mol So, 255.1 No. of sulphur atoms in one molecule = =7.97 =8 32 So, sulphur exists as S8 in the solution. 27. A solution .of urea in water freezes at 0.400 C. What will be the boiling of the same solution if the depression and elevation constants for water are 1.86 deg kg mol-1 and 0.5 12 deg kg mol-1 respectively? Solution: We have the relationships, K b .n 2 K f .n 2 Tb ….(i) and Tf …(ii) w1 w1 where, n2 is the number of moles of the solute, and w1 is the mass of solvent in kg. Dividing Eq. (i) by Eq. (ii) K T b b = T f or b= T Then, K Tf K b f 0.4 0.512 = o o C = 0.11 C 1.86 Kf Boiling point of the urea solution = 100o C + 0.11oC = 100. 11oC 28 .The normal freezing point of nitrobenzene, C6H5 NO2 is 248.82 K .A 0.25 molal solution of certain in nitrobenzene causes a freezing point depression of 2 degree. Calculate the value of Kf for nitrobenzene. Solution: Normal freezing point of nitrobenzene =278.82 K Concentration of solution =0.25 molal =0.25 mol/kg Freezing point depression Tf = 2K K f ? We know that, Tf =K f m BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 16
  • 17. ADDITIONAL PROBLEMS ON SOLUTIONS Tf -1 2K = Kf m =8 K kg mol 0.25 mol/kg 29. Find the (i) boiling point, and (ii) freezing point of a solution containing 0.520 g glucose (C6H12 O6) dissolved in 80.2 g of water for water. Kf =1.86 K/m, and Kb =0.52 K/m. = 100oC = 0.0oC = 0.52 g =180 mol 80.2 Mass of water =80.2 g = kg 1000 K n b 2 We know = T where, n2 is no. of moles of solute (w2/ w b M2) w1 is the mass of solvent in kg 1 Solution: Normal boiling point of water Normal freezing point of water Mass of glucose Molar mass of glucose (C6H12O6  = Tb 0.52 0.52 0.52 (w 2 /M2 ) = K = 0.019K = 0.02K w1 180 (80.2/1000) So, Boiling point of solution = (373+ 00.2) K =0.067 K K (w /M ) 1.86 0.52 2 2 Similarly  = f T = 0.067 K 180 (80.2/1000) w f 1 So, Freezing point of solution = (273 K – 0.067 K) 30. Ethylene glycol (HoH2C- CH2OH) is used as an antifreeze for water to be used in car radiators in cold places .How much ethylene glycol should be added to 1 kg of water to prevent it from freezing at -10o? Molal depression constant of water is 1.86 k kg mol-1 Solution: Mass of water (solvent) w1=1 kg o Tf =10 C K f =1.86 K kg mol -1 w2=? Mass of ethylene glycol required, K f (w 2 / 62) We know, T f = w1 BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 17
  • 18. ADDITIONAL PROBLEMS ON SOLUTIONS 1.86 w 2 10 = 62 1 10 62 1 This gives w = g 1.86 2 = 333.3g 31 A solution containing 4 g of a non- volatile organic solute per 100 cm3 was found to have an osmotic pressure equal to 500 cm Hg. at 27oC. Calculate the molar mass of the solute. Solution: Osmotic pressure = (5000/ 76) atm Temperature, Mass T = (27 +273) =300 K of solute, Volume W= 4 g of solution, V=100 cm3 atm deg-1 mol-1 R = 0.0821 dm3atm deg-1 mol-1 w pv = We know, RT M w This gives, M= V 4 0.0821 300 RT 500 / 76 g/mol = 149.7g/mol 0.1 32. 5 g of a non – volatile non- electrolyte solute is dissolved in water and the solution was made up to 250 cm3. The solution exerted an osmotic pressure equal to 4X105 N M-2 at 298 K Find the molar mass of the solute Solution: Mass of solute, Volume of solution, Osmotic pressure Temperature, Molar mass of solute w=5g = 0.005 kg 3 V= 250 cm =250 10-6 m3 =2.5 10-4 m3 =4 105N m-2 T=298 K M =? R= 8.314 j k-1 mol-1 wRT We know, or M = V 0.005 kg 8.314 JK = 5 -2 4 10 N m -1 m,ol -1 2.5 10 -4 298 K m 3 M=0.1239 kg mol-1 = 123.9 g mol-1 BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 18
  • 19. ADDITIONAL PROBLEMS ON SOLUTIONS 33. Osmotic pressure of a solution containing 7.0 g protein in 100 m L of solution is 20 mm Hg at 37oC Calculate the molecular mass of the protein .(R = 0.0821 litre atmosphere deg-1 mol-1 ) Solution: Mass of protein, w=7.0 g Volume of the solution, V = 100 mL = 0.1 L Molecular mass of the protein = M Temperature, T= 37o = (37+273) k=310 K 20 Osmotic pressure, =20 mm Hg 760 atm we have, wRT M= 7.0 0.0821 310 = 7.0 0.0821 310 760 = =67700 g / mol (20/760) 0.1 pV 20 0.1 34. At 298 K, 100 cm2 of a solution containing 3.02 g of an unidentified solute exhibits an osmotic pressure of 2.55 atmosphere. What is molecular mass of solute? (R=0.0821 Ll atm mol-1 K-1) Solution: Given: T= 298 K Volume of the solution Mass of Solute, Osmotic pressure, V=100 cm3 =0.1 L W= 3.02 g =2.55 atm w The osmotic pressure is given by, V=n RT = wRT So. M = pV M RT 3.02 0.0821 298 = 2.55 0.1 = 67700 g / mol 35. A 5% solution of cane sugar C12 H22 O11 is isotonic with 0.8 77 % of solute A Calculate the relative molar mass of A. Assume density of solutions to be 1 g cm-3 Solution: Conc. of sugar solution =5% Mass of solution =100g So, Mass of sugar in solution = 5 g Relative molar of sugar = (12 12+22+11 16) =144+22+176=342 We know, nv = nRT BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 19
  • 20. ADDITIONAL PROBLEMS ON SOLUTIONS or, n = w RT RT= V M V 5 So, and, 342 Sugar = RT V A= 0.877 RT M V Since, the two solutions are isotonic, hence, 0.877 M or, Sugar = RT So, 5 V 0.877 342 M= A, = RT 342 V = 60g/mol 5 36. Calculate the molecular mass of a substance 1.0 g of which on being dissolved in 100 g of solvent gave an elevation of 0.307 K in the boiling point. ( Molal elevation constant, Kb =1.84 K/m). K b n2 Solution: We know, Tb = where n2 is the no. of moles of solute; w1 is w1 the mass of the solvent in kg. Mass of solute 1.0 g = n2 = Molar mass of the solute M and w1 =100 g =0.1kg Substituting these values, one can write, 1.84K kg mol 0.307K = -1 0.1Kg 1.84K kg mol -1 1.0 g M 1.0g 1.84 1.0 So, M = g/ mol= 59.9 g/ mol 0.1Kg 0.307 K 0.1 0.307 37. The density of water is room temperature is 0997 g cm3 Calculate the molarity of pure water. Solution: The density of water is 0.997 g cm3. Thus, BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 20
  • 21. ADDITIONAL PROBLEMS ON SOLUTIONS Mass of 1 cm3 of water =0.997 g Therefore, Mass of 1 dm3 =1000 cm3 of water =0.9997 g cm-3 1000 cm3= 997 g Molar mass of water= 2+16=18 g mol-1 No. of moles of water per dm3  55.39 Therefore, Therefore, the molarity of water is 55.39 mol L-1 38. A solution is 25 percent water 25 percent ethanol and 50 percent acetic acid by mass. Calculate the mole- fraction of each component. Solution: Let us consider 100 g of the solution .Then, mw=25 g Mass of water, Mass of ethanol, m eth =25g Mass of acetic acid macet=50 g Molar mass of water, ethanol and acid are, 18 g/ mol, 46 g / mol, and 60 g/ mol respectively Then, 25g No. of moles of water, nw= = 1.39 18 g/mol 25g No. of moles of ethanol, n eth= No. of moles of acetic acid n acet= Total no of moles in solution 60g/mol =(1.39 +0.54+0.83) =2.76 46g/mol 50g = 0.54 =0.83 1.39 So, Mole –fraction of water, Mole- fraction of ethanol, Mole- fraction of acetic acid Xw= 2.76 = 0.50 0.54 =0.20 2.76 0.83 =0.30 Xacet = 2.76 Xeth = 39. Concentrated sulphuric acid has a density of 1.9 g/ ML and is 99% H2SO4by weight Calculate the molarity of H2SO4 in this acid. Solution: Let us consider 1 litre of H2SO4Then Mass of 1 litre of this acid =1000 mL 1.9 g/mL =1900 g Since, the given acid is 99% pure, hence BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 21
  • 22. ADDITIONAL PROBLEMS ON SOLUTIONS 1900 99 Actual mass of H2SO4 in 1 litre sample = = 1881g 100 Molar mass of H2SO4=(2+32+64)g mol =98 g mol-1 -1 1881g So, No of moles of H2SO4 in 1 litre of the sample = 98 g mol = 19.19 mol -1 So, the given H2 SO4 sample is 19.9 molar. 40. Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid Calculate the volume of the solution which contained 23 g of HNO3. Density of the conc. HNO3 Solution is 1.41 g cm-3. Solution: Let, Mass of conc. HNO3 sample =100g So, and Mass of HNO3 in 100 g of sample = 69 g Mass of HNO3 in 100 g of sample = 31 g Density of conc. HNO3 sample =1.41 g cm-3 Mass So, Volume of 100 g of the HNO3 sample = Dwnsity 2 100g = 1.41g 3 Thus, 69 g of HNO3 is contained in .70.92 cm of conc. HNO3 cm-3 = 70.92cm 70.92 1g “ “ 23g “ “ 69 70.92 “ 23 “ =23.6 cm3 69 Thus, 23.6 cm3 conc. HNO3 sample contained 23 g HNO3 41. A Solution containing 4.2 grams of an organic the molar mass of the organic compound in Kb of acetone =1.71 k kg mol-1 Solution: Mass = organic compound, Mass of acetone,(Solvent). W 2 =4.2 g W 1=50g ΔTb=1.8 Kb=1.71 k kg mol-1 If the molar mass of the solute is M, then BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 22
  • 23. ADDITIONAL PROBLEMS ON SOLUTIONS M= 1000 K b w 2 1000 1.71 4.2 -1 -1 = gmol =79.8 g mol 50. 1.8 w1 VTb 42. Calculate the molality of a KCI solution in water such that the freezing point is depressed by 2 K . (Kf for water = 1.86 K kg mol-1) Solution: For KCI, the freezing point depression is given by, Tf =iK f m This gives, m= Tf iK f = 2 -1 -1 mol kg =0.54mol kg 2 1.86 43. A decinormal solution of NaCI exerts an osmotic pressure of 4.6 atm at 300 K. calculate its degree of dissociation. (R = 0.082 L atm K-1 mol-1) Solution: No. of moles of NaCI per litre of solution =0.1 Osmotic pressure, =4.6 atm Temperature, T=300 K Had NaCI not dissociated then normal =CRT =0.1 0.082 300atm But =2.46atm obs = 4.6 atm As per dissociation Observedemagnitudeofacolligatuveproperty i = Normalmagnitudeofacolligativeproperty pobs= = 4.6atm pnormal 2.46atm = 1.87 For the the dissociation of an electrolyte producing n ions, i 1 a  n1 1.87101  21 So,Prcentage dissociation =100 0.87 1 0.87 = 100 0.87 =87 % 44. The degree of dissociation of Ca (NO3)2in a dilute aqueous solution containing 7.0 g of the per 100g of water at 100oC 70per cent. If the vapour pressure of water at 100oC is 760 mm , them calculate the vapour pressure of the solution’ Solution: Each molecule of calcium nitrate, (Ca(NO3)2 given 3 ions on dissociation in solution e.g., BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 23
  • 24. ADDITIONAL PROBLEMS ON SOLUTIONS Ca(NO 3)2Ca So, 2+ + 2NO 3 n = 3. For electrolytes which-dissociate into ions in Solution, i-1 = n -1 70 i = (ν - 1) + (3 - 1) + 1 = 2.4 100 1= Normal molar mass of Ca(N03)2 = (40 + 2 x 14 + 6 x 16) g/mol =164 g/mol From the Raoult's law or, o p -p n2 = o i p For dilute solution, n2 << n1.So. 760 - P  760  2 =  /M   i w 2 = /M w n1 +n 2 2.4 (7/164) = (1000/18) 2.4 7 18 = 0.0184 164 100 1 1 760- p = 760 x o.0184 mm Hg = 14 mm Hg p = (760-14) mm Hg = 746 mm Hg 45. Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (Kf (water) = 1.86 K kg mol-l) Solution: Mass of water, w1 = 1 kg Mass of KCl Per kg of water.= w2g Molar mass of KCI = (39 + 35.5) g mol-1 = 74.5 g mol-l We know, For KCI, 2 = w 1 ΔTf=I Kf m =i Kf  w  f 2 /745 1 iK n  i=2  T w 2 745 = 2 74.5 2 f iK 1.86 = 74.5 1.86 g = 40.05g f BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 24
  • 25. ADDITIONAL PROBLEMS ON SOLUTIONS 46. Calculate the amount of sodium chloride which must be added to 1000 mL of water so that its freezing point is depressed by 0.744 K For water Kf =1.86 K/ M Assume density of water to be one g mL-1 Mass of NaCI , w2=? Volume of water =1000 mL ΔTf= 0.744 K Kf (Water) =1.86 K/m Density of water= 1 g mL-1 So, Mass of water, w1=1000 mL 1 g mL-1=1000 g =1 kg w i K f w2 According to the definition,  = iK m = i k  2 / 58.5 T  58.5 w      1 f f f =  w1  T 58.5 1 0.744 58.5 f w = = = 11.7g iK 2 1.86 2 f So Mass of NaCI required = 11.7 g Solution: 47. Calculate the amount of sodium chloride (electrolyte) which must be added to one kilogram of water so that the freezing point is depressed by K Given : Kf for water = 1.86 K kg mol-1 Solution: Mass of sodium chloride. w2=? Molar mass of sodium chloride. M2 =58.5 g/mol Mass of water w1=1 kg Kf=1.86 K kg mol-1 For NaCI I=2 So,  = i K m = iK T f f f w2 M2  w1 = 2 1.86 w2 58.5 1  T 58.5 3 58.5 f w = = g = 47.18g or, 2 1.86 2 1.86 2 48.The freezing point depression. of o.1 m NaCl solution is 0.372 C. What conclusion would you draw about its molecular state? Kf for water is 1.86 K kg mol-1 Solution: Molality ·of NaCl solution, For ionic substances , we have or m = 0.1 ΔTf, =0.3 72 C Kf = 1.86 K kg mol-1 ΔTf= i K1m 0.372 = i X 1.86 X 0.1 0.372 i= = 2 1.86 0.1 BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 25
  • 26. ADDITIONAL PROBLEMS ON SOLUTIONS The value of i=2, indicates that the solute NaCI in solution is completely dissociated giving two ions, i.e., NaCI Na+ (aq) + Cl-(aq). 49. The vapour pressure of a pure liquid A is 40 mm Hg at 310 K The vapour pressure of this liquid in a solution with liquid B is 32 mm Hg . Calculate the molefraction of A in the solution if it obeys the Raoult’s law. o Solution: Vapour pressure of pure A, P A = 40 mm Hg Vapour pressure of A solution, PA = 32 mm Hg o According to the Raoult’s law, PA = PA XA PA 32 mm Hg  0.8 o 40 mm Hg A PA 50. A solution of 12.5 g urea in 170 g urea in 170 g of water gave a boiling point elevation of 0.63 K Calculate the molar mass of urea, taking Kb=0.52 K/m. Then, X = = Solution: Given: Mass of urea, of urea, w2=3.5 g Mass of water, w1=170 g=0.17 kg Elevation of boiling point, ΔTb=0.63 K Molar mass of the solute (urea) is given by, K w b 2 = 0.52 12.5 g/ mol = 60.7 mol-1 M= w VT 0.17 0.63 1 b BANGALORE INSTITUTE OF COACHING II PUC CHEMISTRY STUDY MATERIAL Page 26