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10, , Chapter, , Solutions, Solutions, SECTION - A, Objective Type Questions, 1., , The normality of 10% (weight/volume) acetic acid is, (1) 1 N, , (2) 10 N, , (3) 1.7 N, , (4) 0.83 N, , Sol. Answer (3), , ⎛w⎞, 10% ⎜ ⎟  10 g CH3COOH in 100 mL of solution., ⎝V⎠, w, N = M × n-factor  M  V  n, w, (L), , CH3 COOH = 24 + 4 + 32, ⇒ 60 g, n-factor = 1 (CH3 COOH), , 10  1000, 1, , 60  100, ,  1.66 nearest option is (3), 2., , 20 ml of 0.5 M HCl is mixed with 30 ml of 0.3 M HCl the molarity of resulting solution is, (1) 0.8 M, , (2) 0.53 M, , (3) 0.38 M, , (4) 0.83 M, , Sol. Answer (3), M1V1 + M2V2 = M3V3, 0.5 × 20 + 0.3 × 30 = M3[V1 + V2], 10 + 9 = M3[20 + 30],, 3., , M3 =, , 19, = 0.38, 50, , 20 ml of 0.2 M Al2(SO4)3 is mixed with 20 ml of 6.6 M BaCl2, the concentration of Cl– ion in solution is, (1) 0.2 M, , (2) 6.6 M, , (3) 0.02 M, , (4) 0.06 M, , Sol. Answer (2), BaCl2 , 6.6 M, , Ba+2, , +, , 2Cl–, , 6.6 M, , 2 × 6.6  Initial molarity of Cl–, , V1 of (Cl–) = 20 mL, , So, M1V1 = M2V2, 2 × 6.6 × 20 = M2 × [20 + 20]  Mixed with 20 mL of Al2(SO4)3, M2 , , It increases volume but can't effect concen–, –, tration of Cl because it does not have Cl, , 2  6.6  20,  6.6, 40, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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36, 4., , Solutions, , Solution of Assignment, , The expression relating molarity (M) of a solution with its molality (m) is, (where d = density, MB mol. wt. of solute), , 1000 M, (1) m  1000d  M M, B, , 1000 M, (2) m  1000d – M M, B, , (3) m , , 1000d  MMB, 1000 M, , (4) m , , 1000d – MMB, 1000 M, , Sol. Answer (2), m, , 5., , 1000 M, 1000 d  M MB, , If solubility of any gas in the liquid at 1 bar pressure is 0.05 mol/lit. What will be its solubility at 3 bar pressure,, keeping the temperature constant ?, (1), , 0.05, mol/lit, 3, , (2) 0.15 mol/lit, , (3) 0.05 mol/lit, , (4) 1.0 mol/lit, , Sol. Answer (2), p = KH × solubility (x), 1 = KH × 0.05, , ....(i), , 3 = KH × x, , ....(ii), , 1 0.05, , 3, x, , x = 0.05 × 3, 6., ,  0.15 mol/litre, , The boiling points of C6H6, CH3OH, C6H5 NH2 and C6H5NO2 are 80oC, 65oC, 184oC and 212oC respectively., Which of the following will have highest vapour pressure at room temperature?, (1) C6H6, , (2) CH3OH, , (3) C6H5NH2, , (4) C6H5NO2, , Sol. Answer (2), B.P. , , 1, V.P., , If B.P. is low means liquid can easily boil so, liquid forms vapours easily - vapours increases = vapour pressure, increases among all compounds boiling point of CH3OH is min. B.P. = 65ºC so it has max. V.P., 7., , The highest temperature at which vapour pressure of any liquid can be measured is, (1) Critical temperature, , (2) Boyle’s temperature, , (3) Boiling point of the liquid, , (4) Kraft temperature, , Sol. Answer (3), Boiling point of the liquid is the tempreature at which vapour pressure of liquid equals to atmospheric pressure., 8., , If solute and solvent interactions are more than solute - solute and solvent - solvent interactions then, (1) It is ideal solution, , (2) It is non-ideal solution with positive deviation, , (3) It is non-ideal solution with negative solution, , (4) Can't predicted, , Sol. Answer (3), Solute-Solvent interactions > Solute-solute or solvent-solvent interaction, , interactions are high, So, bonds cannot easily break thats why vapours decreases So, V.P. decreases negative deviation., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 9., , Solutions, , 37, , Which of the following is correct about a solution showing positive deviation?, (1) Vapour pressure observed will be the less than that calculated from Raoult’s law, (2) Minimum boiling azeotrope will be formed, (3) Hmix < 0, (4) Vmix < 0, , Sol. Answer (2), In positive deviation solution, Solute-solvent interactions < Solute-solute or solvent - solvent interaction, , Bond can break easily  Vapours increases V.P. increases, V.P.  High that's why B.P. = Low  min. boiling azeotrope., Hmix > 0, Vmix > 0, 10. If C2H5OH and H2O solution is example of non-ideal solution then which graphical representation is correct?, , V.P, , V.P, , V.P, , (2), , (1), , (3), , Mole fraction, , Mole fraction, , (4), , Mole fraction, , V.P, Mole fraction, , Sol. Answer (2), C2H5 – OH + H2O solution is an example of negative deviation because they can form H-bond so V.P. decreases of solution., solution, , negative deviation from solution, , V.P, , Mole fraction, 11. If P0 and P are the vapour pressure of solvent and solution n1 and n2 are the moles of solute and solvent, then, ⎛ n2 ⎞, (1) P0  P ⎜, ⎟, ⎝ n1  n2 ⎠, , ⎛, ⎞, (2) P  P0 ⎜ n2 ⎟, ⎝ n1  n2 ⎠, , ⎛, ⎞, (3) P0  P ⎜ n1 ⎟, ⎝ n1  n2 ⎠, , (4) Pº = P × n1, , Sol. Answer (2), P = Pºx, , Pº = solvent, , ⎛ n2 ⎞, P = Pº ⎜ n  n ⎟, ⎝ 1, 2 ⎠, , n2, Moles of solvent, x = mole fraction of solvent  moles of solute + moles of solvent  n  n, 1, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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38, , Solutions, , Solution of Assignment, , 12. Which of the following statement is true?, (1) Molarity of solution is independent of temperature, , (2) Molality of solution is independent of temperature, , (3) Mole fraction of solute is dependent on temperature (4) The unit of molality is mol dm–3, Sol. Answer (2), (1) M , , w, Mw  V(L), , M, , 1, Volume, , M – Depends on volume so it also depends on temperature. PV  nRT, , w, (2) Molality = M  w, w, solvent (kg), Molality does not depend on volume so it cann't depend on temperature., (3) Mole fraction also can't depend on volume so don't depend on temperature, , w, mol ⎛, mol ⎞, ⇒, not  3 ⎟, ⎜, (4) m  M  w, kg ⎝, dm ⎠, w, solvent(kg), 13. Among the following mixtures, dipole-dipole as the major interaction, is present in, (1) Benzene and CCl4, , (2) Benzene and C2H5OH, , (3) CH3COCH3 and CH3CN, , (4), , KCl and water, , Sol. Answer (3), Benzene  CCl4, (1) , non polar not dipole, , (2) Benzene C2H5 OH, , Non polar, not dipole, , (3) CH3 – C – CH3 + CH3 – CN, , O, , Polar covalent, dipole, , Polar covalent, dipole, , dipole-dipole, ,  H2 O  not dipole, (4) KCl, Ionic, Ionic, , 14. For an ideal solution with PA  PB , which of the following is true?, (1) A(l) = A(V), , (2) A(l) > A(V), , (3) A(l) < A(V), , (4) No relationship in their mole fraction, , Sol. Answer (3), , PA  PB  Vapour pressure of (A) is more than (B) so mole fraction of (A) is more in vapour phase than, liquid phase. Because vapours are more formed, A(l) < A(v), 15. An azeotropic solution of two liquids has a boiling point lower than either of the boiling points of the two liquids, when it, (1) Shows negative deviation, , (2) Shows positive deviation, , (3) Shows no deviation, , (4) Is unsaturated, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 39, , Sol. Answer (2), , Positive deviation  shows, minimum B.P. azeotrope, V.P. = High  B.P. = minimum, 16. The vapour pressure of a solution having 2.0 g of solute X (molar atomic mass = 32 g mol–1) in 100 g of CS2 (vapour, pressure = 854 torr) is 848.9 torr. The molecular formula of the solute is, (1) X, , (2) X2, , (3) X4, , (4) X8, , Sol. Answer (4), p  p,  xA ,, p, (Solute), , xA , , nA, n,  A (For dilute solution), nA  nB, nB, , 854  848.9 nA WA  MB, , , 854, nB MA  WB, , WA = weight of solute, WB = molar mass of solute, , 5.1, 2  76, ⇒, 854, MA  100, , MB = molar mass of solvent = CS2 = 12 + 32 × 2  76 g, , MA, , , , (Molar mass, of solute), , 2  76  854,  253, 100  5.1, , molar mass 253, Number of atoms = atomic mass  32  7.90  8, , So formula of solute is X 8, 17. Two solutions of a non-electrolyte are mixed in the following manner. 480 mL of 1.5 M first solution + 520 mL, of 1.2 M second solution. What is the molarity of the final solution?, (1) 1.344 M, , (2) 2.70 M, , (3) 1.20 M, , (4) 1.50 M, , Sol. Answer (1), M1V1 + M2V2 = M3V3, 1.5 × 480 + 1.2 × 520 = M3 [V1 + V2], M3 , , 1.5  480  1.2  520 720  624 1344, , ,  1.344 M, 480  520, 1000, 1000, , 18. A pressure cooker reduces cooking time for food because, (1) Cooking involves chemical changes helped by a rise in temperature, (2) Heat is more evenly distributed in the cooking space, (3) Boiling point of water involved in cooking is increased, (4) The higher pressure inside the cooker crushes the food material, Sol. Answer (3), Pressure cooker reduces cooking time for food because B.P. of water involved in cooking is increased due, to which water absorbed more or more heat and transfer it to food by which food can easily cook., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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40, , Solutions, , Solution of Assignment, , 19. Equal moles of benzene and toluene are mixed the V.P. of benzene and toluene in pure state are 700 and, 600 mm Hg respectively. The mole fraction of benzene in vapour state is, (1) 0.7, , (2) 0.47, , (3) 0.50, , (4) 0.54, , Sol. Answer (4), Total moles of benzene and toluene let 1 mole, 1, 1, 1, 1, , Toluene =, Total =, +, =1, 2, 2, 2, 2, , So, benzene =, , pbenzene,  700, , pToluene, , pbenzene, Mole fraction of benzene in vapours y ⇒ p,  600, total, , xbenzene , , x toluene, , 1, 2, , 1 1, , 2 2, , , , 1, 2, , , , 1, 2, , 1, , , 1 1 2, , 2 2, , , pbenzene, xbenzene, , (P x)benzene  (p x)toluene, , ⇒, , 700 , , 1, 2, , 1, 1, 700   600 , 2, 2, , ⇒, , 350,  0.53, 350  300, , 20. The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the, solvent. The mole fraction of solute in solution is 0.2, what would be the mole fraction of solute if decrease, in vapour pressure is 20 mm of Hg?, (1) 0.8, , (2) 0.6, , (3) 0.4, , (4) 0.2, , Sol. Answer (3), , p  p, p, ,  x A (solute), , ( p)1,  (x A )1 ,, p, 10, , (p)1 = 10 (xA)1 = 0.2, ,  0.2, , ...(1), , 20,  (x A )2, p, , ...(2), , p0, , (p2) = 20, (xA)2 = ?, Dividing (1) by (2), 10,  0.2, p, ,, 20,  (x A )2, p, (x A )2 , , 10, 0.2, , 20 (x A )2, , 0.2  20, ⇒ 0.4, 10, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 41, , 21. Calculate the mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene, and toluene having a mole fraction of toluene 0.5. (The vapour pressure of pure Benzene is 119 torr and that, of toluene is 37 torr at the same temperature), (1) 0.5, , (2) 0.75, , (3) 0.625, , (4) 0.237, , Sol. Answer (4), , , y toluene, , (Vapour phase), , p toluene, (px)toluene, 37  0.5, 18.5, 18.5, , ⇒, , ,  0.237, p total, (px)toluene  (px)benzene, 37  0.5  119  0.5, 18.5  59.5, 78, , xbenzene  x toluene  1, xbenzene  0.5  1, xbenzene  1  0.5  0.5, , 22. An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and, ethanol are 2.619 kPa and 4.556 kPa respectively, the composition of vapour (in terms of mole fraction) will, be, (1) 0.635 MeOH, 0.365 EtOH, , (2) 0.365 MeOH, 0.635 EtOH, , (3) 0.574 MeOH, 0.326 EtOH, , (4) 0.173 MeOH, 0.827 EtOH, , Sol. Answer (2), yMeOH , , (Vapour ), , yEtOH , , pMeOH, 2.619, 2.619, ⇒, ,  0.365, pMeOH  pEtOH, 2.619  4.556 7.175, , pEtOH, 4.556, 4.556, ⇒, ,  0.635, pEtOH  pMeOH, 2.619  4.556 7.175, , 23. The vapour pressure of a dilute aqueous solution of glucose is 750 mm Hg at 373 K. Calculate the mole fraction, of solute. (The vapour pressure of pure water is 760 mm Hg at 373 K), (1) 0.013, , (2) 1.3, , (3) 0.13, , (4) 1.1, , Sol. Answer (1), , p  p,  x A (solute) ,, p, , 760  750,  xA, 750, 10,  xA, 750, , xA = 0.013, 24. Two liquids having vapour pressures P10 and P20 in pure state in the ratio of 2 : 1 are mixed in the molar ratio, of 1 : 2. The ratio of their moles in the vapour state would be, (1) 1 : 1, , (2) 1 : 2, , (3) 2 : 1, , (4) 3 : 2, , Sol. Answer (1), P10 : P20 , , 2, 1, , n1 1, , n2 2,  P1  P01x1, P2  P20 x 2, , x1 , , n1, 1, 1, , , n1  n2, 1 2 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 43, , 26. In the case of osmosis, solvent molecules move from solution having, (1) Higher vapour pressure to lower vapour pressure, (2) Higher concentration to lower concentration, (3) Lower vapour pressure to higher vapour pressure, (4) Higher osmotic pressure to lower osmotic pressure, Sol. Answer (1), In osmosis solvent molecules more from their higher concentration to this lower concentration through semi, permeable membrane, , pure solvent, , i.e.,, , S.P.M., solute, low (V.P.), , (Vapour pressure), , Of pure solvent is more than solution i.e. more from higher vapour pressure to lower vapour pressure., 27. Equimolar solution of non-electrolyte in the same solvent have, (1) Same boiling point and same freezing point, , (2) Different boiling point and different freezing point, , (3) Same boiling point but different freezing point, , (4) Same freezing point but different boiling point, , Sol. Answer (1), , Tb  Tb  Tb, Tf  Tf  Tf, For same solvent (given), , Tb = constant, Tf = constant, So Tb  Tb  m, Tf  Tf  m, ,  m = same (given), , So Tb & Tf also becomes same, 28. In the phenomenon of osmosis through the semipermeable membrane, (1) Solvent molecules pass from solution to solvent (2) Solvent molecules pass from solvent to solution, (3) Solute molecules pass from solution to solvent (4) Solute molecules pass from solvent to solution, Sol. Answer (2), In osmosis solvent molecules pass from solvent to solution through semipermeable membrane, 29. The osmotic pressure of 0.1 M sodium chloride solution at 27°C is, (1) 4.0 atm, , (2) 2.46 atm, , (3) 4.92 atm, , (4) 1.23 atm, , Sol. Answer (3),  = (CST) × i, , NaCl  Na  Cl–, (i 2), , T = 273 + 21 = 300, , = (0.1 × 0.0821 × 300) × 2  4.92 atm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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44, , Solutions, , Solution of Assignment, , 30. A solution containing 8.6 g L–1 of urea is isotonic with a 5% solution of unknown solute. The molar mass of, the solute will be, (1) 348.9 g mol–1, , (2) 174.5 g mol–1, , (3) 87.3 g mol–1, , (4) 34.89 g mol–1, , Sol. Answer (1), , ⎛ w ⎞, Urea  8.6 g/L = i.e., ⎜, ⎟, ⎝ volume ⎠, Molar mass = NH2CONH2  14 × 2 + 4 + 12 + 16  60, isotonic  [1 = 2], C1ST = C2ST, C1 = C2, W, W, , Mw  V Mw  V, 8.6, 5  1000, ⇒, 60, Mw  100, 5  60  10,  348.8 g / mol, 8.6, Erata : 5% solution of unknown solute but it must be written 5% weight/volume means 5 g of solute present, in 100 mL of solution, Mw , , 31. Which of the following physical properties is used to determine, the molecular mass of a polymer solution?, (1) Relative lowering of vapour pressure, , (2) Elevation in boiling point, , (3) Depression in freezing point, , (4) Osmotic pressure, , Sol. Answer (4), In relative lowering of vapour pressure elevation in B.P. depression in freezing point get minimum for high, molecular masses but osmotic pressure can't lowered as much for high mol. masses, 32. In a 0.2 molar aqueous solution of a weak acid HX, the degree of ionization is 0.3. The freezing point of the, solution will be nearest to, (1) 0.48°C, , (2) –0.48°C, , (3) –0.36°C, , (4) –0.26°C, , Sol. Answer (2), M = 0.2 (concentration),  = 0.3, Tf = ?, , –, , , HX , H X, , t=0, , 1, , 0, , 0, , t = t1, , 1–, , , , , , ⎡Tf (H2O)  0C⎤, ⎢, ⎥, ⎣⎢ K f  1.86 ⎦⎥, , Total moles at, equilibrium = 1 –  +  +  = 1 + , , number of moles after dissociation, initial moles, 1 , i⇒, 1, Tf = i [kf × m], i, , Tf  Tf ⇒ (1   )[1.86  0.2], 0°C – Tf = 1 + 0.3 [1.86 × 0.2], –Tf  1.3 × 1.86 × 0.2, Tf = –0.48°C, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 45, , 33. The boiling point of water at 735 torr is 99.07°C. The mass of NaCl added in 100 g water to make its boiling, point 100°C is (kb = 0.51 K kg mol–1), (1) 10.68 g, , (2) 5.34 g, , (3) 2.67 g, , (4) 26.7 g, , Sol. Answer (2), Tb = i(k b × m) i(NaCl = 2 (Na+ + Cl– ), Mw = NaCl = 58.5, , ⎡, ⎤, w  1000, Tb  Tb  2 ⎢0.51, ⎥, M  w (solvent)  k f ⎥⎦, ⎣⎢, 100 – 99.07  2  0.51, , 0.93 , , w, , w  1000, 58.5  100, , 2  0.51 w  10, 58.5, , 0.93  58.5, 54.405, ,  5.3 g, 2  0.51 10 2  0.51 10, , 34. If van't Hoff factor, i = 1, then, (1) It is dissociation, , (2) It is association, , (3) Both (1) & (2), , (4) Neither dissociation nor association, , Sol. Answer (4), i, , number of particles after dissociation association, initial moles, , i > 1 = Dissociation, i < 1 = Association, i = 1 = neither dissociation nor association, 35. Which of the following equimolar solution have highest vapour pressure?, (1) Glucose, , (2) NaCl, , (3) K2SO4, , (4) K4Fe(CN)6, , Sol. Answer (1), Vapour pressure , , 1, number of ions / particles, , Glucose does not dissociate into ions so it, , 1 ⎞, has minimum B.P. so maximum V.P. or ⎛⎜ B.P. , ⎟  i (von' t Hoff factor), V.P., ⎝, ⎠, 36. When NaCl is added to aqueous solution of glucose, (1) Freezing point is lowered, , (2) Freezing point is raised, , (3) Freezing point does not change, , (4) Variation is freezing point can't be predicted, , Sol. Answer (1), NaCl is added to aq. solution then V.P. of solution decrease so freezing point is lowered., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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46, , Solutions, , Solution of Assignment, , 37. If is the degree of dissociation of Na2SO4. The van’t Hoff factor used for the calculation of molecular mass, is, (1) 1 + 2, , (3) 1 + , , (2) 1 – 2, , (4) 1 – , , Sol. Answer (1), Na2SO4  2Na  SO4–2, t = 0 (initial), , 1, , 0, , 0, , t = t1, , 1–, , 2, , , , i, , 1–   2  , ⇒ 1  2, 1, , 38. The relationship between the values of osmotic pressure of 0.1M solutions of KNO3(P1) and CH3COOH(P2) is, (1) P1 > P2, , (2) P2 > P1, , (3) P1 = P2, , P1, P2, (4) P  P  P  2P, 1, 2, 1, 2, , Sol. Answer (1),  = i(CRT), If C, T = constant, i, For KNO3 (i) greater than CH3COOH because CH3COOH is a weak acid which shows less dissociation, and KNO3 shows more dissociation thats why its (i) increases so  of KNO3 is more., 39. The relationship between osmotic pressure (P) at 273 K when 10 g glucose (P1), 10 g urea (P2) and 10 g, sucrose (P3) are dissolved in 250 ml of water is, (1) P1 > P2 > P3, , (2) P2 > P1 > P3, , (3) P2 > P3 > P1, , (4) P3 > P2 > P1, , Sol. Answer (2),  = i(CRT), , all are non-electrolytes (i = 1), , ⎛n, ⎞,   i ⎜ RT ⎟, ⎝V, ⎠, R = constant, , T = Constant, , w, So,   Molar mass, , , 1, Molar mass, , W = 10 g = constant, i = non-electrolyte = 1, Glucose, Urea, C6H12O6, NH2CONH2, Molar Mass  180,  60,   P1, P2, , Sucrose, C12H22O11,  342, P3, , ⎡Molar mass  Sucrose  Glucos e  urea, now ⎢   P  P  P, 2, 1, 3, ⎣, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 47, , 40. What is the molality of C 2 H 5 OH in water solution which will freeze at –10 o C? [Mol. wt. of, C2H5OH = 46, Kf for water = 1.86], (1) 6.315 m, , (2) 63.15 m, , (3) 3.540 m, , (4) 5.3 m, , Sol. Answer (4), Tf = i(kf × m), , i = C2H5OH non-electrolyte = 1, , 0 –(–10) = 1(1.86 × m), , Tf  Tf  Tf, , m, , 10, ⇒ 5.3, 1.86, , , , Tf of water = 0°C, , 41. Which of the following pairs of solutions can we expect to be isotonic at the same temperature?, (1) 0.1 M NaCl and 0.1 M Na2SO4, , (2) 0.1 M urea and 0.1 M NaCl, , (3) 0.1 M urea and 0.2 M MgCl2, , (4) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4, , Sol. Answer (4),  = iCRT, when i, C, T are same for two or more than two solutions then their osmotic pressure are also same that, are known as isotonic, (1) NaCl  i = 2 (Na+ + Cl–), , Na2SO4  2Na  SO4–2  i  3 different so not isotonic, (2) Urea i = 1 non-electrolyte ⎤, ⎥ different so not isotonic, NaCl i = 2 [Na+ + Cl– ] ⎦, , Urea i = 1, , (3), , 2+, , MgCl2 i = 3 [Mg, , (4), , ⎤, ⎥ different so not isotonic, + 2Cl ]⎥⎦, –, , Ca(NO3 )2  Ca 2  2NO3–, Na2 (SO 4 )  2Na, , , ,  SO 4–2, , ⎡i  3 ⎤, ⎢i  3 ⎥, ⎣, ⎦, , C = same, T = same so  = same isotonic, 42. Which of the following aqueous solution has maximum freezing point?, (1) 0.01 M NaCl, , (2) 0.005 M C2H5OH, , (3) 0.005 M MgI2, , (4) 0.01 M MgSO4, , Sol. Answer (2), Tf = i(kf × m), , Tf – Tf  i(k f  m), i = minimum, Tf  max imum, , (i) NaCl  electrolyte i = 2, (2) 0.005 M C2H5OH non electrolyte, i = 1 minimum, (3) MgI2  electrolyte i = 3, (4) MgSO4  electrolyte i = 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 49, , (i) decrease  so Tf increase, Because Tf = i(kf × m), , Tf  Tf  i(k f  m), So, Tf , , 1, i, , 46. An aqueous solution freezes at –0.36°C. Kf and Kb for water are 1.8 and 0.52 respectively then value of boiling, point of solution at 1 atm pressure is, (1) 101.04°C, , (2) 100.104°C, , (3) 0.104°C, , (4) 100°C, , Sol. Answer (2), (Tf)solution = –0.36°C, Kf = 1.86, Kb = 0.52°, at 1 atm, Tf  0C , Tb0  100C, , Tf – Tf  i [1.86  m], , ...(i), , Tb – Tb  i [0.52  m], , ...(ii), , Tf  Tf  i[1.86  m], , Tb  Tb  i[0.52  m], , 0  (–0.36) 1.86, , Tb  100, 0.52, 0.36  0.52,  Tb – 100, 1.86, , Tb = 100.10°C, 47. The osmotic pressure of decimolar solution of urea at 27°C is, (1) 2.49 bar, , (2) 5.0 bar, , (3) 3.4 bar, , (4) 1.25 bar, , Sol. Answer (1), Decimolar, M, , 1, , T = 27°C + 273  300 K, 10, ,  = i(CRT) For urea i  1,   1, , 1,  0.0821 300, 10, ,  = 2.4 bar, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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50, , Solutions, , Solution of Assignment, , 48. Arrange the following aqueous solutions in the order of their increasing boiling points, (i) 10–4 M NaCl, , (ii) 10–3 M Urea, , (iii) 10–3 M MgCl2, , (iv) 10–3 M NaCl, , (1) (i) < (ii) < (iv) < (iii), , (2) (ii) < (i) = (iii) < (iv), , (3) (i) < (ii) < (iii) < (iv), , (4) (iv) < (iii) < (i) = (ii), , Sol. Answer (1), (i) 10–4 M NaCl   iCRT, (NaCl  Na  Cl– ) ⇒ 2  10 –4 RT, (ii) 10–3 M Urea  = 1 × 10–3 RT, (iii) 10–3 M MgCl2  = 3 × 10–3 RT, Mg+2 + 2Cl–, (i = 3), (iv) 10–3 M NaCl,  = 2 × 10–3 RT, Now i increases  increases = Tb increases  (Tb – Tb ) increases = Tb increases, i.e. Tb  i, (i) < (ii) < (iv) < (iii), 49. When a saturated solution of KCl is heated it becomes, (1) Unsaturated, , (2) Supersaturated, , (3) Remains saturated, , (4) Attains equilibrium, , Sol. Answer (1), When a saturated solution of KCl is heated then solvent (H2O) becomes evaporated now it lefts only KCl i.e., solution becomes unsaturated., 50. Molal elevation constant of a liquid is, (1) The elevation in b.p. which would be produced by dissolving one g of solute in 100 g of solvent, (2) The elevation in b.p. which would be produced by dissolving 1g. solute in 100 g of solvent, (3) Elevation in b.p. which would be produced by dissolving 1 gm of solute in 1000 g of solvent, (4) Elevation in b.p which would be produced by dissolving 1 mole of solute is 1000 g of solvent, Sol. Answer (4), Kb = molal elevation constant, Tb = i(Kb × m), K b , , Kb , , Tb, im, , Tb, moles, i, mass of solvent (in kg), , when i = 1, moles = 1, mass of solvent = 1 kg = 1000 g, then [Kb = Tb], Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 51, , SECTION - B, Objective Type Questions, 1., , 100 ml of 1 M NaOH is mixed with 50 ml of 1 N KOH solution. Normality of mixture is, (1) 1 N, , (2) 0.5 N, , (3) 0.25 N, , (4) 2 N, , Sol. Answer (1), Nmixture , N1V1, (NaOH), , N1V1  N2 V2, 100  50 150, ⇒, , ⇒ 1N, V1  V2, 100  50 150, ,  1 100  100, , N2V2 = 1 × 50 = 50, 2., , Mass of NaCl required to prepare 0.01 m aqueous solution in 1 kg water is, (1) 0.01 g, , (2) 0.585 g, , (3) 58.8 g, , (4) 5.88 g, , Sol. Answer (2), M, , moles, w, , mass of solvent (kg) Mw  W(kg), , 0.01 , , w, 58.5  1, , [NaCl (Mw = 58.5], , w = 0.01 × 58.5, w  0.585 g, 3., , Two solutions marked as A and B are separated through semipermeable membrane as below. The phenomenon, undergoing, , 0.01 M NaCl, solution, , 0.1 M NaCl, solution, , A, , B, SPM, , (1), , Na+, , moves from solution A to solution B, , (2) Both Na+ and Cl– moves from solution (A) to solution (B), (3) Both Na+ and Cl– moves from solution (B) to (A), (4) Solvent molecules moves from solution (A) to (B), Sol. Answer (4), Solvent molecules moves from lower concentrated solution to higher concentrated solution (osmosis), , 0.01 M NaCl, solvent, A, , 0.1 M NaCl, B, , S.P.M, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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52, 4., , Solutions, , Solution of Assignment, , Correct observation, , I, , II, , III, , 0.1 M Urea, , 0.1 M NaCl, , 0.1 M CaCl2, , (1) Vapour pressure of solution I is lowest, , (2) Relative lowering of vapour pressure is maximum in III, , (3) Freezing point is maximum for III, , (4) Boiling point is minimum for II, , Sol. Answer (2), , Urea NaCl, 0.1 M 0.1 M, i⇒1 2, , CaCl2, 0.1 M, 3, , max relative lowering of vapour, pressure freezing point is minimum, , p,  i(x), p, i increases = (p) increases = relative lowering of vapour pressure increases, 5., , An aqueous solution of sugar is taken in a beaker. At freezing point of solution, (1) Crystals of sugar separated, , (2) Crystals of glucose and fructose are separated, , (3) Crystals of ice separated, , (4) Mixture of ice and some sugar crystals separated, , Sol. Answer (3), At freezing point of solution water get freeze but sugar cannot so water converted into ice so crystals of ice, separated., 6., , 15 g urea and 20 g NaOH dissolved in water. Total mass of solution is 250 g. Mole fraction of NaOH in the, mixture, (1) 0.036, , (2) 0.62, , (3) 0.5, , (4) 0.4, , Sol. Answer (1), moles of NaOH, mole fraction of NaOH  moles of NaOH  moles of urea  moles of H O, 2, w, 20, 1, moles of NaOH = M  40 ⇒ 2, w, , [NaOH = 23 + 16 + 1  40], , w, 15, moles of urea  M ⇒ 60, w, w, 215, moles of H2O  M ⇒ 8, w, , ⎡, ⎤, 15  20  x  250 ⎥, ⎢⎣urea, NaOH H2O, Total ⎦, x = 250 – 35 = 215 g, , 1, 2, ⇒ 0.03, Mole fraction of NaOH , 1 15 215, , , 2 60 18, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 7., , Solutions, , 53, , Which of the following concentration terms is temperature independent?, I., , Molarity, , II., , (1) I & II, , Molality, , (2) I & III, , III. Normality, , IV. Mole fraction, , (3) II only, , (4) II & IV, , Sol. Answer (4), PV = nRT, (V  T), P & n = constant, So molarity and normality depends on volume so also depends on temperature. But molality and mole, fraction doesn't depend on volume so doesn't depend on temperature., 8., , Vapour phase diagram for a solution is given below if dotted line represents deviation, , V.P., , A = 1, B = 0, , A = 0, B = 1, , Mole fraction, , Correct observation for this solution, (1) Hmix : +ve, , (2) Smix : +ve, , (3)  Vmix : +ve, , (4) All of these, , Sol. Answer (4), Shown graph is positive deviation so, for positive deviation H > 0  positive, V > 0  positive, S > 0  positive, 9., , A mixture of two liquids A and B having boiling point of A is 70°C, and boiling point of B is 100°C, distills at, 101.2°C as single liquid, hence this mixture is, (1) Ideal solution, , (2) Non ideal solution showing +ve deviation, , (3) Non ideal solution showing –ve deviation, , (4) Immiscible solution, , Sol. Answer (3), A  70C ⎤, distils at 101.2°C i.e., B.P of solution is greater than A/B, B  100C ⎥⎦, , and it shows maximum boiling Azeotrope means, negative deviation [Vapour pressure of this solution is decreased], 10. The phenomenon taking place, , Egg lining, , Water, , (1) Exo-osmosis, , (2) Endo-osmosis, , (3) Reverse-osmosis, , (4) All of these, , Sol. Answer (2), Water (solvent) moves from their higher concentration to their lower concentration through S.P.M [i.e. from, outside to inside] known as endosmosis, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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54, , Solutions, , Solution of Assignment, , 11. 1 molar aqueous solution is _____ concentrated than 1 m aqueous solution, (1) More, , (2) Less, , (3) Equally, , (4) Very less, , Sol. Answer (1), Molarity ⇒, 1, , Moles, ,, Volume, , Molality , , Moles, ,, Volume, , 1, , i.e., moles in 1 litre solution it is, defined for solution so it is more, than molality which is defined for, solvent., , moles, Mass of solvent (in kg), , moles, W(kg), , i.e. moles in 1 kg of solvent, , 12. Osmotic pressure of solution containing 0.6 g urea and 3.42 g sugar in 100 ml at 27°C, (1) 492 atm, , (2) 4.92 atm, , (3) 49.2 atm, , (4) 28.1 atm, , Sol. Answer (2), Total mass of solution = 0.6 + 3.42  4.02 g, , ⎛n⎞,   i ⎜ ⎟ RT, ⎝V⎠, , , ,  i = 1 [Both urea and sugar are nonelectrolyte], , 1 ntotal  RT, V, , Total moles = moles of urea + moles of sugar, W, w, 0.6 3.42,  M  m ⇒ 60  342, w, w, , 1 0.02  100  0.0821 300,  4.92, 100, , 0.01 + 0.01  0.02, 13. Which gas is most soluble in water?, (1) He, , (2) H2, , (3) NH3, , (4) CO2, , Sol. Answer (3), NH3Shows H-bonding with H2O than rest 3-options so it is most soluble in water, , 1 atm, 14., , V.P., H2O, I, II III, , T, Which is having highest elevation in boiling point?, (1) H2O, , (2) Solution I, , (3) Solution II, , (4) Solution III, , Sol. Answer (4), solution, , Tb3  Tb2  Tb1, , V.P., , So, lve, nt, , When nonvolatile solute is added in volatile solvent  vapour, pressure of solvent decreases and B.P. increases, , Tb = boiling point of solution, As vapour pressure decreases  B.P. of solution increases, So, III have minium V.P. = maximum B.P. = Elevation in B.P., , I II, III, , Tp Tp Tp, 1 2 3, , T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 55, , H2O, , 15., , V.P., A, , B, , NaCl, solution, , C, D, , T, Freezing point of solution is marked as, (1) A, , (2) B, , (3) C, , (4) D, , Sol. Answer (2), , V.P., , point at this V.P. of solution =, V.P. of solvent this point is known as, freezing point, , B, T, , 16. van't Hoff factor for acetic acid in aqueous medium at infinite dilution is, (1) 2, , (2) 1, , (3) 1/2, , (4) 3, , III. 0.1 M NaCl, , IV. 0.05 M CaCl2, , (3) III = II < IV < I, , (4) IV > II > III > I, , Sol. Answer (1), At infinite dilution CH3COOH is completely dissociated, –, , , ,  CH3 COO  H, So CH3 COOH , , [i = 2] number of ions, 17. Correct order of freezing point of given solution, I., , 0.1 M glucose, , II., , (1) I < II < III < IV, , 0.2 M urea, , (2) I > II > III > IV, , Sol. Answer (3), Tf = i(Kf × m) , Tf  Tf  i(K f  m), of solution Tf , I., , 1, when concentration is same, i, , 0.1 M glucose Tf  1 × Kf × 0.1 = 0.1 Kf, , (i = 1) non electrolyte, , ⎤, , glucos e  NaCl, , –⎥, , , , III. 0.1 M NaCl  Na  Cl ⎥⎦, , i2, , so,, , Tf NaCl  glucos e, , Tf  2Kf × 0.1 = 0.2 Kf, , (i = 2), II., , i1, , 0.2 M urea  Tf ⇒ i  K f  m ⇒ 1 K f  0.2  0.2 K f, , IV. 0.05 M CaCl2  Tf  3  K f  0.05  0.15 K f, , CaCl2  Ca2  2Cl–, (i3), , NaCl,  Urea, e, ,   CaCl2  Glucos, T 0.1K, same 0.2 K f, , Tf 0.15 K f, , f, , f, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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56, , Solutions, , Solution of Assignment, , 18. Boiling point of 0.01 M AB2 which is 10% dissociated in aqueous medium ( K bH O  0.52 ) as A+2 and B–, 2, , (1) 273.006 K, , (2) 373.006 K, , (3) 0.006 K, , (4) 272.006 K, , Sol. Answer (2), AB2  A 2  2B –, 1, , 0, , 0, , Tb = i(Kb × m), , 1–, , , , 2, , Tb – Tb  i(Kb  m), , i, , 1      2,  1  2, 1,  1, , Tb – 373 = 1.2 [0.52 × 0.01], , 2  10, 100, , Tb – 373 = 0.00624, , i =  1.2, , Tb = 373.006 K, , 19. Vapour pressure diagram of some liquids plotted against temperature are shown below, , P, , A, , B C, D, , E, T, , Most volatile liquid, (1) A, , (2) B, , (3) C, , (4) D, , Sol. Answer (1), , P, , A, , B C, D, , E, T, Vapour pressure of A, B, C, D, A>B>C>D, Vapour pressure of (A) is maximum so, it formed more vapours that's why it is more volatile., 20. At higher altitude, the boiling point of water is lowered because, (1) Atmosphere pressure is low, , (2) Temperature is low, , (3) Atmospheric pressure increases, , (4) Water solidifies to ice, , Sol. Answer (1), At higher altitude the B.P. of H2O is lowered because pressure is low so, H2O easily boils i.e. easily achieve, pressure which is equal to atm pressure., 21. During evaporation of liquid, (1) The temperature of liquid rises, , (2) The temperature of liquid falls, , (3) The temperature of liquid remains uneffected, , (4) The liquid molecules becomes inert, , Sol. Answer (2), During evaporation of liquid  liquid becomes cool i.e. its temperature decreases and temperature of, surrounding increases., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 57, , 22. What is the concentration of NO3– ions when equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed, together?, (1) 0.1 N, , (2) 0.25 M, , (3) 0.05 M, , (4) 0.2 M, , Sol. Answer (3), AgNO3  Ag , 0.1M, , NO3, 0.1M volume  V1, , M1V1 = M2V2, , Equal volume, , 0.1 × V = M2 [V1 + V2], , [V1 = V2 =V], , NaCl, V2, , The NaCl can't effect the concentration of NO3–, , 0.1 × V = M2 [V + V], M2 , , but its volume added in total volume in mixture., , 0.1V,  0.05, 2V, , 23. If any solute ‘A’ dimerises in water at 1 atm pressure and the boiling point of this solution is 100.52°C. If 2 moles, of A is added to 1 kg of water and kb for water is 0.52°C/molal, calculate the percentage association of A, (1) 50%, , (2) 30%, , (3) 25%, , (4) 100%, , Sol. Answer (4), Tb = iKb × m, , dim erises, , (A)2, 2A , , m, , t=0, , 1, , 0, , 2⎞, ⎛ , ⎞⎛, 100.52 – 100  ⎜ 1    ⎟⎜ 0.52  ⎟, 2, 1⎠, ⎝, ⎠⎝, , t = t1, , 1–, , , 2, , ⎞, ⎛, 0.52  ⎜ 1– ⎟  0.52  2 , 2⎠, ⎝, , i  1–  , , , ,  1 – , 2, 2, , 1–, , moles, w solvent  (K f ), ,  1, , 2 2, , , 1,  1 ,, 2, 2, ,  1, , 2 2, ,  = 1 or (100%), 24. Substance A tetramerises in water to the extent of 80%. A solution of 2.5 g of A in 100 g of water lowers, the freezing point by 0.3°C. The molar mass of A is, (1) 122, , (2) 31, , (3) 244, , (4) 62, , Sol. Answer (4), tetra m erises, , , , 4A ,  (A)4, , t=0, , 1, , 0, , Tf = i(Kf × m), , t = t1, , 1–, , , 4, , 2.5  1000 ⎤, ⎡, 0.3  0.4 ⎢1.86 , M  100 ⎥⎦, ⎣, , , , 80,  0.8, 100, , i  1–  , , M = 62 g, , , 4, ,  1– 0.8 , , 0.8, 4, , i = 0.4, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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58, , Solutions, , Solution of Assignment, , 25. K4[Fe(CN)6] is supposed to be 40% dissociated when 1M solution prepared. Its boiling point is equal to another, 20% mass by volume of non-electrolytic solution A. Considering molality = molarity. The molecular weight of A is, (1) 77, , (2) 67, , (3) 57, , (4) 47, , Sol. Answer (1), (1), K 4 [Fe(CN)6 ] , 4K   [Fe(CN)6 ]–4, , t=0, , 1, , 0, , , , (2), , t = t1, , 1–, , 4, , , , 20% mass by volume of non-electrolyte, , i = 1 –  + 4+ = 1 + 4, , , , ,  20 g in 100 ml of solution, , 40,  0.4, 100, , (i = 1), , i = 1 + 4 × 0.4  2.6, Given (Tb)1 = (Tb)2, iKb × m1 = iKb × m2, 2.6  1  1, 2.6 , M, , w, M V, , molality = molarity, , 20  1000, M  100, , 200,  76.9  77 g, 2.6, , SECTION - C, Previous Years Questions, 1., , What is the mole fraction of the solute in a 1.00 m aqueous solution?, (1) 0.0354, , (2) 0.0177, , (3) 0.177, , [Re-AIPMT-2015], (4) 1.770, , Sol. Answer (2), 2., , Which one is not equal to zero for an ideal solution?, , [AIPMT-2015], , (1) P = Pobserved – PRaoult, , (2) Hmix, , (3) S mix, , (4) V mix, , Sol. Answer (3), 3., , The boiling point of 0.2 mol kg–1 solution of X in water is greater than equimolal solution of Y in water. Which, one of the following statements is true in this case?, [AIPMT-2015], (1) Y is undergoing dissociation in water while X undergoes no change, (2) X is undergoing dissociation in water, (3) Molecular mass of X is greater than the molecular mass of Y, (4) Molecular mass of X is less than the molecular mass of Y, , Sol. Answer (2), 4., , Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of Al2(SO4)3 (if all, are 100% ionised)?, [AIPMT-2015], (1) K4[Fe(CN)6], , (2) K2SO4, , (3) K3[Fe(CN)6], , (4) Al(NO3)3, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 5., , Solutions, , 59, , Of the following 0.10m aqueous solutions, which one will exhibit the largest freezing point depression ?, [AIPMT-2014], (2) C6H12O6, , (1) KCl, , (3) Al2(SO4)3, , (4) K2SO4, , Sol. Answer (3), Tf  imk f, , C6H12O6 (i = 1), , KCl (i = 2), , Al2(SO4) (i = 5), K2SO4 (i = 3), , 6., , pA and pB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA, represents the mole fraction of component A, the total pressure of the solution will be [AIPMT (Prelims)-2012], (1) pB + xA (pB–pA), , (2) pB + xA (pA–pB), , (3) pA + xA (pB–pA), , (4) pA + xA (pA–pB), , Sol. Answer (2), PT = pA + pB, = pA x A  pB xB, = pA x A  pB (1– x A ), ,  xA + xB = 1, xB = 1 – xA, , = pA x A  pB – pB x A, PT ⇒ pB  x A [pA – pB ], , 7., , The van't Hoff factor i for a compound which undergoes dissociation in one solvent and association in other, solvent is respectively, [AIPMT (Prelims)-2011], (1) Greater than one and greater than one, , (2) Less than one and greater than one, , (3), , (4) Greater than one and less than one, , Less than one and less than one, , Sol. Answer (4), For dissociation i > 1 eg NaCl  Na   Cl– (number of ions increases), For association i < 1 eg Na   Cl–  NaCl (number of ions decreases), 8., , The freezing point depression constant for water is –1.86° Cm –1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O,, the freezing point is changed by –3.82°C. Calculate the van't Hoff factor for Na2SO4 [AIPMT (Prelims)-2011], (1) 0.381, , (2) 2.05, , (3) 2.63, , (4) 3.11, , Sol. Answer (3), Na2SO4  2Na  SO4–2, , Tf = iKf × m, , ⎡, ⎤, w, 0  (–3.82)  i ⎢1.86 , ⎥, M  w (solvent) ⎥⎦, ⎢⎣, 3.82  i  1.86 , , i, , MNa2SO4  2  23  96 ⇒ 46  96  142 g, , 5  1000, 142  45, , 3.82  142  45,  2.62, 1.86  5  1000, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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60, 9., , Solutions, , Solution of Assignment, , A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86C/m, the freezing point of the, solution will be, [AIPMT (Mains)-2011], (1) –0.36C, , (2) –0.24C, , (3) –0.18C, , (4) –0.54C, , Sol. Answer (2), HX, , (weak acid, ,  H  X –, , Tf = iKf × m, , t=0, , 1, , 0, , , ,  1.3 × 1.86 × 0.1, , t = t1, , 1–, , , , , , Tf = 0.2418°C, , i, , , , 1–     , ⇒ 1 , 1, , 30,  0.3, 100, i = 1 + 0.3 = 1.3, , , Tf  Tf  0.2418C, 0 – Tf = 0.2418°C, Tf = – 0.2418°C, , 10. 200 mL of an aqueous solution of a protein contains its 1.26g. The osmotic pressure of this solution at 300 K is, found to be 2.57 × 10–3 bar. The molar mass of protein will be (R = 0.083 L bar mol–1 K–1) [AIPMT (Mains)-2011], (1) 31011 g mol–1, , (2) 61038 g mol–1, , (3) 51022 g mol–1, , (4) 122044 g mol–1, , Sol. Answer (2), i, , n, RT, V, , 2.57 × 10–3 , , i = 1 protein (non-electrolyte), 1 1.26  0.0821 300, Mw  200  10 –3, , Mw = 61038 g/mol, 11. A solution of sucrose (molar mass = 342 g mol–1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of, water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol–1), [AIPMT (Prelims)-2010], (1) –0.372C, Sol. Answer (1), Tf = i(Kf × m) = 1 1.86 , , (2) –0.520C, , (3) +0.372C, , (4) –0.570C, , 68.5  1000, 342  1000, , Tf  Tf  0.372, 0 – Tf = 0.37, Tf = –0.372°C, 12. An aqueous solution is 1.00 molal in K. Which change will cause the vapour pressure of the solution to increase?, [AIPMT (Prelims)-2010], (1) Addition of NaCl, , (2) Additon of Na2SO4, , (3) Addition of 1.00 molal K, , (4) Addition of water, , Sol. Answer (4), In H2O, KI will get dissociated into ions so (i) increases that's, ⎡ p, ⎤,  ix  increases ⇒ p decreases  V.P. increases ⎥, why ⎢, p, ⎣, ⎦, , Or KI dissociated into H2O  i increases  V.P. increases, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 61, , 13. A 0.0020 m aqueous solution of an ionic compound Co(NH3)5 (NO2)Cl freezes at –0.00732C. Number of moles of, ions which 1 mol of ionic compound produces on being dissolved in water will be (k1= –1.86C /m), [AIPMT (Prelims)-2009], (1) 3, , (2) 4, , (3) 1, , (4) 2, , Sol. Answer (4), Tf = iKf × m, i, , Tf, 0.00732, ,  1.96  2, K f  m 1.86  0.0020, , 14. 0.5 molal aqueous solution of a weak acid (HX) is 20% ionized. If Kf for water is 1.86 K kg mol sup–1, the lowering, in freezing point of the solution is, [AIPMT (Prelims)-2007], (1) –0.56 K, , (2) –1.12 K, , (3) 0.56 K, , (4) 1.12K, , Sol. Answer (4), Tf = iKf × m, , HX  H  X –, , t=0, , 1, , 0, , , , 1.2 × 1.86 × 0.5, , t = t1, , 1–, , , , , , = 1.1 K, , ( = 20%), , i, , 1–     ,  1 , 1, , Lowering in freezing point, , = 1 + 0.2 = 1.2, 15. A solution containing 10g per dm3 of urea (molecular mass = 60 g mol–1) is isotonic with a 5% solution of a nonvolatile solute. The molecular mass of this non-volatile solute is, [AIPMT (Prelims)-2006], (1) 250 g mol–1, , (2) 300 g mol–1, , (3) 350 g mol–1, , (4) 200 g mol–1, , Sol. Answer (2), 50  60, 10, , M1 = 10, , 1 = 2, , Mw , , V1 = 1 dm3, ,  C1 = C2, ,  300 g/mol, , =1L, , 10, 5  1000, , 60  1 Mw  100, , 16. 1.00 g of a non-electrolyte solute (molar mass 250g mol–1) was dissolved in 51.2 g of benzene. If the freezing point, depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by, [AIPMT (Prelims)-2006], (1) 0.4 K, , (2) 0.3 K, , (3) 0.5 K, , (4) 0.2 K, , Sol. Answer (1), Tf = iTf × m  1 5.12 , , 1 1000, 1000  10, ⇒, ⇒ 4K, 25  51.2, 25  100, , i1, , (non  electrolyte, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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62, , Solutions, , Solution of Assignment, , 17. A solution of acetone in ethanol, , [AIPMT (Prelims)-2006], , (1) Shows a negative deviation from Raoult's law, , (2) Shows a positive deviation from Raoult's law, , (3) Behaves like a near ideal solution, , (4) Obeys Raoult's law, , Sol. Answer (2), , CH3 – C – CH3 + CH3 – CH2 – OH, O, CH3 – C – CH3 CH3 – CH2 – OH, O, , CH3 – C – CH3, O, CH3 – CH2 – OH, , Not too much H-bonding than solvent-solvent so attraction are weak that's why PT > PA + PB positive deviation., 18. During osmosis, flow of water through a semi-permeable membrane is, , [AIPMT (Prelims)-2006], , (1) From solution having higher concentration only, (2) From both sides of semi-permeable membrane with equal flow rates, (3) From both sides of semi-permeable membrane with unequal flor rates, (4) From solution having lower concentration only, Sol. Answer (3), During osmosis water can move from their higher concentration to their lower concentration through S. P.M., so it can takes place from both sides with unequal flow rates., 19. The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of solution, obtained by mixing 3 moles of P and 2 moles of Q would be, [AIPMT (Prelims)-2005], (1) 140 torr, , (2) 20 torr, , (3) 68 torr, , (4) 72 torr, , Sol. Answer (4), np = 3, , pP  80, , nQ = 2, , pQ  60, , pT  pP xP  pQ xQ,  80 , , 3, 2,  60   48  24  72 torr, 5, 5, , 20. A solution of urea (mol. mass 60g mol-1) boils at 100.18C at the atmospheric pressure. If kf and kb for water are, [AIPMT (Prelims)-2005], 1.86 and 0.512K kg mol–1 respectively, the above solution will freeze at, (1) –6.54C, , (2) 6.54C, , (3) 0.654C, , (4) –0.654C, , Sol. Answer (4), Tb = 100.18°C, , Tb  Tb  0.512  m, , Tf = ?, , Tf  Tf  1.86  m, , 0.18 0.512, , (–Tf ) 1.86, , –Tf , , 100.18C – 100  0.512  m, 0  Tf  1.86  m, , 0.18  1.86, ⇒ –0.654C, 0.512, , 21. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20C are, 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would, be, AIPMT (Prelims)-2005], (1) 0.549, , (2) 0.200, , (3) 0.786, , (4) 0.478, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 63, , Sol. Answer (4), npen tan e, nhexane, , 1, 4, , , , , ppen, tan e  440, , , phexane,  120, , ypen tan e , , , , ppentene, p total, , pp . xp, pp ., , xp  pn ., , xn, , ⇒, , 440 , , 1, 5, , 1, 4,  120 , 5, 5, , 440 , , , , 88, 88, ,  0.478, 184, 88  96, , 22. The mole fraction of the solute in one molal aqueous solution is, (1) 0.027, , (2) 0.036, , [AIPMT (Prelims)-2005], , (3) 0.018, , (4) 0.009, , Sol. Answer (3), Molality = 1  1 moles in 1 kg of solvent, x solute ⇒, , 1, 1, , ⇒ 0.018, 1000 1  55.5, 1, 18, , 23. Which of the following compounds can be used as antifreeze in automobile radiators?, (1) Methyl alcohol, , (2) Glycol, , (3) Nitrophenol, , (4) Ethyl alcohol, , Sol. Answer (2), Glycol is anti freezing agent which reduces freezing of solution, , 24. Mole fraction of the solute in a 1.00 molal aqueous solution is, (1) 1.7700, , (2) 0.1770, , (3) 0.0177, , (4) 0.0344, , Sol. Answer (3), molality = 1  1 moles of solute present in 1 kg of solvent, mole fraction of solute =, , nsolute, moles of solute, , Total moles, nsolute  nsolvent, , , mass of solvent  1 kg, , 1, 1, ⇒, ⇒ 0.017, 1000, 1  55.5, 1, 18, , (H2O) = 1000 g, , 25. 1 × 10–3 m solution of Pt(NH3)4Cl4 in H2O shows depression in freezing point by 0.0054°C. The structure of, the compound will be (Given Kf = 1.860 km–1), (1) [Pt(NH3)4]Cl4, , (2) [Pt(NH3)3Cl]Cl3, , (3) [Pt(NH3)2Cl2]Cl2, , (4) Pt(NH3)Cl3]Cl, , Sol. Answer (3), Tf = iKf × m, 0.0054 = i[1.86 × 1 × 10–3], , In (3) option [Pt(NH3 )2Cl2 ]Cl2  [Pt(NH3 )2Cl2 ]+2 + 2Cl–, , , i3, , i  2.9  3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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64, , Solutions, , Solution of Assignment, , 26. Which of the following salt has the same value of van’t Hoff’s factor i as that of K3[Fe(CN)6]?, (1) Na2SO4, , (2) Al(NO3)3, , (3) Al2(SO4)3, , (4) NaCl, , Sol. Answer (2), , K 3 [Fe(CN)6 ]  3K   [Fe(CN)6 ]–3, (i 4), , Na2SO4  2Na  SO4–2, , (i = 3), , (ii) Al(NO3 )3  Al3  3NO3–, , (i = 4), , (i), , (iii) Al2 (SO4 )3  2Al3  3SO4–2 (i = 5), (iv) NaCl  Na   Cl–, , (i = 2), , (ii) has same value of (i), 27. At 25°C, the highest osmotic pressure is exhibited by 0.1 M solution of, (1) Glucose, , (2) Urea, , (3) CaCl2, , (4) KCl, , Sol. Answer (3),  = i(CRT), , (1), , Glucose i = 1, , (2), , Urea i = 1, , (3) have maximum (i) so,, , (3), , CaCl2  Ca2  2Cl– (i  3), , It has maximum , , (4), , KCl  K   Cl– (i  2), ,   i if C = constant, , 28. According to Raoult’s law, the relative lowering of vapour pressure for a solution is equal to, (1) Mole fraction of solute (2) Mole fraction of solvent (3) Moles of solute, , (4) Moles of solvent, , Sol. Answer (1), , p,  x A  (mole fraction of solute), p, 29. The concentration units, independent of temperature, would be, (1) Normality, , (2) Weight volume percent (3) Molality, , (4) Molarity, , Sol. Answer (3), , VT, Normality (w/v)% molarity depends on volume so also depends on temperature but molality do not depend on, volume so do not depend on temperature., 30. In liquid-gas equilibrium, the pressure of vapours above the liquid is constant at, (1) Constant temperature (2) Low temperature, , (3) High temperature, , (4) None of these, , Sol. Answer (1), At constant temperature vapours formed by liquid also constant that's why vapour pressure becomes constant., 31. The vapour pressure of CCI4 at 25ºC is 143 mm Hg. If 0.5 gm of a non-volatile solute (mol. weight = 65) is dissolved, in 100 g CCI4, the vapour pressure of the solution will be, (1) 199.34 mm Hg, , (2) 143.99 mm Hg, , (3) 141.43 mm Hg, , (4) 94.39 mm Hg, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 65, , Sol. Answer (3), , p  p,  x (mole fraction of solute), p, 143 – ps, nA, ⇒, 143, nA  nB, 0.5, 143 – ps, 65, ⇒, 0.5 100, 143, , 65 154, , CCl4 = 12 + 35.5 × 4 = 12 + 142 = 154, , 143 – ps, 0.5, , 100  65, 143, 154, 143  ps , , 0.5  154  143, 110.11, , 100  65, 65  100, , ps = 141.3 mm Hg, 32. What is the molarity of H2SO4 solution, that has a density 1.84 gm/cc at 35ºC and contains 98% by weight, of solute?, (1) 18.4 M, , (2) 18 M, , (3) 4.18 M, , (4) 8.14 M, , Sol. Answer (1), , (H2SO4  98 g), , 98% by weight  98 g H2SO4 in 100 g of solution, M, , molar mass, , w, 98  1.84  1000, , ⇒ 18.4, M V, 98  100, , d, , m, V, , V, , m 100, , cc, d 1.84, , 33. A 5% (w/v) solution of cane sugar (mol. wt. = 342) is isotonic with 1% (w/v) solution of a substance X. The, molecular weight of X is, (1) 68.4, , (2) 171.2, , (3) 34.2, , (4) 136.8, , Sol. Answer (1), (5%) of cane sugar  1% of a substance (x), for isotonic 1 = 2, ,  = CRT, , C1 = C2, , n1 n2, , V, V, 5, 1, ⇒, 342, Mw, Mw ⇒, , 342,  68.4 g, 5, , 34. The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to, the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be 20 mm of mercury?, (1) 0.4, , (2) 0.6, , (3) 0.8, , (4) 0.2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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66, , Solutions, , Solution of Assignment, , Sol. Answer (2), , p,  x solute, p, 10,  0.2, p, , ...(i), , 20, x, p, , ...(ii), , (i) 10 p, 0.2, , , (ii) p 20, x, x  0.4, [xsolvent + xsolute = 1], xsolvent = 1 – xsolute = 1 –0.4 = 0.6, 35. If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by 0.216°C, than that of, the pure solvent. The molecular weight of the substance, (Molal elevation constant for the solvent is 2.16°C), is, (1) 10.1, , (2) 100, , (3) 1.01, , (4) 1000, , Sol. Answer (2), Tb  i(Kb × m), ⎡ 2.16  0.15  1000 ⎤, 0.216 ⇒ 1 ⎢, ⎥, Mw  15, ⎣, ⎦, , Mw = 100 g, 36. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte, solid, weighing 2.175 g is added to 39.08 of benzene. The vapour pressure of the solution is 600 mm of Hg., What is the molecular weight of solid substance?, (1) 65.2, , (2) 59.6, , (3) 49.50, , (4) 79.8, , Sol. Answer (1), , p  p,  x A (Solute), p, , 640 – 600, 2.175  78, ⇒, 640, Mw  39.08, , nA, n, p  p, ,  A, p, nA  nB, nB, , 40, 2.175  78, , 640 Mw  39.08, , Benzene (mol. mass) = C6H6  12 × 6 + 6  78, , 0.06 , Mw , , 2.175  78, Mw  39.08, , 2.175  78,  65.11 g, 39.08  0.06, , 37. From the colligative properties of solution which one is the best method for the determination of molecular weight, of proteins and polymers?, (1) Osmotic pressure, , (2) Lowering in vapour pressure, , (3) Lowering in freezing point, , (4) Elevation in boiling point, , Sol. Answer (1), Osmotic pressure is the best method, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Solutions, , 67, , 38. Molarity of liquid HCl, if density of solution is 1.17 gm/cc is, (1) 36.5, , (2) 18.25, , (3) 32.05, , (4) 42.10, , Sol. Answer (3), M, , W, M V, , d = 1.17 g/cc, , M, , 1.17  1000,  32.05, 36.5  1, , i.e., 1.17 g in 1 cm3, Molar mass (HCl) = 1 + 35.5 = 36.5, , 39. A solution contains non volatile solute of molecular mass M2. Which of the following can be used to calculate, the molecular mass of solute in terms of osmotic pressure? (m2 - mass of solute, V - volume of solution,,  - osmotic pressure), ⎛ m2 ⎞, ⎟ VRT, (1) M2  ⎜, ⎝  ⎠, , ⎛ m ⎞ RT, (2) M2  ⎜ 2 ⎟, ⎝ V ⎠ , , ⎛m ⎞, (3) M2  ⎜ 2 ⎟  RT, ⎝ V ⎠, , ⎛m ⎞ , (4) M2  ⎜ 2 ⎟, ⎝ V ⎠ RT, , Sol. Answer (2),  = CRT, , , , n, RT, V, , , , m2RT, M2  V, , ⎛ m ⎞ RT, M2  ⎜ 2 ⎟, ⎝ V ⎠ , , 40. A solution containing components A and B follows Raoult’s law, (1) A - B attraction force is greater than A - A and B - B, (2) A - B attraction force is less than A - A and B - B, (3) A - B attraction force remains same as A - A and B - B, (4) Volume of solution is different from sum of volume of solute and solvent, Sol. Answer (3), Raoults law, PT  PA  PB, , Solution which obey Raoult's law have same A-B attraction force as A-A and B-B attraction, 41. Formation of a solution from two components can be considered as, (i) Pure solvent  separated solvent molecules, H1, (ii) Pure solute  separated solute molecules, H2, (iii) Separated solvent and solute molecules  solution, H3, Solution so formed will be ideal if, (1) Hsoln= H1 + H2 + H3, , (2) Hsoln= H1 + H2 – H3, , (3) Hsoln= H1 – H2 – H3, , (4) Hsoln= H3 – H1 – H2, , Sol. Answer (1), For ideal solution Hsolution = H1 + H2 + H3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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68, , Solutions, , Solution of Assignment, , 42. Camphor is often used in molecular mass determination because, (1) It is readily available, , (2) It has very high cryoscopic constant, , (3) It is volatile, , (4) It is solvent for organic substances, , Sol. Answer (2), Camphor is often used in molecular mass determination because it has very high cryoscopic constant, 43. Which condition is not satisfied by an ideal solution?, (1) mix H = 0, , (2) mix V = 0, , (3) mix S = 0, , (4) Obeyance to Raoult's Law, , Sol. Answer (3), For ideal solution Hmix > 0, Hmix = positive, Vmix > 0, Vmix = positive, Smix > 0, , SECTION - D, Assertion-Reason Type Questions, 1., , A : Solubility of NaCl increases with temperature., R : Dissolution of NaCl is an endothermic process., , Sol. Answer (1), Solubility [for endothermix reaction] increases as temperature increases. [Hsolution (NaCl) = +ve], 2., , A : 10 ml of liquid A mixed with 20 ml of liquid B total volume of solution is 30 ml., R : A and B will form ideal solution., , Sol. Answer (1), For ideal solution, DVmix = 0,  Vf = V1 + V2., 3., , A : Lowering of vapour pressure depends upon concentration of solute., R : Relative lowering of vapour pressure is a colligative property., , Sol. Answer (1), Lowering of V-P depends only upon the concentration of solute, 4., , P0  P5, P0, ,  x solute ., , A : Boiling point of 0.1 M solution of NaCl is higher than that of 0.1 M solution of urea., R : Greater the value of Van't Hoff factor, greater the elevation in boiling point of solution containing non volatile, solute., , Sol. Answer (1), iNaCl = 2, iurea = 1, [Tb  i], 5., , A : Hexane and heptane forms ideal solution., R : H, S and G is zero for such type of solution., , Sol. Answer (3), H = 0, S > 0, G < 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 6., , Solutions, , 69, , A : Solution containing 1 gram equivalent of solute per litre is known as 1 N solution., R : N = M × n-factor., , Sol. Answer (2), Both statements are true., 7., , A : Observed molecular mass of CaCl2 determined by any colligative property is less than ideal molecular mass., R : CaCl2 gets ionised in water as it is a strong electrolyte., , Sol. Answer (1), CaCl2  Ca+2 + 2Cl– and colligative properties depends upon the number of particle in the solution, i = 3., 8., , A : Isotonic solutions must have same effective molarity., R : Effective molarity = M × i, , Sol. Answer (2), Isotonic  1 = 2, iC1RT = iC2RT, [iC1 = iC2], 9., , A : Sum of mole fraction of all components in a mixture is 1., R : Mole fraction is temperature dependent mode of concentration., , Sol. Answer (3), Mole fraction is independent of temperature., 10. A : Hmix and Vmix for an ideal solution is zero., R : A....B interaction in an ideal solution are same as between A.....A and B.....B., Sol. Answer (1), EA – A = EA – B = EB – B., Vmix = 0 = Hmix., 11. A : An azeotropic solution of two liquids has boiling point lower than either of them., R : Solution shows +ve deviation from ideal behaviour., Sol. Answer (1), Fact., 12. A : On increasing temperature vapour pressure of solution increases., R : Vapour pressure of ether is higher than alcohol., Sol. Answer (2), V.P.  temperature., and V-P of ether is higher than alcohol due hydrogen bonding., 13. A : Solubility of gas increases on increasing pressure., R : Solubility of gas decreases on decreasing the temperature., Sol. Answer (3), , Solubility of gas , , 1, Temperature, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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70, , Solutions, , Solution of Assignment, , 14. A : Raoult’s law applicable for dilute solution only., R : Henry’s law is applicable for solution of gas in liquid., Sol. Answer (2), Both statement are true and are different., 15. A : CuSO4.5H2O is a solution of liquid in solid., R : Solution is a homogeneous mixture., Sol. Answer (2), CuSO4.5H2O  5 water molecule is co-ordinated to CuSO4., Solution is always homogenious mixture., 16. A : 1 M solution and 1 molal solution contain same mass of solute., R : 1 M and 1 m aqueous solution are equally concentrated., Sol. Answer (3), 1 molar aqueous solution is less concentrated than 1 molar., 17. A : Boiling point of water at higher altitude is lower than 100°C., R : Boiling point is a colligative property., Sol. Answer (3), Yes, external pressure is lower than 1 atm and elevation in boiling point is a colligative property not boiling, point., 18. A : van’t Hoff factor for benzoic acid in aqueous medium is 2 assuming complete ionisation., R : van’t Hoff factor for 100% ionised solute equal to the number of ions produces., Sol. Answer (1), , COOH, =1, , , –, , COO, , +, , +H, , i 1, (n  2), n 1, , i2, i–1=n–1, , in, For 100% dissociation, van't Hoff factor is equal to number of ions produces., , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456