Solution and Colligative Properties Mock Test

MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1. Three solutions are prepared by adding ‘w’ gm of ‘A’ into 1 kg of water, ‘w’ gm ‘B’ into another 1kg of water and ‘w’ gm of ‘C’ in another 1 kg of water (A,B,C are non electrolytic). Dry air is passed from these solutions in sequence (A  B  C). The loss in weight of solution A was found to be 2gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A,B and C is : (A) MA : MB : MC = 4 : 3 : 5 (B) MA : MB : MC = (C) MC > MA > MB (D)MB > MA > MC 1 1 1 : : 4 3 5 2. Barium ions, CN– and Co+2 form an ionic complex. If this complex is 75% ionised in aqueous solution with Vant Hoff factor (i) equal to four and paramagnetic moment is found to be 1.73 BM(due to spin only) then the hybridisation state of Co(II) in the complex will be : (A) sp3d (B) d2sp3 (C) sp3d2 (D) dsp3 3. The incorrect statement is : (A) Vapour pressure of a liquid always increases by increasing the temperature. (B) Vapour pressure only depends on temperature and not on the nature of the substance. (C) Vapour pressure does not depend on the quality of the liquid taken and the surface area of the liquid. (D) Vapour pressure is not a colligative property and is independent of the concentration of a liquid. 4. Relative decrease in in vapour pressure of an aqueous solution containing 2 moles [Cu(NH3)3Cl]Cl in 3 moles H2O is 0.50. On reaction with AgNO3, this solution will form (A) 1 mol AgCl (B) 0.25mol AgCl (C) 2 mol AgCl (D) 0.40 mol AgCl A solution of x moles of sucrose in 100gms of water freezes at – 0.2°C. As ice separates the freezing point goes down to 0.25°C. How many grams of ice would have separated? (A) 18 grams (B) 20 grams (C) 25 grams (D) 23 grams A sample of air is saturated with benzene ( vapour pressure = 100 mm Hg at 298K) at 298 K, 750mm Hg pressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the system is (A) 2250 torr (B) 2150 torr (C) 2050 torr (D) 1950 torr Available solutions are 1L of 0.1 M NaCl and 2L of 0.2 M CaCl2. Using only these two solutions what maximum volume of a solution can be prepared having [Cl–]= 0.34 M exactly. Both electrolytes are strong (A) 2.5L (B)2.4L (C)2.3L (D) None of these Calculate the osmotic pressure of the solution prepared in the above ques T = 300K, ( R = 0.082 L atm mol–1 K–1) (A) 10.8 atm (B)12.8 atm (C) 5.6 atm (D) None of these 5. 6. 7. 8. 9. Consider equimolal aqueous solutions of NaHSO4 and NaCl with Tb and T'b as their respective boiling point Tb elevations. The value of mLt will be  0 T ' b (A) 1 10. (B) 1.5 (C) 3.5 (D) 2/3 The vapour pressure of benzene, toulene and a xylene are 75 Torr, 22 Torr and 10 Torr at 20°C. Which of the following is not a possible value of the vapour pressure of an equimolar binary/ternary solution of these at 20°C ? Assume all form ideal solution with each other. (A) 48.5 (B) 16 (C) 35 2 3 (D) 53.5 SOLUTION AND COLLIGATIVE PROPERTIES 11. 3.24g of Hg(NO3)2 (molar mass = 324) dissolved in 1000g of water constitutes a solution having a freezing point of –0.0558°C while 21.68 g of HgCl2 (molar mass = 271) in 2000g of water constitutes a solution with a freezing point of –0.0744°C. The Kf for the water is 1.86 K  kg . About the state of ionization of these two solids in water it can be mol inferred that : (A) Hg(NO3)2 and HgCl2 both are completely ionized. (B) Hg(NO3)2 is fully ionized but HgCl2 is fully unionized. (C) Hg(NO3)2 and HgCl2 both are completely unionized. (D) Hg(NO3)2 is fully unionized and HgCl2 is fully ionized. 12. For a solution of 0.849 g of mercurous chloride in 50 g og HgCl2 () the freezing point depression is 1.24°C. Kf for HgCl2 is 34.3. What is the state of mercurous chloride in HgCl2? ( Hg – 200, Cl – 35.5). (A) as Hg2Cl2 molecules (B) As HgCl molecules (C)As Hg+ and Cl– ions (D)As Hg 22 and Cl– ions 13. Assuming that the formation of an ideal solution, determine the boiling point of a mixture containing 1560 g benzene (molar mass = 78) and 1125 g chlorobenzene ( molar mass = 112.5) using the following against an external pressure of 1000 torr. benzene 2200 1800 Vapour 1350 chlorobenzene 1000 Pressure 540 400 300 200 (A)90°C (B)100°C 90 100 110 120 t(c°) (C)110°C (D)120°C SECTION - II : MULTIPLE CORRECT ANSWER TYPE 14. Dry air is slowly passed through three solutions of different concentration , c1, c2, c3 ; each containing (non volatile) NaCl as solute and water as solvent, as shown in the figure. If the vessel 2 gains weight, then Vessel-1 (A) c1 > c2 (B) c1 < c2 Vessel-2 Vessel-3 (C) c1 < c3 (D) c2 > c3 15. In which of the following pairs of solution will have the values of Vant hoff factors can be the same? (A) 0.05 M K4[Fe(CN)6] and 0.10 M FeSO4 (B) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4(NH4)2SO4.6H2O (C) 0.20 M NaCl and 0.10 M BaCl2 (D) 0.05 M FeSO4(NH4)2SO4.6H2O and 0.02 M KCl.MgCl2.6H2O 16. Vapour pressure of solution containing 6g of a non-volatile solute in 180 g water is 20.0 Torr. If 1 mole water is further added vapour pressure increases by 0.02 Torr. Which of the following is true? (A) The molecular weight of solute is 54g/mol (B) The vapour pressure of pure water is 20.22 Torr. (C) Addition of more water in the solution will further raise the vapour pressure of solution. (D) The vapour pressure of pure water is 22.22 Torr. SECTION - III : ASSERTION AND REASON TYPE 17. Statement - 1 When a cell is placed in hypertonic solution, it shrinks. Statement - 2 Reverse osmosis is used for desalination of water. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 18. Statement - 1 The difference in the boiling points of equimolar solution of HCl and HF decreases as their molarity is decreased. Statement - 2 The extent of dissociation decreases steadily with increasing dilution. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 19. Statement - 1 The molar mass obtained for benzoic acid in benzene is found to be nearly 244. Statement - 2 Benzoic acid has the formula HOOC COOH (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 20. Statement - 1 When ‘a’ mL of a 0.1 molal urea solution is mixed with another ‘b’ mL of 0.1 molal glucose solution, the boiling point of the solution is no different from the boiling points of the samples prior to mixing but if ‘a’ mL of 0.1 molal urea is mixed with ‘b’ mL of 0.1 molal HF the boiling point of the mixture is different from the boiling points of the separate samples. Statement - 2 HF is an electrolyte (weak) whereas glucose is a non electrolyte. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. SOLUTION AND COLLIGATIVE PROPERTIES SECTION - IV : COMPREHENSION TYPE Read the following comprehensions carefully and answer the questions. Ideal Solution at Fixed Temperature Consider two liquid ‘B’ and ‘C’ that form an ideal solution. We hold the temperature fixed at some value T that is above freezing points of ‘B’ and ‘C’. We shall plot the system’s pressure P against xB, the overall mole fraction of B in the system : xB = n B,total n total  n lB  n vB n Bv  n lC  n lB Where n lB and n Bv are the number of moles of B in the liquid and vapor phases, respectively. For a closed system xB is fixed, althoud n lB and n Bv may vary.. Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath. To see what the P-versus-xB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figure) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition x lB where x lB at D is equal to the overall mole fraction xB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoult’s law, PB  x Bv PB0 relates the vapour-phase mole fractions to the liquid composition as follows : x 0B  X lB PB0 x l P0 v and x C  C C P P ...(A) Where PB0 and PC0 are the vapour pressures of pure ‘B’ and pure ‘C’ at T, where the system’s pressure P equals the l sum PB + PC of the partial pressures, where PB  n lB , and the vapor is assumed ideal. (n lB  n lC (Only liquid) (First vapor appears) System Constant- T bath x Bv x lB PB*  x Cv x lC PC* ideal solution (Last liquid vaporizes) (Mole fraction of B) ...(B) v v l l Let B be the more volatile component, meaning that PB0  PC0 . Above equation then shows that x B / x C  x B / x C . The vapor above an ideal solution is richer than the liquid in the more volatile component. Equations (A) and (B) apply at any pressure where liquid–vapor equilibrium exists, not just at point D. Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, we reach point F in the figure, where the last drop of liquid vaporizes. Below F, we have only vapor. For any point on the line between D and F liquid and vapor phases coexist in equilibrium. 21. 22. If the above process is repeated for all other compositions of mixture of C and B. If all the points where vapours start converting into liquid are connected and all the points where vapours get completely converted into liquid are connected obtained graph will look like. (1) (2) (3) (4) The equation of the curve obtained by connecting all those points where the vapors of above mixture (all mixtures of different composition are taken) just start forming will be (A) P = P°C + (P°B – P°C) XB (C) P  23. PB0 PC0 X B (PC0  PB0 )  PB0 (B) P = P°B + (P°B – P°C) XB (D) P  PB0 PC0 X lB (PC0  PB0 )  PB0 Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapor pressure of 700 Torrs at 80ºC. It is distilled till 2/3rd of the solution (2/3rd moles out of total moles) is collected as condensate. The composition of the condensate is x'A = 0.75 and that of the residue is x''A = 0.30. If the vapor pressure of the residue at 80ºC is 600 Torrs, find the original compositions of the liquid. (A) XA = 0.5 (B) XB = 0.6 (C) XA = 0.6 (D) XB = 0.3 SOLUTION AND COLLIGATIVE PROPERTIES SECTION - V : MATRIX - MATCH TYPE 24. 25. Match the following : Column I Column II (A) Pyridine + CH3COOH (p) Smixing = +Ve (B) C6H5Cl + C6H6Br (q) Vmixing = +Ve (C) CHCl3 + CCl4 (r) Gmixing = –Ve (D) H2O + HClO4 (s) Hmixing = –Ve Match the following : Column I Column II (A) Relative lowering in vapour pressure (p) negative deviation from ideal behaviour (B) Depression in freezing point (q) Walker and Ostwald Method (C) Hmix < Zero (r) Beckmann thermometer (D) Osmotic pressure (s) Berkeley and Hartley’s method SECTION - VI : SUBJECTIVE TYPE 26. Barium ions, CN– and Co form a water soluble ionic complex with Ba2+ as free cations. For a 0.01 M solution of this complex osmotic pressure is observed to be equal to 0.984 atm and degree of dissociation is found out to be 75%. Then find the coordination number of Co(II) in this complex. +2 ( T = 300K) (R = 0.082 L atm mol-1 K-1). 27. What is the osmotic pressure of a solution (in Torr) made by mixing 20mL of 0.01 M BaCl2 with 30mL of 0.01M NaF at 27°C ? Ksp (BaF2) = 2.4 × 10–5. R = 0.082 lit atm mol–1 K–1, 1 atm= 760Torr. 28. The composition of vapour over a binary ideal solution is determined by the composition of a liquid. If XA and YA are the mole fractions of A in the liquid and vapour , respectively , find the value of XA for which YA–XA has a maximum value.What is the value of pressure at this composition in terms of P°A and P°B. 29. When 2g of non volatile hydrocarbon containing 94.4 percent carbon is dissolved in 100g benzene, the vapour presure of benzene is lowered from 74.66 torr to 74.01 torr. Determine the molecular formulae for the hydrocarbon. 30. A saturated solution of sparingly soluble salt MCl2 has a vapour pressure of 31.96 torr while pure water has a vapour pressure of 32.0 torr at the same temperature. Calculate the solubility of the product of MCl2. MOCK - TEST 1. C 2. D 3. B 4. A 5. B 6. C 7. A 8. B 9. B 10. D 11. B 12. A 14. B,D 15. B,D 16. A,B,C 17. B 18. C 19. C 20. A 21. B 22. A 23. C 24. A  p,r,s ; B p,r ; C  p,q,r ; D  p,r,s 25. Aq ; Br ; Cp ; Ds 13. B 173