Paper I PHYSICS TEST CODE Paper II CHEMISTRY TEST CODE 1371
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1 Physics by Shiv R Goel, (B.Tech, IIT Delhi) WAVES Chemistry by Dr. Sangeeta Khanna, Ph.D. (CCC) MOCK TEST - NAME: ROLL NO. Time: 3 Hr. Date Max. Marks:98 BOOKLET A Paper I PHYSICS TEST CODE 370 Paper II CHEMISTRY TEST CODE 37 Subject-Chemistry Part : SECTION-A (Single Answer is Correct) Q. The vapour pressure of pure A and B are 0.0 and 0.04 bar respectively. Their liquid mixture behaves as an ideal solution. If the mole fraction of B in the liquid phase is 0.50, then its mole fraction in the vapour phase in equilibrium with the liquid phase is equal to: (a.) 0.50 (b.) 0.33 (c.) 0.67 (d.) 0.5 Ans. C Q. The actual freezing point of the solution is Ct. What percentage of the KI is dissociated if it is.4 M with density.5 gm/ml, K f =.86 K Kg/mole [At. Mass of I = 7, K = 39] (a.) 83% (b.) 9% (c.) 73% (d.) 50% Sol. A Mass of solvent = = = molality = = T f = i K f m 4.46 = i.86.3 i =.83 i = i + α α = 0.83 i.e. 83% Q.3 For oxidation reaction: Fe 0.95 O Fe O 3 If molecular weight of Fe 0.95 O be M, then the equivalent weight of Fe 0.95 O will be:
2 (a.) (c.) M M 0.95 (b.) (d.) Sol. D Fe 0.95 o Fe O mole iron in + state mole Fe in Fe in =.85 M M 0.85 Change in oxidation state = 0.85 x-factor = 0.85 Q.4 In a cubic closest packed structure of mixed oxides, the lattice is made up of oxide ions, 0% of tetrahedral voids are occupied by divalent X + ions and 50% of the octahedral voids are occupied by trivalent Y 3+ ions. The formula of the oxide is : (a.) X YO 4 (b.) X 4 Y 5 O 0 (c.) X 5 YO 0 (d.) XY O 4 Sol. Q.5 B x y O 0.4 X 4 y 5 O 0 The standard potentials are shown in the following diagram: I BrO BrO / Br III What is the value of E for the reaction path (III) shown in the above diagram? (a.) 0.97 volt (b.).6 volt (c.).45 volt (d.) 0.6 volt II Sol. D 0 BrO3 + 4e BrO ; E = 0.54V G = -4 F BrO + e Br ; E = 0.45V G = -F = 3 Br e Br ; E.07V G = -F.07 0 Over all reaction BrO3 + 6e Br E =? G = -6F E Q.6 5 ml of 0. M solution of transition metal M reacts completely with 50 ml of 0.0 M KMnO 4 in presence of H SO 4. Which is correct representation of change in oxidation state of M: (a.) M + M + (b.) M + M +3 (c.) M + M +4 (d.) M + M +5 Sol. C 5 x 0. x = 50 x 0.0 x x = 5 0. x = Q. For the reaction: 4KClO 3 3KClO 4 + KCl
3 7 d[kclo 3 ] If = k 4 [KClO3 ] dt d[kclo 4 ] = k 4 [KClO3 ] dt [KCl] = k3[kclo3 ], the correct relation between k, k and k 3 is: dt d 4 (a.) k = k = k 3 (b.) 4k = 3k = k 3 (c.) 3k = 4k = k 3 (d.) None of these d[kclo d[kclo d[kcl] Sol. C 3 ] 4 ] k k Rate = = = or [KClO [KClO [KClO 4 3 ] 4 3 ] 4 = = k 3 ] 4 dt 3 dt dt or 3k =4k =k 3 Q.8 A radioactive isotope has a half life of 5 days. Starting with 4g of the isotope, what will be mass remaining after 75 days? (a.).00 g (b.) 0.50 g (c.) 0.58 g (d.).58 g Sol. B After 3 half life amount left is = 0. 4 = Q.9 Ans. B Q. 0 In some reduction process, the concentration of a substance changes from 0.4M to 0. M in 0 minute and to 0.06 M in 0 minute. What is the order and the half-life of the reaction? (a.) order =, (c.) order = 0, t = 5 minute (b.) t = 0 minute (d.) order =, order =, The decomposition of ozone is believed to occur by mechanism: O 3 O + O (fast) O + O 3 O (slow) When the concentration of O is increased, the rate : (a.) Increases (c.) Remains the same (b.) Decreases t t = = (d.) Cannot be answered 0 minute 0 minute Sol. B For rate expression of this reaction see solved example 3. This is clear from the rate expression that rate decrease with increase in the conc. of O. Q. B A If all the atoms, on the shaded plane are removed then the molecular formula of the solid will be: (a.) A 5 B 7 (b.) A 7 B 5 (c.) AB (d.) A 3 B 4 Sol. C Effective number of atoms of B = 0 Effective number of atoms of A 5 = 4
4 Q. = = + = A :B = : Simplest formula of solid = AB = : Which one of the following pairs of solution can we expect to be isotonic at the same temperature? (a.) 0. M urea and 0. M NaCl (b.) 0. M urea and 0. M MgCl (c.) 0. M NaCl and 0. M Na SO 4 (d.) 0. M Ca(NO 3 ) and 0. M Na SO 4 Sol. D π[ca(no 3 ) ] = icrt = 3 0.RT = 0.3 RT π[na SO 4 ] = icrt = 3 0. RT = 0.3 RT Q. 3 In the following reaction, SO (g) + O (g) SO 3 (g) The rate of formation of SO 3 is 00 g min. Hence, the rate of disappearance of O is: (a.) 50 g min - (b.) 0 g min - (c.) 00 g min - (d.) 00 g min - Sol. B In the reaction, SO (g) + O(g) SO3 (g) mol mol 3g 60 g 60 g of SO 3 is formed by 3 g of O 3 00 g of SO 3 will be formed by 00 or 0 g of O 60 Rate of disappearance of O = 0 g min Q. 4 Bombardment of aluminium by α - particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products X, Y and Z respectively are: (ii) 7 3 Al 5 P + Y (i) Sol. C Q. 5 (a.) proton, neutron, position (b.) neutron, positron, proton (c.) proton, positron, neutron (d.) positron, proton, neutron Al + He 4 Si + X (X = Al + He 5 P + Y (Y = 0 H) H) 5 P 4 Si + Z (Z = + e) If Ca 3 (PO 4 ) and H 3 PO 3 contain same number of P atom then the ratio of oxygen atoms in these compounds respectively is: 4 Si + X 4 Si + Z (a.) 8/3 (b.) /3 (c.) 3 (d.) 4/3 Sol. D Let number of moles of Ca 3 (PO 4 ) and H 3 PO 3 are x and y respectively.
5 Q. 6 x = y Moles of 'O' in Ca3(PO4 ) 8x 4 = = Moles of 'O' in H3PO3 3y 3 A non stoichiometric compound Cu.8 S is formed due to incorporation of Cu + ions in the lattice of cuprous sulphide. What percentage of Cu + compound? ion in the total copper content is present in the (a.) (b.). (c.) 99.8 (d.) 89.8 Sol. B Let x Cu + and (L8 x) Cu + ions are present in the compound Cu.8 S. Compound is electricially neutral. +x + (.8 x) = x = % Cu + = 00 =..8 Q. 7 At 0 K, solubility of a gas in a liquid was measured at different partial pressures. Mole fraction of Gas Partial pressure of Gas (kpa) 8 66 Which of the following graph is correct in accordance with the Henry s law? (a.) (b.) Partial pressure k H= 8000 kpa Partial pressure k H= kpa Mole fraction Mole fraction (c.) (d.) k H= kpa Partial pressure k H= kpa Partial pressure Ans. B Q. 8 Mole fraction Mole fraction Plots showing the variation of the rate constant k with temperature (T) are given below. The plot that follows Arrhenius equation is: (a.) (b.) K K T T (c.) (d.) K K Sol. A Arrhenius equation is: Ee /RT k = Ae T T
6 Rate constant k will increase exponentially with increase in temperature (as shown below) K T SECTION-B (Multiple Answer is Correct) Q. Which of the following are incorrect statements Sol. B,C,D (A) (B) (C) (D) (a.) Spontaneous adsorption of gases on solid surface is an exothermic process as entropy decreases during adsorption (b.) Formation of micelles takes place when temperature is below Kraft Temperature (T k ) and concentration is above critical micelle concentration (CMC) (c.) A colloid of Fe(OH) 3 is prepared by adding excess of NaOH in FeCl 3 solution, the particles of this sol will move towards cathode during electrophoresis (d.) According to Hardy-Schulze rules, the coagulation (flocculating) value of Fe 3+ more than Ba + or Na + G = H - T S < O as S < O so H has to be negative micelles formation will take place above T k and above CMC this sol will be negatively charged & will move toward anode Fe 3+ ions will have greater flocculatibility power so smaller flocculating value ion will be Q. In which of the following pairs of solutions will the values of the vant Hoff factor be the same? (a.) 0.05 M K 4 [Fe(CN) 6 ] and 0.0 M FeSO 4 (b.) 0.0 M K 4 [Fe(CN) 6 ] and 0.05 M FeSO 4 (NH 4 ) SO 4.6H O (c.) 0.0 M NaCl and 0.0 M BaCl (d.) 0.05 M FeSO 4 (NH 4 ) SO 4.6H O and 0.0 M KCl. MgCl.6H O Sol. B,D Number of particles from K 4 [Fe(CN) 6 ] = 6 Number of particles from FeSO 4 (NH 4 ) SO 4.6H O = 5 Number of particles from KCl.MgCl.8H O = 5 Q.3 An electrochemical cell is set up as follows: Pt (H, atm.) 0. M HCl 0. M CH 3 COOH (H, atm.) Pt E.M.F. of the cell will not be zero because (a.) the temperature is constant (b.) the ph of 0. M HCl and 0. M CH 3 COOH is not the same
7 (c.) acids used in the cell have different K a values (d.) E.M.F. of above cell depends on molarities of the acids used Ans. B,C Q.4 Which of the following statement is incorrect? (a.) doping of As in silicon makes it a semi-conductor of p-type (b.) doping of B in silicon make it a semi-conductor of n type (c.) doping of Cd & Se in silicon forms a binary compound which act as semiconductor (d.) doping of Ge with As make it a semi-conductor of n type Ans. A,B Q.5 The vapour of an organic compound requires three times its own volume of oxygen for complete Ans. B,D combustion and produces twice its own volume of carbon dioxide. Which of the following compounds would give these results? (a.) CH 3 CHO (b.) C H 5 OH (c.) CH 3 CO H (d.) C H 4 SECTION-C (Comprehension Type Questions) Comprehension ( Next 3 Questions) The minimum concentration of the electrolyte in millimoles per litre of the solution, required to cause coagulation of a particular sol is called coagulation value of flocculation value of the electrolyte for the sol. The ions carrying opposite charge to that of sol particles are effective in causing coagulation of the sol. Coagulating power of an electrolyte is directly proportional to the valency of the active ions i.e., ions causing coagulation. Coagulation or flocculation is the precipitation of a colloid through induced aggregate of its particles. There are several methods for coagulation of colloidal solution like by repeated dialysis, by adding electrolyte, by electrophoresis, etc. Q. Which of the following will have highest coagulating power for As S 3 colloid? (a.) Al 3+ (b.) Na + (c.) SO (d.) 4 PO Sol. A As S 3 is negative colloid hence Al 3+ will be most effective for its coagulation Q. In the coagulation of positive sol, the flocculation power of Cl 3 4, SO 4, PO 4, [ Fe(CN) 6 ] are in the order of : (a.) (c.) Cl > SO 4 > PO 4 > [Fe(CN) 6 ] (b.) 4 3 [Fe(CN) 6 ] > PO 4 > SO 4 > Cl 4 Cl > PO 3 4 > SO 4 > [Fe(CN) 6 ] (d.) 4 3 Cl > SO 4 > [Fe(CN) 6 ] > PO 4 Sol. B Greater is the charge of ion, more effective is the coagulation of oppositely charged colloid Q.3 Flocculating value of ion depends on: Ans. D (a.) the shape of flocculating ion (b.) the amount of flocculating ion (c.) nature of the charge on the flocculating ion (d.) both, the nature and magnitude of the charge of the flocculating ion Comprehension ( Next 3 questions)
8 Faraday s law of electrolysis are used for quantitative estimation of products of electrolysis. According to which, mass deposited at electrode depends on quantity of electricity passed. Product of electrolysis depends on discharge potential, concentration & nature of electrode. Answer the following questions: Q.4 If same amount of electricity is passed through aqueous solution of AgNO 3 and CuSO 4 and number of Cu and Ag atoms deposited are x and y respectively, then (a.) x = y (b.) x = 3y (c.) x = y/ (d.) x = y Sol. C Ag + + e Ag; Cu + + e Cu Thus, number of Ag atoms will be twice the number of Cu atoms. Q.5 The passage of 5 milliampere of current through molten CaCl for 60 seconds will cause the deposition of N calcium atoms on cathode. The value of N is (a.) (b.) 0 8 (c.) (d.) Sol. D Q = 0.05 (amp) 60(s) =.5 C C deposit Ca atoms = C deposit Ca atoms = = Q.6 The time required for the complete decomposition of two moles of water using a current of Sol. A ampere is (a.) sec (b.) sec (c.) sec (d.) sec H O H + O Electrons involved = = t t = = sec. Comprehension 3 ( Next 3 questions) The rates of reaction between the reactants A and B were studied starting with different initial concentrations. The following data were obtained:.. 3. [A], mol lit [B], mol lit Initial rate, mol lit - s - at 0 K (mol l s ) at K Q.7 The order of reaction with respect to A and B are respectively (a.), (b.), (c.), (d.), 0 Sol. C using (ii) & (iii) from table we get = [0.5].6 0 [0.5] = [0.5] α α = from (i) & (ii) we get α.5 0 = = (0.5) (0.5) β 4 4 α β
9 (0.5) β = 0.5 β = Rate = KI [A] [B] Order wit h respect to A = B = Q.8 The rate constant of the reaction at K when concentrations are in mol L and time in seconds will be Sol. D (a.) (b.) (c.) (d.) K = = 0 3 ( ) (3 0 5 ) = K = Q.9 What is rate of reaction at in nd experiment? (a.) (b.) (c.) 0-3 (d.) 0-3 Sol. B Rate = [5 0-4 ] [6 0-5 ] = SECTION-D Matrix Match Type Questions Q. Match column I with column II. Column-I Column- II (i) (a.) x t (ii) (b.) Rate conc. [A] (iii) (c.) 3 [A]
10 (iv) (d.) 0 Rate [A] 3 Ans. i s; ii p; iii q; iv r Q. Match Column - I with Column - II and select the correct answer using codes given below the lists. Column-I Column- II S S (i) 0.M sucrose 0.M urea (a.) S and S are isotonic (ii) 0.M NaCl 0.M KCl (b.) No osmosis takes place when S and S are separated using semipermable membrane (iii) 0.M Na SO 4 0.M KCl (c.) S is hypertonic to S (iv) 0.M glucose 0.M CuSO 4 (d.) S is hypotonic to S Ans. i AB; ii AB; iii C ; iv D Q.3 Match Column- I with Column II Column-I Decay process Column- II No. of particles emitted (i) (ii) (iii) (iv) Th 38 9 U 35 9 U Pb Pb Pb 90 Th 8 Pb Ans. i AB; ii AB; iii C ; iv D (a) 34 = x = 4x 8 = 4x x = 7 90 = y y = y = 6 (a.) (b.) (c.) (d.) 7α 6β 8α 6α SECTION-E (Integer Type Questions) Q. The rate expression for reaction A(g) + B(g) C(g) is rate = k[a] / [B]. What will be the rate if initial concentration of A and B increase by factor 4 and respectively? Ans. 8 Q. Density of lithium atom is 0.53 g/cm 3. The edge length of Li is 3.5 Å. The number of lithium atoms in a unit cell will be. (Atomic mass of lithium is 6.94) Ans. Q.3 What is the valency of an element of which the equivalent weight is and specific heat is 0.5 Cal/gm? Ans. atomic mass = ; Valency= = 0.5
11 Q.4 The shaded plane represents diagonal plane of symmetry. How many such planes are possible in the cubic system? A B D C Ans. 6 Q.5 Thus conductivity of 0.0 mol/dm 3 aqueous acetic acid at 0K is ohm cm and the limiting molar conductivity of acetic acid at the same temperature is 390 ohm cm mol. The percentage degree of dissociation of acetic acid is 000 Sol. Λ m = K = =9.5, α = = 0.05 M Percentage α = 5
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