CHEMISTRY-15-10- 11th (PQRS) Code-A SOLUTION

REVIEW TEST-5 Class : XI (P,Q,R,S) PAPER CODE : A Time : 3 hour Max. Marks : 228 INSTRUCTIONS 1. The question paper contain 2-parts. Part-A contains 54 objective question, Part-B contains 2 "Match the Column" questions. All questions are compulsory. PART-A (i) Q.1 to Q.45 have only one correct alternative and carry 3 marks each. There is NEGATIVE marking and 1 mark will be deducted for each wrong answer. (ii) Q.46 to Q.54 have More than one are correct alternative and carry 5 marks each. There is NO NEGATIVE marking. Marks will be awarded only if all the correct alternatives are selected. PART-B (iii) Q.1 to Q.9 are "Match the Column" type which may have one or more than one matching options and carry 8 marks for each question. 2 marks will be awarded for each correct match within a question. There is NO NEGATIVE marking. 2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet. 3. Use only HB pencil for darkening the bubble. 4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only. USEFUL DATA Atomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14, Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Pt = 195, PART-A For example if only 'B' choice is correct then, the correct method for filling the bubble is A B C D For example if only 'B & D' choices are correct then, the correct method for filling the bubble is A B C D the wrong method for filling the bubble are The answer of the questions in wrong or any other manner will be treated as wrong. PART-B For example if Correct match for (A) is P, Q; for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling the bubble is P Q R S (A) (B) (C) (D) XI (PQRS) CHEMISTRY (15-10-2006)CODE-A REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.16 For a general atom A having three different existance A, A–, A+1, Mark the correct option. (A) Size order will be A+ > A– > A (B*) E A order will be A+ > A > A– (C) I E order will be A > A– > A+ (D) E N order will be A– > A > A+ [Sol: (B) size order will be A– > A > A+ ; EA order will be A+ > A > A– IE order will be A+ > A > A– ; EN order will be A+ > A > A– so the correction order will be of EA.] Q.17 Alarge number of oil drop samples in Milikan’s oil drop experiment gave the following values of charges q = 2×10–19 coulomb, 3×10–18 coulomb, 4 ×10–19 coulomb, 5 ×10–18 coulomb & no other values. What could be concluded as the charge of e– from above information. (A) 2 × 10–18 coulomb (B*) 2 × 10–19 coulomb (C) 1 × 10–19 coulomb (D) 5 × 10–20 coulomb [Sol: (B) q = 2 × 10–19, 3 × 10–8, 4 × 10–19 & 5 × 10–18 coulombs. So the highest common factor = 2 × 10–19 ] Q.18 Certain amount of phosphorus (P4) was made to react with certain amount of oxygen to give a mixture of P4O6 & P4O10 in the molar ratio of 2 : 1 (P4O6 : P4O10). If none of the reactants remained after the reaction(s) then what was the ratio of mass of P4 : O2 taken initially. [Sol: (A*) 93 : 88 (A) (B) 88 : 93 (C) 3 : 11 (D) 11 : 3 P4 + a 3O2  P4O6 3a a P4 b here + a = 2 5O2  5b P4O10 b b 1  moles of P4 reacted = a + b & moles of O2 reacted = 3a + 5b  mass ratio of P4 : O2 = (a + b)124 (3a + 5b)32 3124 = 11 32 93 = 88 ] Q.19 For a closed (not rigid) container containing n = 10 moles of an ideal gas, fitted with movable, frictionless, weightless piston operating such that pressure of gas remains constant at 0.821 atm, which graph represents correct variation of log V vs log T where V is in lit. & T in Kelvin. (A*) (B) (C) (D) [Sol: (A) PV = nRT  nR  V =   T   10  0.0821  V =   0.821  T = T  log V = log T so the curve is ] Q.20 Calculate total maximum mass (kg) which can be lifted by 10 identical balloon (each having volume 82.1 lit. and mass of balloon & gas = 3 kg) at a height 83.14 m at Mars where g = 5 m/s2 & atmosphere contains only Ar (At. wt.40). At Mars temperature is 10 K and density of atmosphere at ground level is 2 k gm/lit. [Given : e–0.1 = 0.9] (Assume d 0.821 H = d0 eMgh RT to be applicable). (A*) 2000 × 0.81–30 (B) 1970 (C) 2000 × 0.9 –30 (D) 2000 × 0.81 – 3 – Mgh [Sol: (A) d = d0 2 d = 0.821 e RT – 40103583.14 e 8.31410 2 = 0.821 e–0.2 = 2  (0.9)2 0.821 Now at height h = 83.14 m V × d = m1 a4ss4of4 b4al4loo4n4s +2 m4as4s 4of 4ga4s 4fil4le3d + mass of load  82.110  (0.9)2 0.821 = (3 × 10) + m  m = (2000 × 0.81) – 30 Kg ] Q.21 Consider the following nuclear reactions involving X & Y. X  Y + 4 He Y  18 1 If both neutrons as well as protons in both the sides are conserved in nuclear reaction then identify period number of X & moles of neutrons in 4.6 gm of X (A) 3, 2.4 NA (B*) 3, 2.4 (C) 2, 4.6 (D) 3, 0.2 NA [Sol: (B) Y  18 1  Y is 19Y so X is X  Y + 23 11 4 He Then atomic no. of X = 11 mass no. of X = 23 Electronic configuration of X = 1s2, 2s2, 2p6, 3s1  period no. = 3 4.6 moles of X = 23 = 0.2 & moles of neutrons in X = 0.2 × 12 = 2.4 ] Q.22 40 miligram diatomic volatile substance (X2) is converted to vapour that displaced 4.926 ml of air at 1 atm & 300 K. Atomic weight of element X is (A) 400 (B) 40 (C) 200 (D*) 100 [Sol: (D) PV = nRT 40 103 1 × 4.926 × 10–3 = M M = 200  Atomic mass of A = × 0.0821 × 300 200 2 = 100 ] Q.23 Electromagnetic radiations having  = 310.5 Å are subjected to a metal sheet having work function = 12.7 eV. What will be the velocity of photoelectrons with maximum Kinetic Energy.. (A) 0, no emission will occur (B) 2.18 × 106 m/sec (C) 3.1 × 107 m/sec (D*) 3.1 × 106 m/sec [Sol: (D) (KE)max = 1242 31.05 – 12.7 = 27.3 eV  ½ mv2 = 27.3 × 1.6 × 0–19 27.31.6 1019  2  v = 9.11031  v = 3.1 × 106 m/sec. ] Q.24 An U tube containing ideal gas closed at both ends having infinite long columns consists of mercury having initial height difference of 76 cm as shown. Both left column & right column have a hole from which gas is coming out according to rate law as given. [Assume outside the U-tube to be a perfect vaccum at all the instant] Given: ln2 = 0.693 Left Column dP = K P ; K = 0.693 × 10–2 sec–1 dt 1 1 dP Right Column = K P ; K = 1.386 × 10–2 sec–1 dt 2 2 where P = Pressure in the respective column If at t = 0 pressure in Left Column = 1atm then what will be the height difference in the two levels after 200 sec. (A*) 0 (B) 76 (C) 38 (D) 19 [Sol: (A) At t = 0, pressure in left column = 1 atm = 76 cm of Hg  pressure in right column = 76 + 76 = 152 cm of Hg In left column In the right column P dP t P dP t  = –K1  dt P0 0  ln P = –K  P0 × t  P = –  K2 dt 0 ln P = –K × t P0 1 P0 2 ln P = –0.693 ×10–2 × 200 ln 76 P P = –1.386 × 10–2 × 200 152 152 ln 76 = 0.693 × 2 ln P = 1.386 × 2  ln 76 P = 2 × 0.693 = 2 ln2 152 = 4ln 2 = ln24 P  76 = 22  P = P 76 152 16 P = 4 = 19 cm P = 9.5 cm of Hg Thus before 200 sec, pressure in both columns become equal so height difference in the two columns are equal. ] Q.25 A very long rectangular box is divided into n equal compartments with (n–1) fixed SPM (semi permeable membrane) numbered from 1 to (n–1) [n is unknown] as shown. The gases are initially present in only I compartment & can pass through only those SPM whose number is less than or equal to their subscript (like A1 can pass through 1st SPM only, A2 through 1st & 2nd & so on). If initially all gases have same moles & after substantial time ratio of P in 3rd compartment to P in 1st compartment is 5 then what 6 n1 would be the value of n (where PA6 & PAn1 represents partial pressure of A6 & An1 respectively) is (A) 10 (B) 15 (C*) 35 (D) 30 [Sol: (C) Let initially moles of A6 & An – 1 = x x finally moles of A6 = 7 x & moles of An – 1 = n P  P n1 x / 7 = x / n = 5 n = 35 ] Q.26 A mixture of CH4 (15 ml), CO (10 ml) is mixed with just sufficient O2 gas in an eudiometrytube(operating at 1 atm & 500 K) is subjected to sparking and then cooled back to the original temperature of 500K. If Vc1 be volume contraction due to the above process and Vc2 be the contraction after passing the resultant gas(es) through CaCl2 (anhydrous), then Vc1,Vc2 will be (A) 30, 0 (B) 30, 30 (C*) 5, 30 (D) 5, 0 [Sol: (C) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 15 30 15 30 CO(g) + ½ O2(g)  CO2(g) 10 5 10  V = [(15 + 10) + (35)] – [30 + 15+10] = 5ml 1 & V = 30 ml ] 2 Q.27 If in Bohr’s model, for unielectronic atom, time period of revolution is represented as Tn,z where n represents shell no. and Z represents atomic number then the value of T1,2 : T2,1 will be (A) 8 : 1 (B) 1 : 8 (C) 1 : 1 (D*) None of these n3 [Sol: (D) Tn, z  z 2 (1)3 (2)3 T1, 2  (2)2 & T2, 1  (1)2 T1,2  = 1  1 = 1 ] T2,1 4 8 32 Q.28 Assuming Heisenberg Uncertainity Principle to be true what could be the minimumuncertaintyin de-broglie wavelength of a moving electron accelerated by Potential Difference of 6 V whose uncertainty in position 7 is 22 n.m. (A) 6.29 Å (B) 6 Å (C*) 0.629 Å (D) 0.3125 Å h [Sol: p=  p = h  2 h So x × p = 4  x × h 2  = h   = 4 2 4x h2 = 2mev × 1 4x (6.631039 )2  22 = 2(9.11031 )(1.6 1019 )(4  22  7 109 ) = 6.29 × 10–11 m = 0.629 Å. ] Questions No. 29 to 30 (2 questions) Read the following comprehension and answer the question that follow. One of the concerns of various chemists is of selecting an appropriate equation of state which can relate the four parameters (P, V, T and n) & give results close to experimental values at critical conditions. Although Vander Waal equation is easy to use, the experimental value of 'Z' (compressibility factor) at critical condition does not match with the predicted value [Experimental :  0.27, Predicted (Vander Waal)  0.375] For this reason, another equation of state is being used known as  RT  a V RT Dieterici Equation P =   e m  V  b   m  where a, b are dieterici constant, Vm = Molar volume, P = Pressure & T = Temperature which gives better results at Critical Condition. ( ZCritical Condition = 2/e2).  dP   d2P  Also at Critical condition   = 0 &  2  = 0.  dV T   T If PC, VC, TC are critical pressure, volume & temperature and V & T are related as 2a × (V –b)2 = V3RT . C C C C C Q.29 PC, VC, TC in terms of dieterici constant will be respectively (A*) a 4b2e2 , 2b, a 4bR (B) a 27b2 , 3b, 8a 27 Rb (C) a 4b2e2 , 3b, a 4bR (D) a b2e2 , 2b, a 4bR Q.30 If Vr = V (reduced volume), Tr = C T P (reduced temperature) & Pr = C C (reduced pressure) then which of the following represents reduced equation of state for dieterici equation. [All equation dimensionally correct]  2  2    VrTr  1  e Tr V T e2T (A*) Pr e  Vr   =   2 (B) Pr e r r (Vr – b) = r 2 2a  1  e2T 2  1  e2T T (C) P e VrTr V   = r (D) P e VrTr V   = r C r  r 2  2 r  r 2  2 [Sol: ZC = PCVC RTC 2 = e2  PC = 2RTC VCe2 ..........(1) Now 2a(VC – b)2 = VC3 RTC (2) a If we put VC = 2b & TC = then from (2) 4bR 2a(2b – b)2 = (2b)3 Ra 4bR  2a × b2 = 8b3a 4b  2ab2 = 2ab2 And from (1) 2R a a PC = 4bR 2be2 = a 4b2e2 a 29. PC = 4b2e2 , VC = 2b, TC = 4bR 30. P = RT (Vm  b) ea / Vn RT  PrPC = RTrTC (VrVC  b) ea /(VrVC RTrTC ) a / 2bV RT a  Pr a 4b2e2 = RTra e  4bR(2Vrb  b) r r 4bR   Pr = ( Tr e2 /V T  2    V T  Pr = e r r  (2V – 1) = e2Tr  2   V T  e2T  Pr = e r r  (V – ½) = r ] 2 Select the correct alternative. (One or more than one is/are correct) [3 × 5 = 15] Q.4 Following represents the Maxwell distribution curve for an ideal gas at two temperatures T1 & T2. Which of the following option(s) are true? [Given : dN = 4π M 3/2   v exp  Mv2  dv ] N  2ππR   2RT  (A*) Total area under the two curves is independent of moles of gas (B*) If dU1= a Umps1 & dU2 = a Umps2 then A1 = A2 (where a is a constant) (C) Here T1 > T2 and hence higher the temperature, sharper the curve. (D*) The fraction of molecules having speed = Umps decreases as temperature increases. [Sol: Differential area under the curve = 1 dN du = dN N du N So total area = 1 (independent of moles) (B) At T = T1 A =   = 4  3 / 2  (aU )2 – – mps1 (aU )  dN  1  M    mp s M (aU )2 2RT1 mps1  N 1  2RT1  1 e  1  3 / 2  M 3 / 2 3 3 a2 A1 = 4     2RT a (Ump s ) e    1  4 3 a2  M 3 / 2  2RT1  3 / 2 A1 = a e   2RT  M  A1 =  4 a3ea2 1    Now at T = T2 3 / 2 A =   = 4   (aU )2  mps2  (aU ) 2  N 2  2RT  mps2 e 2RT2  mps2 A2 =  2  4 a3ea2  A1 = A2 (C) T1 < T2 (D) The fraction of molecules having speed = Umps decreases as temperature increases.] Q.5 In the following six electronic configuration (remaining inner orbitals are completely filled). Mark the correct option(s). C–I C–II C–III C–IV C–V C–VI (A*) Stability order : C–II > C–I & C–IV > C–III (B*) Order of spin multiplicity : C–IV > C–III = C–I > C–II (C*) C–V violates all the three rules of electronic configuration (D) If C–VI represents Athen A2+ when kept near a magnet faces weak repulsions (acts as dimagnetic). [Sol: (A), (B), (C) (A) C – II is more stable than C–I since C–I is excited state of C – II C–IV is more stable than C–III due to high spin multiplicity (B) spin multiplicity in C–IV = 7 spin multiplicity in C–III = 5 spin mjultiplicity in C–I = 5 spin multiplicity in C–II = 3 so C–IV > C–III = C–I > C–II (C) C–V violates all the three rules of electronic configuration i.e. Hund’s rule, Pauli’e exclusion principle & Aufbau rule. (D) C–IV is A then configuration of A 2+ is so it is highly paramagnetic. ] Q.6 A sample of H2O2 solution labelled as 33.6 volume has density of 264 gm/lit. Mark the correct option(s) representing concentration of same solution in other units.[Solution contains only H2O & H2O2] (A*) Mole fraction of H2O2 in the solution = 0.25 (B) % w/v = 102% (C) MH O = 6 M (D*) 1000 m 2 2 = 54 m [Sol: (A, D) Molarity = 33.6 11.2 = 3M Consider 1 litre solution w + w 2 2 = 264 gm moles of H2O2 = 3  w 2 2 & wH O = 3 × 34 102 gm = 264 – 102 = 162 gm  nH 2O 162 = 18 = 9 3 (A) mole fraction of H2O2 = 3 + 9 = 0.25 (B) % w/v = 102 100 1000 = 10.2 % (C) M 2 2 = 3M (D) m = 31000 = 1000 m ] H 2O2 162 54 MATCH THE COLUMN [2 × 8 = 16] INSTRUCTIONS: Column-I and column-II contains four entries each. Entries of column-I are to be matched with some entries of column-II. One entry of column-I may have one or more than one matching with entries of column-II. Q.3 For any chemical reaction Reactants  products; the net energy change involved in the process is Eproducts – Ereactants . Using this & the following ENERGY diagram representing energy of each collection. Match column I with column II. Column I (Process) Column II | Energy change | (A) EA1 of Mg+(g) (P) 25 (B) IE1of Cl– (g) (Q) 100 (C) IE2 of Mg (g) (R) 200 (D) H for Mg2+(g) + 2Cl–(g)  MgCl2(s) (S) 400 [Ans. (A) Q (B) P (C) R (D) S] [Sol: (A) Mg(g)  Mg+ (g) + e H = 100 kCal/mol  |EA1 of Mg+| = 100 kcal/mol  50 100  (B) Cl(g) + e  Cl–(g) , H =    |IE1 of Cl–| = 25 kcal/mol  = –25 kcal/mol 2  (C) Mg+(g)  Mg2+ + e H = (250 – 50) = 200 kcal/mol (D) Mg2+(g) + 2Cl– (g)  MgCl2(g) |H| = (–150 – 250) = 400 kcal/mol ] Q.4 Column I & column II contain data on Schrondinger Wave–Mechanical model, where symbols have their usual meanings.Match the columns. Column I Column II (Type of orbital) (A) (P) 4s (B) (Q) 5px (C) (, ) = K (independent of  & ) (R) 3s (D) atleast one angular node is present (S) 6dxy [Ans. (A) P, (B) P,Q,S, (C) P, R (D) Q, S] [Sol: (A) no. of radial nodes = 3 so it is 4s (B) no. of radial nodes = 3 so it may be 4s, 5px & 6dxy (C)  (, ) = K (independent of angles) so the orbital will be spherically symmetrical. So it may be = 4s, 3s (D) at least one angular node is present in 5px & 6dxy ]