CHEMISTRY-19-11-11th (PQRS)

REVIEW TEST-6 Class : XI (PQRS) Time : 100 min Max. Marks : 100 INSTRUCTIONS General Remarks: 1. The question paper contain 20 questions and 24 pages. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. 2. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 3. Use of Calculator, Log table and Mobile is not permitted. 4. Legibility and clarity in answering the question will be appreciated. 5. Put a cross ( × ) on the rough work done by you. Name Father's Name Class : Batch : B.C. Roll No. Invigilator's Full Name 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, Useful constant : h = 6.62 × 10–34 J sec. For Office Use ……………………………. Total Marks Obtained………………… Q.No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Marks XI (PQRS & J) CHEMISTRY REVIEW TEST-6 / 5 Q.1 (a) The first ionisation energy of Mg is higher than that of Na, but on the other hand the second ionisation energy of Na is very much higher than that of Mg. Explain (limit your explanation within 40 words). [2] (b) Select the one having lower first ionisation energyin each given pair with explanation (limit your explanation within 20 words) (i) I and I– (ii) Ba and Sr (iii) Be and B [3] [Sol.(a) Along the period ionisation energy increases (IE1 for Mg> IE1 for Na) but in the formation of Na+2 removal of e– takes place from 2p orbital which require more energy than that required for 3s as in case of Mg+. (b) (i) Zeff decreases as negative charge increases therefore ionisation energy decreases (Ans. I–) (ii) As size increases ionisation energy decreases (Ans. Ba) (iii) due to completely filled '2s' subshell Be has higher I.E. than B. (Ans. B) ] Q.2 1 mole of CCl4 vapours at 27°C occupies a volume of 40 lit. If Vander Waals constant are 24.6 L2 atm mol–1 and 0.125 Lmol–1. Calculate compressibility factor under (a) Low pressure region (b) High Pressure region [Take R = 0.082 lit-atm/mol/K] [2.5+2.5] P + a V 1 24.6 [Ans. (a)   V2  = RT  Z = 1 – RTV = 1 – 24.6  40 = 0.975 Pb (b) Z = 1 + RT b = 1 + V 0.125 = 1 + 40 = 1.003 ] Q.3 Select those species from the following having X–X linkage (X-represent central atom) and draw their Lewis Dot structure. [5] (i) H N O (ii) N O (iii) H P O (iv) S O2 2 2 2 2 5 4 2 5 2 7 (v) S O2 2 6 [Sol. Only (i) and (v) have X–X linkage in their structure (i) H2N2O2 (ii) S2O62– ] Q.4 Write all the possible orbitals in each case (upto n = 3) which fulfill following condition [5] Case – I  having only one radial node Case – II  having XY plane as angular node Case – III  having onlytwo maxima if a curve is plotted between radial distribution function vs radial distance Case – IV  both case – I and case – II are fulfilled [Sol. Case – I  2S, 3Px, 3Py, 3Pz Case –II  2Pz, 3Pz , 3dzx, 3dyx Case–III  3S Case –IV  3Pz ] Q.5 Arrange the following in (a) order of increasing melting point Mg, F, Na, Al (b) order of decreasing density Na, Be, B, K (c) order of increasing ionic mobility Mg2+, Ba2+, Cs+, Al+3 [1.5+1.5+2] [Sol. (a) F < Na < Mg < Al (b) K < Na < Be < B (c) Al+3 < Mg2+ < Ba2+ < Cs+ ] Q.6 NO2 partially converts into NO, O2 and N2O4 according to following reaction. NO + O2 NO2 N2O4 Pressure increases from 5 atm to 5.25 atm due to above reactions at constant temperature and the ratio of number of moles of O2 & N2O4 becomes 3/2. Calculate the final partial pressure of each gas in the final state. [5] [Sol. Initial pressure Pi = 5 ; Pf = final pressure = 5.25 2NO2  2NO + O2 5–x–y x x/2 2NO2  N2O4 5–x–y y/2 since Pf = 5.25 atm 5 – x – y + x + x/2 + y/2 = 5.25 5 + x/2 – y/2 = 5.25  x–y = 0.5 (i) since P 2 PN O x = y = 1.5 (ii) x = 1.5 y = 1 PNO = 2.5 PNO = 1.5 NN O = 0.5 P = 0.75 ] Q.7 With the help of EN values [ENA = 1.8, ENB = 2.6, ENC = 1.6, END = 2.8] answer the following questions for the compounds HAO, HBO, HCO, HDO (a) Compounds whose aqueous solution is acidic and order of their acidic strength [1.5] (b) Compounds whose aqueous solution is basic and order of their basic strength [1.5] (c) Comment on the chances of being coloured on the basis of percent ionic character for the compounds CD & AB [2] [Sol. (a) Bases : AOH , COH and order of basic strength AOH < COH (b) Acids : HBO , HDO and order of acidic strength HDO > HBO (c) % Ionic character = 16 | XA–XB | + 3.5 (XA – XB)2 Q.8 Two containers of equal volume containing equal of masses of gas A (MA = 32) & gas B (MB = 16) separately are maintained at temperature T & 2T respectively. (A = 2B) Find (a) ratio of total translational K.E. of gas A& gas B. (b) ratio of the mean free path of gas A & gas B (c) ratio of the collision frequency of gas A& gas B [1.5+1.5+2] [Sol. (a) KEA = nATA = MB  TA = 16  T = 1 KEB n BTB MA TB 32 2T 4  2 N* 1 2 1 (b) A = B B =  = B (Z ) 2 * A A 2 (N* )2 4 1 1 (c) 11 A = B A  =   = ] (Z11)B 2 (N* )2 1 4 2 Q.9 (a) Draw the structure of the following compounds : [2] (i) Hydrogen phosphite ion (ii) Dihydrogen phosphate ion (b) Arrange in order of increasing magnitude of energy released in the formation of adduct (addition product) between BF3 & MH3 where M may be N, P, As. [3] O O ||  || [Sol.(a) (i) H2PO3– H – P – OH | O (ii) H2PO4– O – P – OH | OH (b) Down the group as size increases tendency to form coordinate bond decreases order of increasing energy. AsH3 < PH3 < NH3 ] Q.10 With the help of above information answer the following questions (1) HL.E. of NaAlCl4 will be _ . (2) HH.E. of Al+3(g) ion will be . (3) Hf of NaCl(s) will be . (4) BECl + Heg[Cl– (g)] will be . (5) Heg of [Al+3(g)] will be . [5] [Sol. (i) HL.E = – (90 + 100) = – 190 kJ (ii) HH.E = – 40 (iii) Hy [NaCl(s)] = 90 (iv) 160 – 50 = 110 (v) Heg = – 20 kJ ] Q.11 Acompound of iron & chlorine is soluble in H2O. An excess of silver nitrate was added to precipitate the Cl– ion as AgCl. If a 10 gm of the compound gave 22.6 gm of AgCl. What is the formula of the compound. [Ans. FeCl2] [5] [Sol. FeClx + 10 xAgNO3  xAgCl x 10 + Fe(NO3)x 22.6 56 + 35.5x x = 2 ] 56 + 35.5x 143.5 Q.12 (a) Arrange in order of increasing solubility (i) Na2SO4, K2SO4, Rb2SO4 (ii) MgCO3, CaCO3, BaCO3 [2] (b) Solubilities of the halides of lithium & cesium was found to be in the following order (i) LiF < LiCl < LiBr (ii) CsF > CsCl > CsBr Justify the above observation with proper explanation (limit your explanation within 40 words) [3] [Sol. (a) (i) Rb2SO4 < K2SO4 < Na2SO4 (ii) BaCO3 < CaCO3 < MgCO3 (b) In case of halides of lithium ; lattice energy decreases much faster than hydration energy therefore solubility increases down the group but in case of cesium hydration energy decreases faster than lattice energy therefore solubility decreases down the group.] Q.13 Photon having wavelength 12.4 nm was allowed to strike a metal plate having work function 25 eV. Calculate the (a) Maximum kinetic energy of photoelectrons emitted in eV. (b) Wavelength of electron with maximum kinetic energy in Å. (c) Calculate the uncertainity in wavelength of emitted electron if the uncertainity in the momentum is 6.62 × 10–28 Kg m/sec. [1.5+1.5+2] 1240 [Sol. (a) K.E.max = 12.4 eV – 25 = 75 eV (b)  = = Å = 1.414 Å h (c) P =  2 ; dp = 2 1020 h 2 d  = P = h 6.6210 34 6.62 × 10–28 = 2 × 10–14 m ] Q.14 The maximum amount of energy which can be obtained bya mixing 30 gm of carbon with 48 gm oxygen will be? Given:– C(s) + O2  CO2 H = – 80 kJ/mol 1 C(s) + 2 O2  CO H = – 50 kJ/mol [5] 1 [Sol. C + 2 O2  CO H = – 50 × 2.5 = –125 2.5 1.5 – 0.25 2.5 1 CO + 2 O2  CO2 H = – 30 2.5 0.25 2 – 0.5 energy due to second reaction = –30 × 0.5 = 15 Total energy = – 125 – 15 H = – 140 kJ / mole ] Q.15 Electron present in single electron specie jumps from energy level 3 to 1. Emitted photons when passed through a sample containing excited He+ ion causes further excitation to some higher energy level (Given Z2 En = –13.6 n2 ). Determine (i) Atomic number of single electron specie. (ii) principal quantum number of initial excited level & higher energy level of He+. [2+3] [Sol. Energy of emitted photons can not be greater than 13.6 eV (otherwise He+ will ionise) therefore single electron specie must be hydrogen energy emitted = E3 – E1 = –1.51 + 13.6 = 12.09 For He+ ion this energy corresponds to excitation from 2 to 6. ] Q.16 Find the number of diffusion steps required to separate the isotopic mixture initially containing some amount of H2 gas and 1 mol of D2 gas in a container of 3 lit capacity maintained at 24.6 atm & 27 °C to  w D   2  1 the final mass ratio   H mD2  equal to . [Ans. 4] 2  [5] [Sol. nD2 4 = = 1  1 = 1 nH2 mH2 2 4 2 8 (n +1) = 24.6  3 = 3, n = 2 mol H2 24.6 nH H2 n'H 2 2 n = 2 8 2  , 4 x 2 8 n' = 2 1 = 1 ×  2  22 = 2x/2 x = 4 ] Q.17 With the help of following information compare the reducing power of Na or Cs. [5] Type of enthalpy change(kJ) Na Cs HS.E. 108 78 HI.E. 496 376 HH.E. – 406 – 276 [Sol. Na (s) HTotal  Na+ (aq) HTotal = HSE + HIE + HHE since substance having lower value of Htotal is more reducing therefore reducing power of Cs is more than Na. HTotal (Na) = 198 HTotal (Cs) = 178 ] Q.18 A 500 ml solution was prepared by dissolving 28.5 gm of MgCl2 in water. If the density of MgCl2 solution is 1.2 gm/ml, calculate (a) Molarity of Cl– ion (b) Mole fraction of Mg2+ ion (c) Concentration in ppm of Mg2+ ion [1.5+1.5+2] [Sol. (a) moles of Cl– = 0.6 moles of Mg+2 = 0.3 mass of solvent = 571.5 [Cl–] = 0.6 1000 = 1.2 M 500 (b) XMg+2 = 0.3 0.9 + 31.75 0.3 = 32.65 = 9.18 × 10–3 (c) Mg+2 concentration in ppm = 72 × 106 600 = 1.2 × 104 ppm ] Q.19 A balloon containing 1 mole air at 1 atm initially is filled further with air till pressure increases to 3 atm. The initial diameter of the balloon is 1 m and the pressure at each state is proportion to diameter of the balloon. Calculate (a) No. of moles of air added to change the pressure from 1 atm to 3 atm. (b) balloon will burst if either pressure increases to 7 atm or volume increases to 36 m3. Calculate the number of moles of air that must be added after initial condition to burst the balloon. [2+3] [Sol. P  d 1atm P= kd & k = 1meter (a) P V = nRT kd 1 d3 = nRT 6 4 1 1 4 n 4 = n 2 n2 = 81 moles added = 81–1 = 80 Ans. (b) To achieve 7 atm pressure, diameter must be 7 m whereas to achieve volume 36 m3 diameter required is 6 m therefore balloon will burst when diameter becomes 6m. no. of moles added n = (6)4 – 1 = 1295 moles ] Q.20 (a) A 40 ml sample of a mixture of H2 & O2 was placed in a gas burette at 18 °C and 1 atm pressure. A spark was applied so that the formation of water was complete. The remaining pure gas that diffuses faster than N2 gas under similar condition had a volume of 10 ml at 18 °C and 1 atm pressure. What was the initial mole percent of H2 in the mixture? [Ans. 75%] (b) Air is pumped into the tubes of a bicycle at a pressure of 2 atm. The volume of each tube at this pressure is 2 lit. One of the tube get punctured and the volume of the tube reduces to 0.5 lit. Assuming temperature to be constant at 300 K and air behaving as an ideal gas, calculate the number of moles of air leaked out? [Take Atmospheric Pressure = 1 atm, R = 0.082 lit-atm/mol/K] [Ans. 0.142 mol] [2+3] [Sol. (a) 2H2 + O2  2H2O (l) a b a + b= 40 (i) a–2b – – a–2b = 10 b = 10 ml a = 30 ml (b) 30 mole % = 40 ×100 = 75 % 2  2  0.51 = moles leaked out R  T RT 3.5 RT = 0.142 Ans. ]