CHEMISTRY-17-09- 11th (PQRS) SOLUTION

REVIEW TEST-4 Class : XI Time : 100 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contains 16 questions and 24 pages. All questions are compulsory. 2. Write your answer(s) in the space given at the end of each question. Otherwise Marks will not be awarded. 3. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 4. Use of Calculator, Log table and Mobile is not permitted. 5. Legibility and clarity in answering the question will be appreciated. 6. 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 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 Q.1 A packet of an aritifical sweetner contains 45.75 mg of saccharin which has the structural formula How manymoles & molecules of saccharin are contained in 45.75 mg of saccharin? How many hydrogen atoms? [NA = 6 × 1022] [Ans. 2.5 × 10–4, 1.5 × 1020, 7.5 × 1020] [3] 45.75103 [Sol: Moles of Comp. = 183 = 2.5 × 10–4 Molecules of compound = 2.5 × 10–4 × NA = 2.5 × 10–4 × 6 ×1023 = 1.5 × 1020 No. of hydrogen atoms = 1.5 × 1020 × 5 = 7.5 × 1020 ] Q.2 The subshell that arises after f sub-shell is called g sub-shell. (a) How many g orbitals are present in the g sub-shell (b) In what principal electronic shell (n) would the g sub-shell first occur and what is the total number of orbitals in the shell. [Ans. (a) 9, (b) 5, 25] [3] [Sol. For f subshell l = 3  g subshell l = 4 (a)  total g orbitals in g-subshells = 2l +1 = 2 × 4 + 1 = 9 (b) for principal shell, l = 4, l = n–1  n = 5 total no. of orbitals in shell = n2 = 52 = 25 ] Q.3 For a real gas (mol. mass = 30) if density at critical point is 0.40 g/cm3 and its Tc = 2105 821 K, then calculate Vander Waal's constant a (in atm L2mol–2). [Ans. 1.6875 ] [3] [Sol. Vc = 30  75 cm3mol–1 0.40  b = Vc  25 cm3mol–1  0.025 Lmol–1 3  Tc = 2105 8a 27Rb 8a 821 = 27  0.0821 0.025  a = 1.6875 ] Q.4 The percentage by volume of C3H8 in a gaseous mixture of C3H8, CH4 and CO is 39. When 100 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced.[Ans. 178 ml] [3] [Sol. 100 ml gaseous mixture contain 39 ml C3H8 So, mixing volume of CH4 & CO = (100–39) = 61 ml 1 C3H8 + 5O2 — 3CO2 + 4H2O; CH4 + 2O2 — CO2 + 2H2O; CO + 2 O2 — CO2 61 ml (CH4 & CO) will produce 61 ml CO2; C3H8 will produce = 3 × 39 = 117 ml Total CO2 produce = 117 + 61 = 178 ml ] Q.5 The angular momentum of an electron in a Bohr's orbit of H-atom is 3.1652×10–34 kg-m2/sec. Calculate the wavenumber in terms of Rydberg constant (R) of the spectral line emitted when an electron falls from  8  this level to the ground state.[Use h = 6.6 × 10–34 Js] [Ans. R  9  ] [3]   nh [Sol. Angular Momentum mvr = 2 3.1652×10–34 = n=3 n  6 .6  10 34 2π  1 1    = R. Z2.  2 2     1 2   1 1   8   = R. 12  12  32   R  9  Ans.]     Q.6 Use the Heisenberg's Uncertainity Principle to calculate the uncertainity (in meters) in the position of a honeybee weighing 0.70 g and travelling at a velocity of 0.85 m/s. Assume the uncertainity in the velocity is 0.1 m/s.[Use h = 6.6 × 10–34 Js] [Ans. 7.53 × 10–31 m] [3] h [Sol: x × v = 4πm 6.6 1034 x = (4)3.140.7  0.1103 x = 7.53 × 10–31 m ] Q.7 If Pd v/s P (where P denotes pressure in atm and d denotes density in gm/L) is plotted for H2 gas (ideal gas) at a particular temperature. If  d (Pd) = 10, then find the temperature in °C. [3] [Sol. PM = dRT P2  M   dP d(Pd) P8.21atm 2PM 2  8.21 2 [Ans. 40 K] Pd =  RT   dP = RT  10 ; 0.0821T = 10   T = 40 K ] Q.8 Radiation corresponding to the transition n = 4 to n = 2 in hydrogen atoms falls on a certain alkali metal (work function = 2.0 eV). Calculate maximum kinetic energy (in eV) of the photoelectrons. [4] [Ans. 0.55] [Sol. En = – 13.6 eV; E n2 2 = – 13.6 ; E 22 4 = – 13.6 eV/atom 42 E = E4 – E2 = 2.55 eV Absorbed energy = work function of metal + KEmax 2.55 = 2 + KEmax KE max = 0.55 eV Ans. ] Q.9(a) What is the shortest wavelength of line (in nanometers) in the Lyman series of He+ spectrum [use R = 1.097 × 10–2 nm–1] (b) Which is denser at the same temperature and pressure, dry air or air saturated with water vapour? Justify in brief. [2+2] 100 [Ans. (a) 1.097  4 nm (b) dry air] 1 [Sol. λ = RH  1  n2 1   n2   1 2  1 1  1  λ = 1.097 × 10–2(2)2 1   100  = 1.097  4 nm (b) Dry air is denser; it has a large average molecular mass] Q.10 For the reaction 2NH3(g) — N2(g) + 3H2(g). What is the % of NH3 converted if the mixture diffuses twice as fast as that of SO2 under similar conditions [Ans. 6.25] [5] [Sol. rmix  rSO  2 = 1 ; Mavg = 16 Let the initial moles be one 2NH3  N2 + 3H2 1–2x x 3x Total wt. 17 1 1 Mavg = Total moles ; 16 = 1 2x ; x = 32 ; moles dissociated = 16 % dissociation = (1 16) 100 = 6.25% ] 1 Q.11 Urea (H2N–CO–NH2) is manufactured on large scale by passing CO2(g) through ammonia solution followed by crystallization. CO2 for the above reaction is prepared by combustion of hydrocarbon. If combustion of 236.11 kg of a saturated hydrocarbon (CnH2n+2) produces as much CO2 as required for production of 1000 kg urea, deduce molecular formula of hydrocarbon. 2NH3 + CO2  NH2CONH2 + H2O [Ans. C12H26] [5] [Sol: CnH2n+2  nCO2 POAC for C n × moles of CnH2n+2 = nCO2 ..........(1) Now, 2NH3 + CO2  NH2CONH2 + H2O nCO = nurea n  236.11103  1000 103 14n  2 60 On solving  n = 12 C12H26 ] Q.12(a) Arrange the correct order of increasing wavenumber of the following radiations I.R., U.V., radiowaves, X-rays and visible light (b) Match the column Column I Column II (A) Lyman series (P) Visible region (B) Humphrey series (Q) Ultraviolet region (C) Paschen series (R) Infrared region (D) Balmer series (S) Far infrared region (c) Two flasks Aand B of equal volumes maintained at temperature 300 K and 600 K contain equal mass of H2 and CH4 respectively. Calculate the ratio of total translational kinetic energy of gas in flask A to that of flask B. [2+2+2] [Ans. 4 ] [Sol.(a) X-rays > U.V. > visible light > I.R. > radiowaves (b) AQ, BS, CR, DP 3 3  ω  (c) (KE)A = nH TA =  300 2 2  2  3 3  ω  (KE)A = nCH TB =  600 2 2  16   KEA = 16  300 = 4 Ans. ] KEB 2 600 Q.13 2.5 g of a sample containing Na2CO3, NaHCO3 and some non-volatile impurity on gentle heating loses 12.4 % of its weight. Residue is dissolved in 100 ml water and its 10 ml portion required 15 ml of 0.1 M aqueous solution of BaCl2 for complete precipitation of carbonates. Determine mass % of Na2CO3 in the original sample. [Ans. 42.4% , Na2CO3] 2NaHCO3  Na2CO3 + CO2 + H2O Na2CO3 + BaCl2  2NaCl + BaCO3  [6] [Sol: Let nNa CO = a & nNaHCO = b 2 3 3 2NaHCO3  Na2CO3 + CO2 + H2O  moles of CO2 b produced = 2 b moles of H2O produced = 2  b  44  b 18 total loss in weight =  2  +  2  = 31 b  12.4  2.5 100  = 31b    b = 0.01 b Now in residue moles of Na2CO3 = a + 2 = a + 0.005 Na2CO3 + BaCl2  2NaCl + BaCO3  moles of Na2CO3 = moles of BaCO3 a + 0.005 = 15 × 0.1 × 100 × 10–3 10 a = 0.01  wt. of Na2CO3 = 0.01 × 106 = 1.06 gm 1.06 % Na CO = 100 = 42.4% Ans. ] 2 3 2.5 Q.14(a) At 273 K and 8 atm pressure, the comprassibility factor for a gas is 0.8. Calculate the volume of 1 milli moles of gas at this temperature and pressure. (b) From his exceptional 'mathematical skills', Mr. Gupta was able to obtain the following expression for  (the amplitude of electron wave) by solving Schrodinger equation. However he wants to know those 'r' where probability of finding the electron is zero. Since he is 'exhausted' in his work, help himto find the values of 'r' acting as nodes(  2 = zero). 1  Z 3 / 2     (  1)( 2  8  12 ) e / 2 16 4  a 0  where a0 & Z are the constant in which answer can be expressed &   2Zr a 0 [3+5] [Sol: (a) PV = ZnRT 8 × V = 0.8 (10–3) (0.0821× 273) V = 2.24 ml Ans. 1  Z 3/ 2 (b)  =   σ 1σ 2  8σ 12e–/2  a0  At node,  = 0   3/ 2    1  Z  σ 1σ 2  8σ 12eσ /2  = 0 16  a0    ( – 1) (2 – 8 + 12) = 0  ( – 1) (2 – 8 + 16 – 4) = 0  ( – 1) [( – 4)2 – 4] = 0  ( – 1) [( – 4)2 – 22) = 0  ( – 1) ( – 4+2) ( – 4 – 2) = 0  ( – 1) ( – 2) ( – 6) = 0 so  = 1 or  = 2 or  = 6 2Zr  na0 = 1 or 2Zr na0 = 2 or 2Zr na0 = 6  r = na0 or r = 2Z na0 or r = Z 3na0 Z Ans.] Q.15(a) Calculate the ratio of time periods in first and third orbits of hydrogen atom. (b) The electron in the first excited state (n1 = 2) of H-atom absorbs a photon and is further excited (n2).The Debroglie wavelength of the electron in this excited state is 1340 pm. Find the value of n2. [3+5] [Ans.(a) 1/27, (b) 4] T  n 3  1 3 1 [Sol (a) 1   1  =    T3  n3   3  27 (b) r  n nr   λ 1340 n  2π r1  23.1453   ] Q.16 The apparatus shown consists of three temperature jacketed 1 litre bulbs connected by stop cocks. Bulb A contains a mixture of H2O(g), CO2(g) and N2(g) at 27°C and a total pressure of 547.2 mm Hg. Bulb B is empty and is held at a temperature –23°C. Bulb C is also empty and is held at a temperature of –173°C. The stopcocks are closed and the volumes of lines connecting the bulbs is zero. Given: CO2(g) converted into CO2(s) at –78°, N2(g) converted into N2(s) at –196°C & H2O(g) converted into H2O(s) at 0°C. [Use R = 0.08 atm-litre/mole·K] (a) The stopcock betweenA & B is opened and the system is allowed to come to equilibrium. The pressure in A & B is now 228 mmHg. What do bulbs A & B contain? (b) How many moles of H2O are in system? (c) Both stopcocks are opened and the system is again allowed to equilibrium. The pressure throughout the system is 45.6 mmHg. What do bulbs A, B and C contain? (d) How many moles of N2 are in the system? [8]  547.5 1 [Sol. Total initial no. of moles in bulb A = n  760 =  0.72 1 = = 0.03 After opening the cock: In bulb A total 0.08 300 0.08 300  228   760  n( N  CO2 ) =  = 0.0125 24  228 1 I bulb B n( N CO ) =  760  = 0.3 = 0.015 2 2 0.08 250 0.08 250 n H O = 0.03 – (0.0125 + 0.015) = 0.0025 After opening C  N2 = 45.6 1 760 0.08 300 = 0.0025 N2 in B N2 = 45.6 1 760 0.08 250 45.6 1 = 0.003 n N in C n N = 760 = 0.08100 0.06 0.08100 = 0.0075 ]