CHEMISTRY-13-08- 11th (PQRS)

REVIEW TEST-3 CHEMISTRY Class : XI (PQRS) Time : 90 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain 15 questions. All questions are compulsory. 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 : g = 9.8 m/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) CHEMISTRY REVIEW TEST-3 Q.1 Match the column [4] Column I Column II (A) Atomic theory of matter (P) Rutherford scattering experiment (B) Quantization of charge (Q) Muliken's oil drop experiment (C) Quantization of energy (R) Atomic spectrum (D) Size of nucleus (S) Law of chemical combination Ans. Q.2 Match the description in Column I with graph provided in Column II. For n moles of ideal gas at temperature 'T'. [4] Column I Column II P (A) V vs P (P) (B) P vs V (Q) V (C) V vs P–2 (R) P P (D) V vs log P (S) Ans. Q.3 Calcium and magnisium ion from a 105 litre of sample of hard water was quantitatively precipitated as carbonates and weight of ppt. obtained was found to be 568 gm. If the ppt. lost 264 gm of wt. on strong heating. [2 + 2] (a) Find the hardness of water in ppm (in terms of wt. of CaCO3) (b) Molarity of Ca2+ and Mg2+ ions in hard water. [Sol. 568 gm same  MgCO3  MgO + CO2 CaCO3  CaO + CO2 264  Moles of CO2 lost = 44 = 6 Moles of CaCO3 + MgCO3 = 6 mole Wt. of eq. quantity of CaCO3 = 600 gm 600106 108 = 6 ppm Also (6 – X) 100 + 84 X = 568  X = 2  MgCO3 = 2 mole in 10–5 litres CaCO3 = 4 mole in 10–5 litres [Ca2+] = 4 × 10–5 M [Mg2+] = 2 × 10–5 M ] Q.4 The Vander Waal's constant 'b' of a gas is 88 104 7 L/mol. How near can the centres of the two molecules approach each other? [Use NA = 6 × 1023] [4] [Sol. b = 4 × 4 × r3 × N 3 A 88 104 × 1000 = 4 × 7 4 × 22 × r3 × 6 × 1023 3 7 8.8 7 = 4 × 4 × 22 × r3 × 6 × 1023 3 7 r = 5 × 10–9 cm Distance of closest approach = 2r = 10–8 cm ] Q.5 Calculate the minimum volume of a hot air balloon, which can lift a payload of 1000 kg. The density of air is 1 gm/litre. The temperature of air 300 K and air inside the balloon can be heated upto 400 K. The mass of balloon is 0.2 kg/m3. [4] [Sol. dair = 1 gm/litre = 1 kg/m3 at const. pressure of 1 atm d1T1 = d2T2  d2 = d1T1 T2 = 1 × 300 = 0.75 kg/m3 400 Max. wt. lifted by 1 m3 balloon = 1 – 0.75 – 0.2 = 0.5 kg/m3 1000 = 0.05 × V V = 1000 × 102 = 0.2 × 105 m3 = 2 × 104 m3 Ans. ] 5 Q.6 100 gm of KClO3 when heated gives 14.90 gm KCl and 41.55 gm KClO4 [2+2] 3 KClO3  KCl + 2 O2 ...(1) 4KClO3  KCl + 3KClO4 ...(2) (a) Find the weight of KClO3 remaining indecomposed. W Weight of KClO used in 1st Reaction (b) Ratio 1 = 3 nd W2 Weight of KClO3 used in 2 Reaction 3 [Sol.(a)KClO3  KCl + 2 O2 –x x (3/2)x 4KClO3  KCl + 3KClO4 –4y y 3y moles of KClO3 used = x + 4y mole 14.90 14.90 gm KCl = 74.5 = 0.2 mole x + y = 0.2 ...(1) 41.55 41.55 gm KClO4 = 138.5 = 0.3 mole 3y = 0.3  y = 0.1 mole x = 0.1 mole moles of KClO3 used = 0.1 + 0.4 = 0.5 1 wt. of KClO3 remain undecomposed = 100 – 2 × 122.5 = 100 – 61.25 = 38.75 gm (b) x  1  4y =  4    Q.7 Find the mole fraction of N2 and O2 in air at 8314 meter height from surface of earth on the basis of following assumptions. [5] I. Composition of air on the surface of earth is 20% O2 and 80% N2 by moles and P = 1 atm. II. Temperature of atmosphere is constant at 300 K upto 8314 meter height. III. Take acceleration due to gravity to be constant and equal to 10 m/s2. IV. No turbulance in air to make the composition of gases uniform [Use e1/15   1.07]  28 g  h 103  [Sol. PN = P exp   2 N2  RT    32  g  h 103  PO = P exp   2 O2  RT  PN 8 (32  28)g·h 103  8  4 10  8314 103  2 = P 2 exp  2   4  8.314  300  2   = exp   2  8.314  300  = 4exp 30  = 4exp   = 4 × (1.07)2 = 4.58   15  XO2 = 1 5.58 = 0.179  0.18 17.9% 82.1% ] Q.8 A container having very small orfice contained Helium gas at pressure of 2000 mmHg. The Helium gas leaked slowly and it took 5000 second, when the pressure of He gas dropped to 1000 mmHg. How much time it will take for pressure of methane gas to drop from 2000 mmHg to 1500 mmHg if filled in same container under identical conditions. Consider effect of variation of pressure on rate of diffiusion. Use equation ln [Given log 3 = 0.5 & log 2 = 0.3] [Sol. For Helium Pi = (constant) × t [5] Pf ln Pi = ·t Pf  ln 2000  = 1000  k·(5000) 2  k = 2ln2 5000 For Methane ln Pi = ·t Pf  ln 2000  = 1500  2ln2 4  5000 ·t  4    3  = ln2 2  5000 ·t    t = (5000 × 2)  2ln2  ln3  ln2  2  log3   t = 10000   log2   t = 3333.33 sec ] Q.9 In an experiment 0.15 gm of a biological sample containing amino acid glycene (NH2CH2COOH) was treated with nitrous acid [5] NH2CH2COOH + HNO2  HO–CH2–COOH + H2O + N2 The nitrogen gas thus produced is collected over water at 300 K, (vapour pressure(aqueous tension) of H2O at 300 K is 28 torr) at a total pressure of 700 torr. The volume of gas measured as 15.2 ml. Find the percentage of glycene in biological sample. [Sol. moles of Amino acid(n) = (15.2 103)(672)  273 760  22.4  300 = 2 × 273 × 10–6 = 5.46 × 10–4 moles Wt. of Amino acid = n × (24 + 32 + 5 + 14) = n × 75 n  75 % = 0.15 100 = 2 273 75106 102 15102 = 273 × 10–1 = 27.3% ] Q.10 15 ml of an oxide of nitrogen was taken in a endiometer tube and mixed with hydrogen till the volume was 65 ml. On sparking the resulting mixture occupied 20 ml. To the mixture, 10 ml of oxygen was added and on explosion, again the volume fell to 22.5 ml. Find the formula of the oxide of nitrogen that was originally admitted to eudiometer tube. Both explosion led to formation of H2O(l). [5] [Sol. Let the oxide of nitrogen = NxOy x y NxOy  2 N2(g) + 2 O2(g) 15x 15 2 15y 2 H2 + 1 2 O2  H2O(l) 15y  15y 2   0   2 Contracted volume in I explosion 15x 15y 15y  15 – 2 – 2 + 3    = 65 – 20 2  15x 15y – 2 = 30 ...(1) Contracted volume in II explosion = 30 – 22.5 = 7.5 ml  Vol. of O2 used (in II explosion) = 2.5 ml Vol. of H2 used (in II explosion) = 5 ml Vol. of H2 used in I explosion  Using eq. (1) 15y 2 × 2 = 45  y = 3 Thus, x = 2 ] Q.11 The compressibility factor for nitrogen at 220 K and 800 atm is 1.90 and at 380 K and 200 atm is 1.10. A certain mass of N2 occupies a volume of 1 dm3 at 220 K and 800 atm. Calculate volume occupied by same quantity of N2 gas at 380 K and 200 atm. [5] PV [Sol. Z = nRT 1 800 1.90 = n  0.0821 220 800 n = 1.90  R  220 V 200 Z = 1.10 = n  R  380 1.10 = V  200 1.90  R  220 800  R  380 1.10  800  380 V = 200 1.90  220 = 4 litre ] Q.12(a) A 50 cc sample of H2O2 liberates 5.08 gm of I2 from acidified KI solution. Find the volume of O2 gas liberated by complete decomposition of 250 cc of same H2O2 solution at STP. HCl + K I + H2O2  I2 + KCl + H2O H2O2  H2O + O2 (b) What volume of M/10 and M/30 solutions of H2SO4 should be mixed to prepare 1 L of certain H2SO4 solution whose 50 ml is completely neutralized by 10 ml of 0.5 M NaOH solution. [3 + 2] [Sol.(a) Required balanced reaction 2HCl + 2KI + H2O2  I2 + 2KCl + 2H2O 1 H2O2  H2O + 2 O2 Vol. of H2O2 = 50 ml Liberate amount of I2 from 50 ml H2O2 = 5.08 gm From balancing reaction No. of moles of H2O2 = No. of moles of I2 50 M1 × 1000 5.08 = 254  M = 5.08 1 254 1000 × 50  Liberated moles of O 1 1 = [Moles of H O ] = [M × V ] = 1 5.08 [ 1000 × × 250 ] 2 2 = 0.05 2 2 2 1 2 2 254 50 1000  Vol. of O2 liberated at STP = 22.4 × 0.05 = 1.12 L (b) 10 ml 0.5M NaOH = 5 m moles 2.5 m mole H2SO4 50 ml  2.5 m mole  Molarity of H2SO4 required = 2.5/50 = 1/20 M V · 1 1 10 + V · 1 2 30 = 1 V1 + V2 20 2 2V + V = V + V 1 3 2 1 2 V = 1 V  3V = V 1 3 2 1 2 V1 + V2 = 1000  4V1 = 1000  V1 = 250 ml V2 = 750 ml 250 ml of M H SO and 750 ml of M H SO ] 10 2 4 30 2 4 Q.13 1023 gas molecules each of mass 10–25 kg are taken in a container of volume 1 dm3. The root mean square speed of gas molecules is 1 km sec–1. Find [2 + 3] (a) Pressure exerted by gas molecules (b) Temperature of gas molecules [Sol.(a) PV = 1 n·m· c2 3 n·m·c2 P = 3V 1023 1025  (103)2 = 3103 = 1 × 1023 × 10–25 × 106 × 103 = 1 3 3 × 107 = 0.333 × 107 = 3.33 × 106 Pa (b) 1000 = 103 =  10 3 8.314  T 25  6.02310 1/ 2 23  106 1025  6.0231023 T = 3 8.314 6.023 = 3 8.314 × 104 = 2414 K ] Q.14 'n' moles of a gas 'X' was trapped in a gas jar over surface of a liquid 'A' as shown in figure. The liquid 'A' rose in inner column (of total length 85 cm above completely filled outer vessel liquid surface) by 20 cm. When same number of moles of gas 'X' is trapped in same apparatus over liquid 'B', the liquid B rose by length 22 cm in inner column. [4+2+2] From the given data at 300 K, calculate (a) Pressure of gas over liquid 'A' (b) Pressure of gas over liquid 'B' (c) Density of liquid 'B'. (Use: Atmospheric pressure = 760 mmHg, Density of Hg = 13.6 gm/cc) S.N. density (g/cc) V.P. (aq. Tension) (mmHg) hinner-houter (cm) Liquid 'A' 6.8 30 20 Liquid 'B' — 40 22 [Sol. Liq. 'A' P(atmosphere) = Pgas + aq. tension + PL.C 6.8 200 [Ans. 4.32 gm/cc] 760 = Pgas + 30 +  Pgas, 1 = 630 mmHg 13.6 In Liq. 'B', since no. of moles of gas is same and under isothermal conditions Pgas, 2 = (Pgas,1)(l1) l2 65 = 630 × 63 = 650 mmHg  760 = 650 + 40 + (B ·lB ) Hg  B = (70)(Hg ) 220 = 4.32 gm/cc ] Q.15 For the following problem carefully examine the figure and information provided with figure which describes set up of a experiment under isothermal conditions. [3+5] The figure shows initial conditions of experiment with frictionless pistons A and B held in shown position by mechanical stoppers. If the mechanical stopper holding A and B as shown in figure is removed (a) Pressure developed in each compartment in final state. (b) What will be the final positions of A and B (with respect to far left end of container) [Sol.(a)In final state Pressure developed in first compartment (Pfinal) = Pressure developed in other two compartments (100 × 2 ) × P 4.5 final 10 = 3 × 60 Pfinal = 4.5 atm (b) PAr · 60 · A = nAr · RT ....(1) [A = Area of cross-section of cylinder] PNe · 20 · A = nNe · RT (2) PHe · 20 · A = nHe · RT (3) Putting values of PAr = 10/3 atm, PNe = 7.5 atm & PHe = 5 atm and dividing eq. (1) by (2) nAr n Ne = 200 150 Dividing eq. (1) by (3) n Ar n He = 200  n 100 Ar : nNe : nHe = 2 : 1.5 : 1 In the final condition pressure throughout the container will be same, hence ratio of length of three parts will be same as ratio of moles.  Length of the compartment containing Argon will be 100 × 2 = 44.44 cm 4.5  Length of the compartment containing Neon will be 100 × 1.5 4.5 = 33.33 cm  Length of the compartment containing Helium will be 100 ×  Position of A w.r.t. left end = 44.44 cm. Position of B w.r.t. left end = 77.77 cm. 1 4.5 = 22.22 cm