CHEMISTRY-13-08- 11th (PQRS) SOLUTION

REVIEW TEST-3 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. 6. Write your answer(s) in the box given in the last of each question. Otherwise Makrs will not be awarded. Name Father's Name Class : Batch : B.C. Roll No. Invigilator's Full Name USEFUL DATA Atomic weights: Mg = 24, Cl = 35.5, O = 16, H = 1, N = 14, I = 127, K=39, C = 12, Ca = 40; He = 4. Useful constant : g = 10 m/sec2, NA = 6 × 1023, R = 8.314 J/(mol-K) = 0.082 lt-atm/(mol-K) 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) Millikan's oil drop experiment (C) Quantization of energy (R) Planck's Quantum Theory (D) Size of nucleus (R = RoA1/3) (S) Law of chemical combination Ans. [Ans. (A) – S, (B) – Q, (C) – R, (D) – P] 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) P (B) V vs V (Q) (C) V vs P–2 (R) P  P  (D) log  V  vs log P (S)   Ans. [Ans. (A) – S, (B) – R, (C) – Q , (D) – P] 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. MgCO3  MgO + CO2 CaCO3  CaO + CO2 [Ans. (a) 6 ppm, (b) [Mg2+] = 2 × 10–5 M; [Ca2+] = 4 × 10–5 M] [Sol.(a) consider 105 litre hard water sample  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 (b) 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 L/mol. How near can the centres of the two 7 molecules approach each other? [Use NA = 6 × 1023] [4] [Sol. b = 4 × 4 × r3 × N 3 A 88 104 × 1000 = 4 × 4 × 7 3 22 × r3 × 6 × 1023 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] [Ans. 2 × 104 m3] [Sol: Let the volume of balloon is Vm3  V × damb = mass of balloon + mass of gas filled + payload (1) Here damb = 1gm/lt = 1kg/m3 P ×M = 1 × R × 300 (2) & P ×M = dhot × R × 400 (3) Divide (2) by (3) dhot = 3 kg/m3 4  mass of balloon = (0.2 V) kg  3 V  mass of gas filled =    kg  & payload = 1000 kg from (1) (V × 1) = 0.2 V + (0.75V) + 1000 V = 1000 = 2 × 104 m3 ] 0.05 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 41.55 [Ans. 38.75 gm; 1:4] [Sol: (a) moles of KClO4 = 138.5 = 0.3  moles of KClO3 used in (2) reaction = 0.4 moles of KCl formed from (2) reaction = 0.1 14.9 total moles of KCl formed = 74.5 = 0.2 therefore, moles of KCl from (1) reaction = 0.2 – 0.1 = 0.1  moles of KClO3 used in (1) reaction = 0.1 total moles of KClO3 used = 0.4 + 0.1 = 0.5 wt. of KClO3 decomposed = 0.5 × 122.5 = 61.25 gm  wt. of KClO3 remaining underdecompoased = 100 = 61.25 = 38.75 gm (b) ratio w1 w2 = wt. of wt. of KClO3 used in (1) reaction KClO3 used in (2) reaction = 0.1  122.5 = 1 ] 0.4  122.5 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 non–uniform [Use e1/15  1.07] [Ans. 82.1%, 17.9%]   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 15  = 4 × (1.07)2 = 4.58     1 y 2 = 5.58 = 0.179  0.18 and y = 1 – 0.18 = 0.82 ] 2 Q.8 A container having very small orifice 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 Pi = t Pf [5] [Given log 3 = 0.5 & log 2 = 0.3] [Sol. For Helium ln Pi = ·t Pf  ln 2000  = 1000  K ·(5000) 2  k = 2ln2 5000 For Methane ln Pi = Pf K ·t  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 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. (15.2 103)(672)  273 [Sol. moles of Amino acid(n) = 760  22.4  300 = 2 × 273 × 10–6 = 5.46 × 10–4 moles Wt. of Amino acid = n × (24 + 32 + 5 + 14) = n × 75 = (5.46 × 10–4 × 75) gm 5.46 104  75 % of glycene is sample = 0.15 × 100 = 27.3% ] Q.10 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] [Ans. 4 litre] 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.11 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 1 2 [Sol.(a) PV = 3 N·m·vrms N·m·v 2 P = rms = 3V = 3.33 × 106 Pa (b) 1000 = 1023 1025  (103)2 3103 = 1 × 1023 × 10–25 × 106 × 103 = 1 3 3 × 107 = 0.333 × 107  3  8 .314  T  1 / 2 103 =  10  25  6  10 23  10 6  10  25  6  10 23 T = 3  8 .314 6 = 3 8.314 × 104 = 2403 K ] 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. (1) 2HCl + 2KI + H2O2  I2 + 2KCl + 2H2O (2) 2H2O2  2H2O + 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. H2SO4 + 2NaOH  Na2SO4 + 2H2O [3 + 2] [Ans.(a) 1.12 lit, (b) 250 ml of M/10, 750 ml of M/30] [Sol.(a) In (1) n = n  2 2 2 nH 2O2 = 5.08 254 = 0.02  moles of H2O2 in 50 cc = 0.02 250 & moles of H2O2 in 250 cc = 0.02 × 50 = 0.1 In (2) n 2 2 2 = n  2 0.1 2 = n 2  n = 0.05 2 V (at STP) = 22.4 × 0.05 = 1.12 lt. 2 (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 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 M M 250 ml of 10 H2SO4 ; 750 ml of 30 H2SO4 ] Q.13 15 ml of an oxide of nitrogen was taken in a eudiometer 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 having N2 and O2 gases. Find the formula of the oxide of nitrogen that was originally admitted to eudiometer tube. Both explosion led to formation of H2O(l) only. x y [Ans. N2O3] [5] [Sol. NxOy  2 N2(g) + 2 O2(g) 15x 15 2 15y 2 H2 + 1 2 O2  H2O(l) 15y  15y 2   0   2 15x 15y 15y  15 – 2 – 2 + 3    = 45 2  15x 15y – 2 = 30 Last step— const. 7.5  O2 used = 2.5 H2 used = 5 15y  2 × 2 = 45  y = 3 Thus, x = 2 ] Q.14 For the followingproblem carefullyexamine the figure and information provided withfigure which describes set up of a experiment under isothermal conditions. [5+3] The figure shows initial conditions of experiment with frictionless pistonsAand B held in shown position by mechanical stoppers.(Thickness of pistons is negligible). 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: For Ar P1V1 = P2V2  10  60 3 = P × x Px = 200 (1) For Ne P1V1 = P2V2  7.5 × 20 = P × (y – x) P(y – x) = 150 (2) For He P1V1 = P2V2  5 × 20 = P (100 – y) P(100 – y) = 100 (3) On solving equations (1), (2) and (3) P = 4.5 atm 400 x = 9 cm 700 & y = 9 cm. Q.15 '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. (mmHg) hinner-houter (cm) Liquid 'A' 6.8 30 20 Liquid 'B' — 40 22 [Sol. Liq. 'A' P(atmosphere) = Pgas + V.P. + PL.C 6.8 200 760 = 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 = 4.33 gm/cc ]