CHEMISTRY-04-06- 11th (PQRS) SOLUTION

REVIEW TEST-1 Class : XI (PQRS) Time : 90 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain two parts. Part-A contains short question and Part-B contains subjective questions. All questions are compulsory. Paper contains 10 questions in Part-A and 9 questions in Part-B. 2. You are advised NOT to spend more than 30 minutes in any case for part-A. 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 : Roll No. Batch Class : XI 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, Ti = 48, Ba = 137, R = 0.082 (lt-atm)/(K-mol). For Office Use ……………………………. Total Marks Obtained………………… Part-A Part-B 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-1 PART A Q.1 Rearrange the following (I to IV) in the order of increasing masses [3] (I) 0.5 mole of O3 (II) 0.5 gm-molecule of N2 (III) 3.011 × 1023 molecules of O2 (IV) 5.6 litre of CO2 at STP [Ans. IV < II < III < I] [Sol: (I) n O (II) n N = 0.5,  = 0.5 ,   = 48 × 0.5 = 24 g  = 28 × 0.4 = 14g (III) n O 3.0111023 = 6.022 1023 = 0.5, O  32  0.5 = 16g (IV) nCO = 5.6 22.4 = 0.25,  CO = 44× 0.25 = 11g (IV) < (II) < (III) < (I) ] Q.2 Calculate number of neutrons in 17 g of H2O2. [Ans. 8 × NA] [3] [Sol: n = 17 = 0.5 H2O2 34 No. of H2O2 molecules = 0.5 NA  1 molecule of H2O2 contains = 16 neutrons  0.5 NA ––––––"––––––––– = 16 × 0.5 NA  no. of neutrons = 8NA ] Q.3 The hydrated salt Na2CO3.xH2O undergoes 63% loss in mass on heating and becomes anhydrous, calculate the value of x [3] [Ans. 10] [Sol: Na2CO3 · xH2O % of H2O = % loss in wt. = 63  18x 100 = 63 106  18x  x = 10 ] Q.4 The vapour density of a mixture containing NO2 & N2O4 is 27.6. Calculate mole fraction of N2O4 in the mixture. [3] [Ans. 0.2 ] [Sol: NO  N O  ––– M = y  46 + y  92 = 2 × 27.6 (1) & yNO y = 1 (2) On solving y 2 4 yNO = 0.2 = 0.8 ] Q.5 Suppose two elements X and Y combine to form two compounds XY2 and X3Y2 when 0.05 mole of XY2 weighs 5 g while 3.011 × 1023 molecules of X3Y2 weighs 90 g. Calculate the atomic masses of X and Y. [3] [Ans. 40 and 30] [Sol: Let atomic weight of x = a Let atomic weight of y = b wt xy2 moles = molar mass  0.05 = 5 a  2b a + 2b = = 100 (1) x3y2 moles = wt molar mass 3.0111023 6.022 1023 90 = 3a  2b 3a + 2b = 180 (2) On solving a = 40 ; b = 30 ] Q.6 What is the molarity of SO2 ion in aqueous solution that contain 34.2 ppm of Al (SO ) [Assume 4 2 4 3 complete dissociation & density of solution 1 g/ml) [3] [Ans. [ SO2 ] = 3 × 10–4 M] [Sol. Al2(SO4)3  2Al3+ + 3SO42– wt. of Al2(SO4)3 in 106 gm solution.  n Al (SO ) = 34.2 342 = 0.1  moles of SO32– formed = 3 × n Al (SO ) molrity of SO42– = 0.3 V(lt )  0.3 103 = 3 × 10–4 M (d = 1 g/ml) ] Q.7 Concentration of aq. NaOH solution is 3.0 Molal and it's density is 1.1 gm/ml. What is the Molarity of the solution? [3] 1000  M [Sol: m = 1000d  MM B 1000M  3 = 1000(1.1)  M(40) M = 2.946 M Ans ] Q.8 What volume of 90% alcohol by weight (d = 0.80 g/cm3) must be used to prepare 150 cm3 of 30% alcohol by weight (d = 0.90 g/cm3) [3] [Sol: Alcohol (concentrated)  Alcohol (diluted)  (90% w , d = 0.8 g/ml) (d = 0.9 g /ml, 30% /w) mass of alcohol before dilution = mass of alcohol after dilution  V × 0.8 × 0.9 = 150 × 0.9 × 0.3 V = 56.25 cm3 ] Q.9 1.44 gram of Titanium (Ti) reacted with excess of O2 and produced x gram of compound Ti1.44O. Find the value of x. [3] [Sol: Ti  Ti1.44O POAC for Ti nTi  1.44  ncomp 1.44  48 = 1.44 x 1.44(48) 16  x = 1.773 gm ] PART B Q.10(i) Find the value of x in Nicotine, C10H14Nx? (Given: mol. mass of Nicotine is 162 amu.) (ii) If an atom of 12C had been assigned a relative value of 24.0 a.m.u., what would be the atomic weight of hydrogen relative to this mass. [2 + 2] [Ans. (i) 2, (ii) 2 ] [Sol: (ii) wt. of one atom of H = 1  wt. of one atom of 12C 12 = 1  24 = 2 ] 12 Q.11 (i) How many millilitres of a 0.45 M BaCl2 aq. solution contain 15.0 g of BaCl2? (ii) Consider the balanced equation for the formation of 1 mole of Fe2(CO3)3 2Fe(NO3)3 + 3Na2CO3  Fe2(CO3)3 + 6NaNO3 How many oxygen atoms are on each side of the equation? [2 + 2] [Ans. (i) 160 ml (ii) 27NA ] [Sol: (i) BaCl2 moles of BaCl2 = V = 160 ml 15 137  71 0.45 V = 1000 (ii) 2Fe(NO3)3 + 3NO2CO2  Fe2(CO)3 + 6NaNO3 ] Q.12 An aq. solution of H2SO4 contain 196 g acid per litre; solution has density 1.24 g/ml. Calculate (i) wt % of H2SO4 in solution (ii) mole % of H2SO4 in solution [4] [Sol: In 1 litre solution, H SO = 196 g consider 1 litre solution – mass of solution = 1000 × 1.24 = 1240 g mass of H2SO4 = 196 g 196 100 (i) wt % of H2SO4 = 1240 = 15.81 % 196 (ii) moles of solute (H2SO4) = 198 = 2 mass of solvent = 1240 – 196 = 1044 g 1044 moles of solvent = 18 = 58 mole % of H2SO4 = 2 100 2  58 = 3.33 % ] Q.13 (i) In the decomposition of impure KClO3 9.8 g of it gave 1.49 g of KCl and 0.03 mol of O2 gas. Calculate percent purity of KClO3 with the help of law of conservation of mass. [Ans.25%] (ii) Phosphorus (V) chloride reacts with water to give phosphoric acid and hydrogen chloride according to the following equation (not balanced) PCl5 + H2O  H3PO4 + HCl In an experiment 0.36 mole of PCl5 was reacted to 2.88 mole of water (a) Which reactant was the limiting reagent. (b) Calculate the theoretical yields (in moles) of H3PO4 and HCl. [2+3] [Sol: (i) KClO3  KCl + O2 mass of KClO3 = mass of KCl + mass of O2 = 1.49 + (0.03 × 32) = 2.45 g (ii) moles of H3PO4 formed = 0.36 moles of HCl formed = 5 × 0.36 = 1.8 ] Q.14 A mixture of NH4NO3 & (NH4)2HPO4 showed the mass percent of nitrogen to be 30.40%. What is the mass ratio of the two components in the mixture? [5] [Sol: NH4NO3 & (NH4)2HPO4 Let mass of NH4NO3 = x g ––––––– '' –––––––– = y g mass of N in NH NO = 28 x 4 3 80 mass of N in (NH ) HPO = 28 y 4 2 4 132 total mass of N in the mixture = N 28 x + 80 28 y 132 % of N in the mixture = total wt × 100 30.4 =  28 x   28 132    100 x  y  30.4 (x + y) = 35 x + 21.21 y 30.4x + 30.4 y = 35 x + 21.21 y 9.19 y = 4.6 x x  9.19 = 2 y 4.6 mass of NH4 NO3  2 ] mass of NH4 2 HPO4 Q.15 In the preparation of iron from haematite (Fe2O3) by the reaction with carbon Fe2O3 + C  Fe + CO2 (a) Balance the equation (b) How much 80% pure iron could be produced from 120 kg of 90% pure Fe2O3 [5] [Sol: (a) 2Fe2O3 + 3C  4Fe + 3CO2 90 (b) wt. of Fe2O3 taken = 120 × 100 kg = 180 kg nFe2O3  nFe 2 4 108 103  160  2 Fe = 56  4 Fe = 75.6 kg  wt. of iron sample formed = 75.6 0.8 = 94.5 kg Ans. ] Q.16(i) The actual mass of the atomic mass unit is 1.66 × 10–24 g using this value, calculate the mass in grams of 10 atoms of 12C. (ii) A gaseous compound is composed of 85.7% by mass carbon and 14.3% by mass hydrogen. It's density is 2.28 g/litre at 300K and 1.0 atm pressure. Determine the molecular formula of the compound. [2+4] [Ans. (i) 19.92 × 10–23 g (ii)C4H8] [Sol: (i) mass of 1 atom of carbon = 12 amu  mass of 10 atoms of carbon = 10 × 12 amu = 120 × 1.66 × 10–24 g = 2 × 10–22gm (ii) mass% moles of elements emipirical formula 85.7 C = 85.7 nC = 12 =7.14 C7.14H14.3 H = 14.3 nH = 14.3 CH2 Now PM = dRT 1 × M = 2.28 × 0.082 × 300 M = 56  n = 56 14 = 4  molecular formula = (CH2) × 4 = C4H8 ] M Q.17(i) Calculate volume of water to be added to 300 ml of 5 M 15 . HNO3 solution so that its new strength becomes (ii) It is known that when 1 mole of carbon reacted with 1 mole of oxygen (C + O2  CO2), 1 95 Kcal heat liberated & when 1 mole carbon reacted with 0.5 mole of O2 (C + 2 O2  CO), 24 Kcal heat liberated. When 12.0 g of carbon reacted with oxygen to form CO & CO2, 80.8 Kcal of heat liberated and no carbon remained. Calculate the mass of oxygen which reacted. [2 + 4] [Ans. (i) 600 ml (ii) 28.8 gm] [Sol: (i) M1V1 = M2V2  1  300  1 300  V   5 15 H 2O VH O = 600 ml (ii) (i) C + O2  CO2 1 mol 1 mol qrel = 95 kcal (ii) C + ½O2  CO 1 mol 0.5 mol qrel = 24 kcal 12 Total moles of C given = 12 = 1 Let moles of C taking part in (1) reaction = x  moles of C taking part in (2) reaction = 1 – x total heat liberated = 80.8 k cal  heat released in (1) + heat released in (2) = 80.8 kcal  (x × 95) + (1 – x) × 24 = 80.8  x = 0.8  moles of O2 reacted in (1) = 0.8 0.2 & moles of O2 -------(2) = 2 = 0.1  total moles of O2 reacted = 0.9  total mass of O2 reacted = 0.9 × 32 = 28.8 gm. ] Q.18 8KI + 5H2SO4 100%yield 4K2SO4 + 4I2 + H2S + 4H2O H2SO4 + 2NaOH 100%yield  Na2SO4 + 2H2O I2 + 2Na2S2O3 50%yield 2NaI + Na2S4O6 50% yield I2 + 2KCl  2KI + Cl 100 ml of a sample of dilute H2SO4 of specific gravity 1.47 is treated with a 10 g sample of impure KI. The final mixture had to be treated with 20 ml of 4% (w/V) NaOH solution to neutralize left H2SO4. Iodine liberated was trapped and divided into two parts. One part on passing through 20 ml. Na2S2O3 solution yielded 0.3 g of NaI and the other part of I2 sample on treatment with excess KCl yielded 0.497 g of Cl2.Calculate, (i) molarity of Na2S2O3 solution. (ii) % (w/W) of H2SO4 sample (iii) % purity of KI sample [3+3+3] [Ans. (i) 0.2 M, (ii) 2 %, (iii) 53.12 %] [Sol: (1) 8KI + 5H2SO4 100% 4K2SO4 + 4I2 + H2S + 4H2O (2) H2SO4 + 2NaOH 100% Na2SO4 + 2NaI + 2H2O (3) I2 + 2Na2S2O3 50% 2NaI + Na2S4O6 (4) I2 + 2KCl 50% 2KI + Cl2 (i) observed wt. of NaI = 0.3 g 0.3 calculate wt of NaI = 0.5 = 0.6 g 0.6 moles of NaI (calculated) = 150 = 0.004  n Na S O 2 M  20  1000  2 = n NaI 2 0.004 = 2 M = 0.2 Molar (ii) moles of I2 reacted in (3) reaction = moles of Na 2S2O3 2 I2 3  20  0.2 2 1000  n 3 = 0.002 Now in (4) reaction observed wt. of Cl2 = 0.497 g  calculated wt. of Cl2 0.497 = 0.5 = 0.994 gm 0.994 moles of Cl2 formed calculated = 71 = 0.014  moles of I2 reacted in (4) reaction = nCl2 = 0.014  total moles of I2 reacted = 0.002 + 0.014 = 0.016 In (1) nH SO 5  nI2 4  n = 5 × n = 5 × 0.016 H2SO 4 4 = 0.02 I2 4  moles of H2SO4 reacted in (1) = 0.02  moles of H2SO4 taken – moles of H2SO4(left) = 0.02  n H SO (taken) – n H SO (reacted in (2)) = 0.02 (1) In (2) reaction n H SO = From (1) n NaOH = 2 20  4 100  40  2 = 0.01 n H SO (taken) = 0.02 + 0.01 = 0.03  Molarity of H SO taken = 0.031000 = 0.3 M 2 4 100  Molarity = 10  pd M B 10  p1.47 0.3 = 98 p = 2%     %  w  of H2SO4 = 2%   (iii) In (1) equation n nH SO (reacted in 1) KI = 8 2 4 5  KI 166 8 0.02 = 5 % KI = wKI = 5.312 gm 5.312 100 = 53.12 % Ans. ] 10