CHEMISTRY-04-06- 11th (PQRS)

REVIEW TEST-1 Class : XI 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 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 [Sol. I 0.5 mol O3 = 24 g O3 II 0.5 gm molecule of N2 = 14 g N2 3.0111023  III 6.0231023 32 = 16 g O2 5.6 IV 22.4  44 g CO2 = 11 g CO2 IV < II < III < I Ans. ] Q.2 Calculate number of neutrons in 17 g of H2O2. [3] [Sol. 0.5 mol H2O2 has 8 mol neutrons  8NA Ans. ] Q.3 The hydrated salt Na2CO3.xH2O undergoes 63% loss in mass on heating and becomes anhydrous, calculate the value of x [3] [Sol. Na2CO3.xH2O (s)  Na2CO3(s) + xH2O (g) Let 100g of Na2CO3.xH2O be present 100x  106  18x mol of H2O formed  63 = 100x 106  18x (18) 6678 + 1134 = 1800 x 666x = 6678 x = 10.027 Ans.] Q.4 The vapour density of a mixture containing NO2 & N2O4 is 27.6. Calculate mole fraction of N2O4 in the mixture. [3] [Sol. Let 1 mol of mixture have x mol N2O4  2 × 27.6 = x(92) + (1–x) 46 x = 0.2 Ans.] 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] 5 [Sol. Mol. wt. of XY2 = Mol. wt. of X2Y3 = 0.05 = 100 90 3.0111023 ×6.023×1023 = 180 Let molar mass of X and Y are a & b respectively  a + 2b = 100 2a+3b = 180 a = 40; b = 30 Ans.] Q.6 Average atomic mass of magnesium is 24.31 a.m.u. This magnesium is composed of 79 mole % of 24Mg & remaining 21 mole % of 25Mg and 26Mg. Calculate mole % of 26Mg. [3] [Sol. Let mol% of 26Mg be x (21 x)25  x(26)  79(24)  100 = 24.31 x = 10% Ans.] Q.7 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] mass of Al2 (SO4 )3 [Sol. mass of water × 106 = 34.2 1 Lt. solution contains 1000 g of water  In 1 Lt. solution mass of Al2 (SO4)3 = 34.2 1000 = 34.2 mg 106 34.2103 molarity of Al2(SO4)3 = 342 M = 10–4 M Al (SO ) (aq)  2Al3+(aq) + 3SO2 (aq) 2 4 3 4 10–4 M 2×10–4 M 3×10–4 M 3 × 10–4 M Ans.] Q.8 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] [Sol. 3 mol NaOH in 1000 g water 3 mol NaOH in [1000 + 40 × 3] g solution 1000  40 3 3 mol NaOH in  3 1100  L solution 1000 120 molarity = 1100 3300 = 1120 = 2.94 M Ans. ] Q.9 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. Let V ml of alcohol be required  mass of alcohol is same in both solutions 90  100 ×0.8 ×V = 30 100 × 0.9 × 150 V = 56.25 mL Ans.] Q.10 1.44 gram of Titanium (Ti) reacted with excess of O2 and produced x gram of non-stoichiometric compound Ti1.44O. Find the value of x. [3] [Sol. Ti + O2  Ti1.44O or by balancing reaction 1.44 48 mol x 48 (1.44) 16 mol 1.44 Ti + 1/2O2  Ti1.44O 1.44 P.O.A.C. on Ti  48 = x = 1.77 g Ans.] 1.44x 48(1.44) 16 PART B Q.11(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] [Sol. (i) 10 × 12 + 14 × 1 + x ×14 = 162  x = 2 Ans. (ii) New atomic mass of hydrogen = 2 amu Ans.] Q.12 (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] [Sol. (i) Let volume is V ml  V  15  1000  ×0.45 = 137  2  35.5   V = 160. 26 ml Ans. (ii) 2Fe(NO3)3 + 3Na2CO3  Fe2(CO3)3 + 6NaNO3 No. of O-atom each side = 18 + 9 = 27 Ans. ] Q.13 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. Assume 1000 ml solution wt. of solution = 1000 × 1.24 gm 196 (i) % wt. of H2SO4 = 1000 1.24 ×100 = 15.81 % Ans. (ii) Mole % of H SO nH SO = 2 4 ×100 2 4 nH SO  nH O 196   98    = 196   1240 196  ×100   98  18  2 = 2  58 ×100 = 3.33 % Ans.] Q.14 (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. (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) 2KClO3  2KCl + 3O 9.8 gm 1.49 gm 0.03 ×32 = 0.96 gm Total mass of product = 1.49 + 0.96 = 2.45 gm Then the mass of KClO3 (pure)= 2.45 gm 2.45 % purity of KClO3 = 9.8 ×100 = 25 % Ans. (ii) PCl5 + 4H2O  H3PO4 + 5HCl (Balanced reaction) 0.36 mol 2.88 mol Li2 (a) PCl5 is the L.R. (b) Theoretical yield of H3PO4 = 0.36 mole Ans. Theoretic yield of HCl = 5 × 0.36 = 1.8 mole Ans. ] Q.15 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. x gm y gm NH4 NO3  (NH4 )2 HPO4 –––––––– Total( xy)gm x  2 14  y  2 14 28  4  48 218 1 31 64 ×100 = 30.4 x  y 28x  28y  80 132 x  y ×100 = 30.4 0.35x + 0.212y = 0.304 x + 0.304 y  0.046 x = 0.092 y x y = 2 Ans. ] Q.16 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] 3C [Sol. (a) Fe2O3 + 2   2Fe + 3 CO 2 2 or 2F2O3 + 3C  4Fe + 3CO2 Ans.  120 1000  90 (b) no. of moles of Fe2O3 =  2  56  48  100   Amount of 80% pure iron = 120 1000 0.9  2 56 = 94500 gram Ans. 2 56  48 0.8 = 94.5 kg Ans. ] Q.17(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] [Sol. (i) 1 amu = 1.66 × 10–24 gm Mass of 10 atoms of 12C = 10×12 amu = 10×12×1.66×10–24 = 1.992 ×10–22 gm Ans. PM dRT 2.28 0.0821 300 (ii) d = RT  M = P = 1 = 56.15 gm/mol E.F. : 85.7 : 14.3 = 7.14 : 14.3 = 1 : 2 12 1 E.F.  CH2 56.15 M.F. = (CH2)n where n = 12  2 = 4 M.F. C4H8 Ans. ] M Q.18(i) Calculate volume of water to be added to 300 ml of 5 M becomes 15 . HNO3 solution so that its new strength (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] [Sol. (i) Let V ml of H2O is added 300  1 = mol of HNO = V × 1 1000 5 3 1000 15  V = 900 ml water added = 900 – 300 = 600 ml Ans. (ii) Let x mole of carbon is reacted with O2 and form x mole of CO2 C + O2  CO2, heat liberated = 95 Kcal x x x 1 C + 2 O2  CO, heat liberated = 24 Kcal 1–x (1–x)/2 1–x x × 95 + (1–x) × 24 = 80.8  71x + 24 = 80.8 x = 0.8 mol  Mol of O2 1 x used = x + 2 1 0.8 = 0.8 + 2 = 0.8 + 0.1 = 0.9 mol  Mass of O2 reacted = 0.9 × 32 = 28.8 gm Ans.] Q.19 8KI + 5H2SO4 100%yield 4K2SO4 + 4I2 + H2S + 4H2O 100% yield H2SO4 + 2NaOH  Na SO + 2H O 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 [6] 0.3 [Sol. (i) no. of mole of NaI = 23 127  0.002 100 0.004 no.of mole of Na2S2O3 required  0.002 × 50 = 0.2 M Ans. (ii) Initial mass of H2SO4  100 ×1.47  147gm  Molarity = 20 0.02 ×1000 mass of unreacted H2SO4  moles of NaOH used  2 100 ml  4 g 4 20 ml  100 ×20  0.8 gm 0.8 nNaOH = 40 = 0.02  0.01  0.497   0.3  1 Total moles of I2 formed   71  × 2 +  150  × 2 × 2      0.014 + 0.002  0.016 5 Moles H2SO4  4 × 0.016  0.02 Total initial moles of H2SO40.02+0.010.03Mass of H2SO4 = 0.03 × 98 = 2.94 gm w 2.94 % W = 147 × 100 = 2 % Ans. 8 (iii) Moles of KI reacted with H2SO4  5 × 0.02  0.032 Mass of KI  0.032 × (127 + 32)  5.312 g 5.312 % purity  10 × 100  53.12 % Ans.]