CHEMISTRY-24-09- 11th (J-Batch)

REVIEW TEST-3 Class : XI (J-Batch) Time : 100 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contains 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, He = 4, Ba = 137 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(J) CHEMISTRY REVIEW TEST-3 Q.1 Calculate rms velocity of a dust particle of mass 8.31×10–12 g at 329.3°C assuming kinetic theory of gases is applicable. [3] [Ans. 1.73 × 10–3 m/sec.] [Sol. u rms 3 8.314 (329.3 + 273) = 8.311015  6.0231023  {1.5 Marks} = 1.73 × 10–3 m/s  {1.5 Marks} ] Q.2 A solution containing 2 moles each of Ba(OH)2, NH4Cl and KOH was divided into two equal parts and treated with V1L and V2L of HCl and NaOH respectively for complete neutralization. Calculate ratio V1 : V2 if molarity of HCl & NaOH solutions is in ratio 1 : 2. [3] [Ans. 6 : 1] [Sol. Each half has 3 mol OH– & 1 mol NH4+. If M' & 2M' are molarities of HCl and NaOH M'V1 = 3 & (2M')V2 = 1  { 1 Mark } Dividing, V1 2V2 = 3  1 V1 = 6 V2 1  { 2 Marks } ] Q.3 6 g of an organic compound on treatment with NaOH gave NH3 which when passed through 1 200 ml of 2 M H2SO4 completely neutralized it. Calculate % of nitrogen in the organic compound. [3] [Sol. Organic Compound NaOH NH3 [Ans. 46.67%] 2NH3 + H2SO4  (NH4)2SO4 0.2 mol 0.1 mol  { 1 Mark } % of N = 0.2 14 6 × 100 = 46.67%  { 2 Marks } ] Q.4 A tube of uniform cross-section of length 100 cm is divided into two parts by a weightless and frictionless piston. One part contains 4 moles of hydrogen at 2 atm at equilibrium and other part contains 1 mole of nitrogen at the same temperature. Assume volume of piston to be negligible. (a) Calculate the length of each compartment if the tube was placed horizontally. (b) The tube is then held at angle of 45° with the horizontal keeping the nitrogen end upwards. Find the length of each compartment. [1.5+1.5] [Ans. Both cases H2 end is 80 cm and N2 end is 20 cm] [Sol.(a) Since P & T are same in each compartment  V  n  V l 2 = 2 V l 2 2 n 4 = 2 = n 1 2 Length of H2 compartment, lH = 4 5 × 100 = 80 cm 1    { 1.5 Marks } Length of N2 compartment, lN = 5 × 100 = 20 cm  (b) Length of H2 compartment, lH2 = 80 cm  Length of N2 compartment, lN = 20 cm   { 1.5 Marks }  ] Q.5 A container holds 3L of N2(g) & liquid water at 300 K at a pressure of 1 atm. The water in 1 container is instantaneously split into H2 & O2 by electrolysis : H2O(l) H2(g) + 2 O2(g). After the completion of reaction, the total pressure was 1.698 atm. Calculate mass of water present in container. Aqueous tension of water at 300 K is 0.04 atm. [5] [Ans. 1.08 g] [Sol. Partial pressure of N2 = 1 – 0.04 = 0.96 atm  { 2 Marks } Let x atm be partial pressure of H2 after decomposition of H2O. Thus, x x + 2 + 0.96 = 1.698  x = 0.492 atm  { 2 Marks } 0.492  3 Moles of H2O = moles of H2 = 24.6 = 0.06 Mass of H2O = 0.06 × 18 = 1.08 g  { 1 Mark } ] Q.6 A 7.38 L vessel containing N2 at 1 atm and 27°C is connected to a 8.21 L vessel containing N2 at 4 atm and 127°C. Keeping both vessels at their original temperatures, calculate (a) Final pressure (b) Amount of N2 in each vessel. [5] [Ans. P = 2.36 atm 9.84 L  0.71 mol 8.21 L  0.59 mol ] [Sol. In 7.38 L vessel — moles of N2 = In 8.21 L vessel — 1 7.38 24.6 = 0.3 moles of N2 = 4  8.21 0.0821 400 = 1  { 1 Mark } Let x mol N2 be in 7.38 L vessel & (1.3 – x) mol in 8.21 L vessel & let P atm be final pressure  P × 7.38 = x (24.6)  { 1 Mark } & P × 8.21 = (1.3 – x) × 0.0821 × 400  { 1 Mark } Dividing, 7.38 8.21 = 24.6 x (1.3  x)  0.0821 400  x = 0.71 mol.  { 1 Mark } i.e. 0.71 mol in 7.38 L vessel & 0.59 mol in 8.21 L vessel P = 2.36 atm  { 1 Mark } ] Q.7 A common method of working of temperature scales is by assigning temperatures for steam point (Boiling point of water) and ice point (melting point of water) or absolute zero. A temperature scale commonly used by engineers is the Rankine (R) scale on which absolute zero is 0 R and steam point is 671.4 R (a) Calculate ice point on Rankine scale. (b) Express 1800 R in Kelvin units. [5] [Ans. (a) 491.4 R, (b) 1000 K] [Sol.(a) Let TR be ice point on Rankine scale  671.4  TR 0  TR 373  273 = 0  273  { 1.5 Marks }  TR = 491.4 R  { 1 Mark } (b) 671.4  491.4 1800  491.4 373  273 TK  273  { 1.5 Marks }  TK = 1000 K  { 1 Mark } ] Q.8 5 moles of a tetra-atomic non-linear gas 'A' at 10 atm & T K are mixed with 10 moles of another T gas B at 3 K & 5 atm in a closed, rigid vessel without energy transfer with surroundings. If final 5T temperature of mixture was 6 K, what is atomicity of gas B? Assume all modes of energy are completely manifested. [5] [Ans. One] [Sol. CV of gas A = 3 2 R + 3 2 R + 6R = 9R  { 1 Mark } Let CV of gas B = x Heat lost by gas A = 5 × 9R × T  5T   { 1 Mark }    6   5T  T  Heat gained by gas B = 10 × x ×   6   { 1 Mark }   T   T  3  45R  6  = 10x  2   x = 2 R  { 1 Mark }      gas is monoatomic  { 1 Mark } ] Q.9(a) Find molality of Ca2+ and NO in 2 M Ca(NO3)2 aqueous solution of density 1.328 g/mL. (b) Also find mole fraction of solvent in solution. [3+2] [Ans. (a) [Ca2+] = 2 molal [NO3–] = 4 molal, (b) 0.965] [Sol.(a) In 1 L of solution, mass of solute = 2 × 164 = 328 g In 1 L solution mass of solvent = 1328 – 328 = 1000 g  molality of Ca(NO3)2 = 2 molal  { 1 Mark } molality of Ca2+ = 2 molal  { 1 Mark } molality of NO3– = 4 molal  { 1 Mark } (b) Mole fraction of water = 1000 /18 6 + 1000 18 = 0.902  { 2 Marks } ] Q.10 Compressbility factor Z vs P plot for four real gases A, B, C & D are shown at 300 K (a) Out of gases A and B whose molecules are larger? Justify. (b) If slope of curve for B is 0.02 atm–1, calculate Van der Waals constant b for gas B. (c) Arrange Boyle temperatures (T) for gases A, B, C & D in increasing order. [2+2+2] [Ans. (a) A > B, (b) 0.492 L/mol, (c) TA < TB < TC < TD ] [Sol.(a) For gas A & B; a is negligible bP so, Z = 1 + RT b slope of Z vs P curve = RT  { 1 Mark } Since slope is more for A  bA > bB  A is a larger molecule than B  { 1 Mark } b (b) RT = 0.02  { 1 Mark }  b = 0.02 × 24.6 = 0.492 L/mol  { 1 Mark } (c) TA < TB < TC < TD  { 2 Marks } ] Q.11 20 g mixture of Na2CO3, NaHCO3 and NaCl on heating to moderate temperatures produced 1.12 L of CO2 at NTP. 8 g of the same mixture in a different experiment required 80 ml of 1 M HNO3 for neutralization. Calculate mass % of each component in mixture. [6] [Ans. 26.5 % Na2CO3, 42% NaHCO3, 31.5% NaCl] [Sol. Let a & b be moles of Na2CO3 & NaHCO3 in 20 g mixture. 2NaHCO3  Na2CO3 + H2O + CO2 b moles b/2 moles b  2 = 1.12 22.4  b = 0.1 mol = mol. of NaHCO3  { 1 Mark } In 8g of mixture, moles of Na2CO3 = 0.4 a & moles of NaHCO3 = 0.04 Na2CO3 + 2HNO3  2NaNO3 + H2O + w2 0.4a 0.8a NaHCO3 + HNO3  NaNO3 + H2O + CO2 0.04 0.04 Total moles of HNO3 = 0.8a + 0.04 = 0.08  a = 0.05 moles = mol. of Na2CO3  { 2 Marks } Mass% of Na2CO3 = 0.05106 20 × 100 = 26.5%  { 1 Mark } Mass% of NaHCO3 = 0.1 84 20 × 100 = 42%  { 1 Mark } Mass% of NaCl = 100 – 26.5 – 42 = 31.5%  { 1 Mark } ] Q.12 A gaseous mixture containing N2 & H2 in mole ratio 1 : 3 is used to carry out the reaction in a sealed container. N2 (g) + 3H2(g)  2NH3 (g) The total pressure was 0.8 atm at 25°C before the reaction. Determine final pressure at 323°C assuming 60% yield of reaction. [6] [Ans. 1.12 atm] [Sol. At 298 K total pressure = 0.8 atm At 596 K total pressure = 1.6 atm i.e. initial pressure of N2 = 0.4 atm & initial pressure of H2 = 1.2 atm  { 1 Mark }  { 1 Mark } N2 + 3H2  2NH3 t = 0 0.4 1.2 t =  0.4(1–0.6) 1.2(1–0.6) 2 × 0.4 × 0.6  { 2 Marks } Total pressure = 0.16 + 0.48 + 0.48 = 1.12 atm  { 2 Marks } ] Q.13 Calculate (a) Number of nitrogen atoms in 160 amu of NH4NO3. (b) Number of gram-atoms of S in 490 kg H2SO4. (c) Grams of Al2(SO4)3 containing 3.2 amu of S. [6] [Ans. (a) 4, (b) 5000 moles, (c) 1.89 × 10–23 g] [Sol.(a) 160 amu has 2 molecules of NH4NO3  4 atoms of N  { 2 Marks } 490 103 (b) moles of H2SO4 = 98 = 5000 moles of S = 5000  { 2 Marks } (c) 96 amu of S is in 342 amu Al2(SO4)3  342  3.2 amu of S is in   96  × 3.2 amu of Al2(SO4)3  342  i.e. 3.2 ×    96  × 1.66 × 10–24 g = 1.89 × 10–23 g  { 2 Marks } ] Q.14 Calculate compressibility factor (Z) for 0.02 moles of a Van der Waals gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible. Given: RT = 20 L-atm-mol–1 (T = Temperature of gas), a = 1000 atm-L2-mol–2 [6] [Ans. 0.01]  1000  (0.02)2  [Sol.  0.1+  V = 20 × 0.02    { 2 Marks }  0.1 V2 – 0.4 V + 0.4 = 0  V2 – 4V + 4 = 0  V = 2 L  { 2 Marks } PV Z = RT 0.1 2 = 20 = 0.01  { 2 Marks } ] Q.15 Three ideal gas samples in separate equal volume containers are taken and following data is given: Pressure Temperature Mean free paths Mol. wt. Gas A 1 atm 1600 K 0.16 nm 20 Gas B 2 atm 200 K 0.16 nm 40 Gas C 4 atm 400 K 0.04 nm 80 Calculate ratio (A : B : C) of following for the three gases. (a) Collision frequencies (Z11). (b) Number of collision by one molecule per sec. (Z1). (c) Average velocities. [3+3+2] [Ans. (a) 1 : 4 : 16, (b) 4 : 1 : 4, (c) 4 : 1 : 1 ] [Sol. Let A, B & C be collision diameters of A, B & C  = kT i.e.    A (a) Z11  : B : C =  { 1 : : Mark } = 4 : 1 : 2  A : B : C = 16 1 20 (1600)3/ 2 : : = 1 : 4 : 16  { 2 Marks } (b) Z1   { 1 Mark } 161  A : B : C = 201600 : : (c) uavg  = 4 : 1 : 4  { 2 Marks }  A : B : C = 1600 : : 20 = 4 : 1 : 1  { 2 Marks } ]