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XI(J) CHEMISTRY REVIEW TEST-2 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 Q.1 In a mass spectrograph experiment, He+ ion was found to strike at distance of x cm. Calculate distance where H2O+2 ion would strike if (a) Potential difference 'V' and magnetic field 'B' are kept constant. (b) Velocity of ions & magnetic field are constant [2+2] [Ans. (a) 3x/2 (b) 9x/4] [Sol.(a) r = r + r 2 3 He = r +2 3 H2O r +2 = x H2O 2 mu (b) r = qB m r q r + 4 9 He = r +2 9 H2O r +2 = x ] H2O 4 Q.2 Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of N2 & 12.7 g of I2 at temperature T1 (K). The other container (II) is completely evacuated. The container (I) is heated to T2 (K) while container (II) is maintained at T2/3 (K). The valve is opened. Calculate masses of N2 in both containers after a very long time. I2 sublimes at T2 (K) and volume of vessel (I) is half that of vessel (II). Neglect vapour pressure of I2 (s). [4] [Ans. 0.4 g in (I) & 2.4 g in (II)] [Sol. Let x mole of N2 diffuse to vessel II P = final pressure of N2 P (2V) = xR (T2/3) P (V) = (0.1 – x) R T2 2 = x 3(0.1 x) 0.6 – 6 x = x x = 0.6/7 i.e. mass = 2.4 g II has 2.4 g N2 & I has 0.4 g of N2 ] Q.3 H2O2 + 2KI 40%yield I2 + 2KOH H2O2 + 2KMnO4 + 3H2SO4 50%yieldK2SO4 + 2MnSO4 + 3O2 + 4H2O 100 ml of H2O2 sample was divided into two parts. First part was treated with KI and KOH formed required 200 ml of M/2 H2SO4 for neutralisation. Other part was treated with KMnO4 yielding 6.74 lit. of O2 at STP. Using % yield indicated find volume strength of H2O2 sample used. [4] [Ans. 50.4 volumes] [Sol. mol of H2SO4 = 0.1 mol of KOH = 0.2 mol of H O = 0.2 × 1 = 0.25 mol of O2 = 2 6.74 22.4 2 2 = 0.3 0.4 mol of H2O2 = 0.3 3 0.5 = 0.2 Total mol of H2O2 = 0.45 0.45 Molarity of H2O2 = 0.1 = 4.5 M Vol. strength = 11.2 × 4.5 = 50.4 volumes ] Q.4 Rate of diffusion of ozonized oxygen is 0.4 5 times that of pure oxygen. Find (a) Percentage by mass of ozone in the ozonized sample (b) Degree of dissociation of oxygen assuming pure O2 in the sample initially. [3+1] [Ans. 60%, 0.6] [Sol. rmix r 2 = = 0.4 32 = 0.8 M Mavg 2 O2 l 3 O3 avg = 40 g/mol t = 0 1 2 t = 1 – 3 32 1 ( / 3) = = 0.6 40 mol. of O2 mol. of O3 1 1 = 2 / 3 3 0.4 = 0.4 = 1 ; mass of O2 mass of O3 32 2 = 48 = 3 mass % of O3 = 5 × 100 = 60% ] Q.5 An open vessel at 27°C is heated. Assuming that volume of vessel remains constant. Calculate (a) Temperature to which vessel was heated till (3/5)th of air in it has been expelled. (b) Fraction of molecules escaped out when vessel is heated to 900 K (c) Temperature at

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XI(J) CHEMISTRY REVIEW TEST-2 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 Q.1 In a mass spectrograph experiment, He+ ion was found to strike at distance of x cm. Calculate distance where H2O+2 ion would strike if (a) Potential difference 'V' and magnetic field 'B' are kept constant. (b) Velocity of ions & magnetic field are constant [2+2] [Ans. (a) 3x/2 (b) 9x/4] [Sol.(a) r = r + r 2 3 He = r +2 3 H2O r +2 = x H2O 2 mu (b) r = qB m r q r + 4 9 He = r +2 9 H2O r +2 = x ] H2O 4 Q.2 Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of N2 & 12.7 g of I2 at temperature T1 (K). The other container (II) is completely evacuated. The container (I) is heated to T2 (K) while container (II) is maintained at T2/3 (K). The valve is opened. Calculate masses of N2 in both containers after a very long time. I2 sublimes at T2 (K) and volume of vessel (I) is half that of vessel (II). Neglect vapour pressure of I2 (s). [4] [Ans. 0.4 g in (I) & 2.4 g in (II)] [Sol. Let x mole of N2 diffuse to vessel II P = final pressure of N2 P (2V) = xR (T2/3) P (V) = (0.1 – x) R T2 2 = x 3(0.1 x) 0.6 – 6 x = x x = 0.6/7 i.e. mass = 2.4 g II has 2.4 g N2 & I has 0.4 g of N2 ] Q.3 H2O2 + 2KI 40%yield I2 + 2KOH H2O2 + 2KMnO4 + 3H2SO4 50%yieldK2SO4 + 2MnSO4 + 3O2 + 4H2O 100 ml of H2O2 sample was divided into two parts. First part was treated with KI and KOH formed required 200 ml of M/2 H2SO4 for neutralisation. Other part was treated with KMnO4 yielding 6.74 lit. of O2 at STP. Using % yield indicated find volume strength of H2O2 sample used. [4] [Ans. 50.4 volumes] [Sol. mol of H2SO4 = 0.1 mol of KOH = 0.2 mol of H O = 0.2 × 1 = 0.25 mol of O2 = 2 6.74 22.4 2 2 = 0.3 0.4 mol of H2O2 = 0.3 3 0.5 = 0.2 Total mol of H2O2 = 0.45 0.45 Molarity of H2O2 = 0.1 = 4.5 M Vol. strength = 11.2 × 4.5 = 50.4 volumes ] Q.4 Rate of diffusion of ozonized oxygen is 0.4 5 times that of pure oxygen. Find (a) Percentage by mass of ozone in the ozonized sample (b) Degree of dissociation of oxygen assuming pure O2 in the sample initially. [3+1] [Ans. 60%, 0.6] [Sol. rmix r 2 = = 0.4 32 = 0.8 M Mavg 2 O2 l 3 O3 avg = 40 g/mol t = 0 1 2 t = 1 – 3 32 1 ( / 3) = = 0.6 40 mol. of O2 mol. of O3 1 1 = 2 / 3 3 0.4 = 0.4 = 1 ; mass of O2 mass of O3 32 2 = 48 = 3 mass % of O3 = 5 × 100 = 60% ] Q.5 An open vessel at 27°C is heated. Assuming that volume of vessel remains constant. Calculate (a) Temperature to which vessel was heated till (3/5)th of air in it has been expelled. (b) Fraction of molecules escaped out when vessel is heated to 900 K (c) Temperature at

- 1. XI(J) CHEMISTRY REVIEW TEST-2 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 Q.1 In a mass spectrograph experiment, He+ ion was found to strike at distance of x cm. Calculate distance where H2O+2 ion would strike if (a) Potential difference 'V' and magnetic field 'B' are kept constant. (b) Velocity of ions & magnetic field are constant [2+2] [Ans. (a) 3x/2 (b) 9x/4] [Sol.(a) r = 2 qB Vm 2 r q m 2 2O H He r r = 3 2 2 2O H r = x 2 3 (b) r = qB mu r q m 2 2O H He r r = 9 4 2 2O H r = x 4 9 ] Q.2 Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of N2 & 12.7 g of I2 at temperature T1 (K). The other container (II) is completely evacuated. The container (I) is heated to T2 (K) while container (II) is maintained at T2/3 (K). The valve is opened. Calculate masses of N2 in both containers after a very long time. I2 sublimes at T2 (K) and volume of vessel (I) is half that of vessel (II). Neglect vapour pressure of I2 (s). [4] [Ans. 0.4 g in (I) & 2.4 g in (II)] [Sol. Let x mole of N2 diffuse to vessel II P = final pressure of N2 P (2V) = xR (T2/3) P (V) = (0.1 – x) R T2 2 = ) x 1 . 0 ( 3 x 0.6 – 6 x = x x = 0.6/7 i.e. mass = 2.4 g II has 2.4 g N2 & I has 0.4 g of N2 ]
- 2. Q.3 H2O2 + 2KI yield % 40 I2 + 2KOH H2O2 + 2KMnO4 + 3H2SO4 yield % 50 K2SO4 + 2MnSO4 + 3O2 + 4H2O 100 ml of H2O2 sample was divided into two parts. First part was treated with KI and KOH formed required 200 ml of M/2 H2SO4 for neutralisation. Other part was treated with KMnO4 yielding 6.74 lit. of O2 at STP. Using % yield indicated find volume strength of H2O2 sample used. [4] [Ans. 50.4 volumes] [Sol. mol of H2SO4 = 0.1 mol of KOH = 0.2 mol of H2O2 = 2 2 . 0 × 4 . 0 1 = 0.25 mol of O2 = 4 . 22 74 . 6 = 0.3 mol of H2O2 = 5 . 0 3 3 . 0 = 0.2 Total mol of H2O2 = 0.45 Molarity of H2O2 = 1 . 0 45 . 0 = 4.5 M Vol. strength = 11.2 × 4.5 = 50.4 volumes ] Q.4 Rate of diffusion of ozonized oxygen is 5 4 . 0 times that of pure oxygen. Find (a) Percentage by mass of ozone in the ozonized sample (b) Degree of dissociation of oxygen assuming pure O2 in the sample initially. [3+1] [Ans. 60%, 0.6] [Sol. 2 O mix r r = avg M 32 = 5 4 . 0 avg M 32 = 0.8 Mavg = 40 g/mol O2 l 3 2 O3 t = 0 1 t = 1 – 3 2 40 32 = 1 ) 3 / ( 1 = 0.6 3 2 mol. of O mol. of O = 3 / 2 1 = 4 . 0 4 . 0 = 1 ; 3 2 mass of O mass of O = 48 32 = 3 2 mass % of O3 = 5 3 × 100 = 60% ]
- 3. Q.5 An open vessel at 27°C is heated. Assuming that volume of vessel remains constant. Calculate (a) Temperature to which vessel was heated till (3/5)th of air in it has been expelled. (b) Fraction of molecules escaped out when vessel is heated to 900 K (c) Temperature at which half of the air escapes out. [1.5+1.5+1] [Ans. (a) 477°C (b) 2/3 (c) 327°C] [Sol.(a) 5 3 1 R 300 PV = ) R ( T PV T = 300 × 2 5 = 750 K = 477°C (b) Let x be fraction escaped R 300 PV (1 – x) = R 900 PV x = 2/3 (c) R 300 PV (1 – 0.5) = TR PV T = 327°C ] Q.6 Heat released when 1 mole of acid HX neutralizes 1 mol of base BOH is 3 30 KJ. To a solution containing HX of molarity 10 M another solution containing base BOH was slowly added. A plot between heat liberated (J) and volume of BOH (ml) added was obtained as shown. (a) Calculate molarity of base. (b) Calculate volume of original HX solution [2.5+2.5] [Ans. (a) 0.011 M (b) 66.66 ml] [Sol.(a) Let V ml of base be added & M be molarity of base Moles of base = MV × 10–3 Heat released = MV × 10–3 × 3 30 × 103 J = ( 3 30 MV) J Slope = 3 30 M = 3 1 M = 1/90 = 0.011 M (b) 346.4 = 3 30 × 10 1 × V V (where V is vol. of HX in ml) V = 66.66 ml ] Q.7 200 g of an oleum sample was dissolved in excess water. To this solution 1 kg of impure Ba(NO3)2 was added causing formation of BaSO4 precipitate. The precipitate formed was filtered, washed & dried. Its weight was 582.5 gm. Find % labelling of oleum sample. [5] [Ans. 122.5%] [Sol. moles of BaSO4 = moles of H2SO4 = 2.5 moles Let x be % labelling of oleum moles of SO3 = 18 100 x ×2
- 4. moles of H2SO4 = 2 80 18 100 x 200 98 1 2 × 18 100 x + 98 200 – 49 80 18 100 x = 2.5 x = 122.5% ] Q.8 400 ml of 0.2 × 10–3 M Ca(NO3)2 solution is mixed with 400 ml of 10–4 M NaHCO3 solution. If density of resulting solution is 1 g/ml. Calculate (a) ppm of Ca(HCO3)2 in final solution (b) Molality of Na+ in solution (c) Molarity of Ca2+ in solution [2+2+1] [Ans. (a) 4.05 ppm (b) 5 × 10–5 mol/kg (c) 10–4 M] [Sol. Ca(NO3)2 (aq) + 2NaHCO3 (aq) Ca(HCO3)2 (aq) + 2NaNO3 (aq) t = 0 80 × 10–6 mol 40 × 10–6 mol t = 60 × 10–6 mol 0 20 × 10–6 mol 40 × 10–6 mol Mass of final solution = 1 × 800 = 800 gm (a) ppm of Ca(HCO3)2 = 800 162 10 20 6 × 106 = 4.05 ppm (b) Molality of Na+ = 800 10 40 6 × 1000 = 5 × 10–5 mol/kg (c) Molarity of Ca2+ = 3 6 10 800 10 80 = 10–4 M ] Q.9 A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density (1/3)rd that of mercury. Initially h was 228 cm. Through a small hole in the bulb gas leaked causing pressure decrease as dt dp = – kP. If value of h is 114 cm after 7 minutes. Calculate value of k in units of hour–1. [Use: ln (4/3) = 0.28 & density of Hg = 13.6 g/ml] [5] [Ans. 2.4 hr–1] [Sol. P = P0e–kt P0 = 3 228 + 76 = 152 cm Hg At t = 7 min. P = 3 114 + 76 = 114 cm Hg 152e–kt = 114 k = t 1 ln 114 152 = 7 1 × 0.28 = 0.04 min–1 = 0.04 × 60 hr–1 = 2.4 hr–1 ]
- 5. Q.10 Calculate minimum number of balloons each of volume 82.1 lit. required to lift a mass of 1 kg to a height of 831 m. Given: molar mass of air = 29 g/mol, temperature is constant at 290 K and mass of each balloon is 50 g. [Use e–0.1 = 0.9, pressure at sea level = 1 atm, acceleration due to gravity (g) = 10 m/s2] [5] [Ans. 25 balloons] [Sol. Density of air at sea level, d0 = 290 0821 . 0 1 29 = 821 . 0 1 g/lit Density at 831 m = d = d0e–Mgh/RT = 821 . 0 1 × 290 31 . 8 831 10 10 29 – 3 e = 821 . 0 1 × e–0.1 = 821 . 0 9 . 0 g/lit Let n be number of balloons 103 × g + n × 50g = 821 . 0 9 . 0 × 82.1 × g × n 1000 + 50 n = 90 n n = 25 balloons ] Q.11 A mixture 1,2–Dipropene and hydrogen gas was placed in a rigid steel container at a constant temperature of 18°C. Initial pressure of mixture was 10 atm. Sparking the mixture caused hydrogenation reaction C3H4 (g) + 2H2(g) C3H8 (g) causing pressure to decrease by 6 atm. Excess O2 was then added and on sparking the pressure further decreased by11 atm. Calculate decrease in pressure by treating final mixture with aqueous KOH solution. [6] [Ans. 12 atm] [Sol. If C3H4 is limiting reagent C3H4 + 2H2 C3H8 t = 0 P 10–P t = 0 10–3P P 10 – 3P + P = 4 P = 3 atm On sparking, C3H8 + 5O2 3CO2 + 4H2O 3 atm 15 atm 9atm H2 + 2 1 O2 H2O 1atm 2 1 atm Pressure decrease = (3 + 15 – 9) + (1 + 0.5) = 10.5 atm 11 atm Case II: H2 is limiting reagent C3H4 + 2H2 C3H8 t = 0 P 10–P t = P– 2 P 10 0 2 P 10 P = 4 atm
- 6. C3H4 + 4O2 3CO2 + 2H2O 1 atm 4atm 3 atm C3H8 + 5O2 3CO2 + 4H2O 3 atm 15atm 9 atm (1 + 4 – 3) + (3 + 15 – 9) = 11 atm Pressure of CO2 = 12 atm Pressure decreased with KOH solution = 12 atm ] Q.12 0.2 gm of an anhydrous organic acid gave on combustion 0.04 g water and 0.195 g of CO2. 0.5 g of its silver salt leaves on ignition 0.355 g silver. In another experiment 100 ml of 0.2 M acid solution was neutralised by 400 ml of 10 1 M caustic potash solution. What is molecular formula of acid? [6] [Ans. C2H2O4] [Sol. moles of H = 2 18 04 . 0 moles of C = 44 195 . 0 moles of O = 16 1 44 195 . 0 12 9 04 . 0 2 . 0 C : H : O = 0.0044 : 0.0044 : 0.0089 = 1 : 1 : 2 Empirical formula = CHO2 Let n be basicity of acid 100 × 0.2 × n = 400 × 10 1 n = 2 Let H2X be acid and Ag2X its silver salt Ag2X 2Ag 0.5 g 0.355 2 M 216 2 5 . 0 = 108 355 . 0 M = Mol. wt. of acid 214 M 1 = 108 355 . 0 M = 90.2 n = 45 2 . 90 2 Molecular Formula = C2H2O4 ]
- 7. Q.13 A vessel of volume 246 lit. contains 1.8 g of water and N2 gas at a total pressure of 0.41 atm. If half the contents of vessel are transferred to another evacuated vessel of volume 24.6 lit. Calculate (a) Final total pressure in both vessels (b) Relative humidity in both vessels. [Assume constant temperature of 300 K. Aqueous tension of water is 15.2 mmHg. ] [3+3] [Ans. In 246 lit vessel : T P = 0.205 atm and RH = 25% In 24.6 lit vessel : T P = 2.02 atm and RH = 100% ] [Sol. Moles of H2O = 0.1 mol Partial pressure of H2O = 246 300 082 . 0 1 . 0 = 0.01 atm Aqueous tension of water = 760 2 . 15 = 0.02 atm All water is in vapour form moles of N2 in vessel = 6 . 24 246 4 . 0 = 4 moles In each vessel there are 2 moles of N2 & 0.05 moles of water vapour. In 246 lit. vessel total pressure P = 246 6 . 24 05 . 2 = 0.205 atm In 24.6 lit vessel, partial pressure of water = 6 . 24 6 . 24 05 . 0 = 0.05 atm Since this pressure is greater than aqueous tension so, partial pressure of water = 0.02 atm(rest water in liquid form) Pressure exerted by N2 = 6 . 24 6 . 24 2 = 2 atm Total pressure = 2.02 atm RH of 2nd vessel = 100% RH of 1st vessel = 02 . 0 005 . 0 × 100 = 25% ] Q.14 A tube of uniform cross-section of 0.5 cm2 and length 118 cm is divided into three compartments by using two frictionless pistons each of mass 101.32 gm. The compartments I, II & III are filled by equal masses of CH4, O2 and He gases respectively as shown. The tube is then inverted at angle of 30° with horizontal as shown. If pressure in CH4 compartment was 1 atm find volumes of compartments I, II and III. Assume volume of piston is negligible and temperature is constant. [Use g = 10 m/s2] [6] [Ans.VI = 9 ml, VII = 5 ml, VIII = 45 ml] [Sol. Let P1, P2 & P3 be pressures in I, II & III compartments in atm units
- 8. Then, P2 – P3 = 3 4 3 10 32 . 101 10 5 . 0 ) 2 / 1 ( 10 10 32 . 101 atm = 0.1 P1 – P2 = 0.1 Q P1 = 1 atm P2 = 0.9 atm & P3 = 0.8 atm Let w gm of each gas be present 1 × V1 = 16 w RT T 0.9 × V2 = 32 w RT T 0.8 × V3 = 4 w RT T where V1, V2, V3 = Vol. of compartment 1, 2 & 3 V1 : V2 : V3 = 16 1 : 8 . 28 1 : 2 . 3 1 = 1.8 : 1 : 9 V1 = 8 . 11 8 . 1 × 59 ml = 9 ml V2 = 8 . 11 1 × 59 = 5 ml V3 = 8 . 11 9 × 59 = 45 ml ] Q.15 1.8 g of MBrx (M is a metal) when heated with a stream of HCl gas was completely converted to chloride MClx which weighed 1 g. Specific heat of metal is 0.11 cal/g. Calculate percent error in molecular weight of metal bromide as determined from specific heat data. [6] [Ans. 0.7 %] [Sol. MBrx + xHCl MClx + xHBr Atomic mass of metal = 11 . 0 4 . 6 = 58.2 x 80 2 . 58 8 . 1 = x 5 . 35 2 . 58 1 0.8(58.2) = (80 – 35.5 × 1.8)x x = 2.89 x = 3 Let M be exact atomic wt. of metal ) 3 ( 80 M 8 . 1 = ) 3 ( 5 . 35 M 1 M = 8 . 0 3 ) 8 . 1 5 . 35 80 ( M = 60.3 Exact mol. wt. of MBr3 = 60.3 + 240 = 300.3 Mol. wt. of MBr3 from specific heat data = 58.2 + 240 = 298.2 % error = 37 . 300 2 . 298 37 . 300 0.7% Ans. ]