CHEMISTRY-09-07- 11th (PQRS) SOLUTION

XI(PQRS) CHEMISTRY SOLUTIONS REVIEW TEST-2 USEFUL DATA Atomic weights: Cl = 35.5, O = 16, H = 1, He = 4, P = 31, Ag = 108, N = 14, S = 32, Na = 23, C = 12, Fe = 56, K = 39, Br = 80. R = 0.0821 (lt-atm)/(K-mol) PART-A Each question is of 3 marks. Q.1 10 gm of H2 gas at 300 K would (a) occupy how much volume at 5 atm pressure. (b) exert how much pressure if it occupies 10 lit. volume. [Sol. (a) 5 × V = 5 × R × 300 V = 24.63 lit. (b) P × 10 = 5 × R × 300 P = 12.315 atm ] Q.2 Determine the % labelling of oleum in a sample containing 60% w/w "free" SO3. Also calculate % w/w of "combined" SO3 in oleum. [Sol. Mass of free SO3 = 60 gm in 100 gm oleum Mass of water which can combine = 13.5 % labelling = 113.5 % Mass of "combined" SO3 = 40  80 = 32.65 ] 98 Q.3 Calculate the molar mass of diacidic organic lewis base if 12 gm of chloroplatinate salt on ignition produced 5 gm residue. [Sol. B H2PtCl6  Pt, POAC for Pt, n BH PtCl = n Pt 12 MB + 410 5 = 195 ; MB = 58 ] Q.4 Calculate the number of steps required to enrich a sample containing SO2 and O2 in molar ratio 4 : 1 to 1 : 8 mass ratio. [Sol.    2    SO 1 = 4 ,    2    SO = 32 = 16 1 1  2 i 16 1  2 f 64 f = 1 4 = 64 ; f1 = f = (f1)x  x  64 =      n = 12 ] = 2x/2 Q.5 Calculate the intercept on y-axis and slope of curves plotted between (a) log10 P v/s log10 T (b) P/T v/s T for an ideal gas having 2 moles in a closed rigid container of volume 8.21 lit. (P = Pressure in atm & T = Temp in K, log10 2 = 0.300) [Sol. (a) log P v/s log T slope = 1 nR Intercept on y-axis = log10 V 2  0.0821 P (b) T v/s T curve nR Intercept = V 2  0.0821 = log10 8.21  –1.70 = 8.21  2 × 10–2 slope = 0 ] Q.6 Hey congrats! you have been selected by the NASA for its "Mission to Mars". You are supposed to bring a sample of air from Mars in a 1 lit. rigid container. Which can bear 5 atm pressure difference. You filled air at 5 atm pressure but then realised that temperature is changing. Then calculate the rate (no.of moles/day) at which moles must be leaked so that container did not explode if time taken for return trip was 4 days. Given: TMars = 50K, PEarth surface = 1 atm, TEarth surface = 300 K. Assuming all the other parameters same and temperature increases gradually from Mars to Earth. [Sol. 5 × 0.821 = n ×0.0821 × 50 n = 1 5 1 50 = 6 n  300 1 n = 5 = 0.2 0.8 rate = 4 = 0.2 moles /day ] Q.7 A compound containing C, H and Br was quantitatively analysed. In Liebeig method mass of anhydrous CaCl2 and KOH was found to increase by 3.6 gm and 13.2 gm respectively and when same amount of sample was analysed for bromine by Carius method 37.6 gm of pale yellow ppt. was obtained then calculate the empirical formula of compound. [Sol. C  CO2 H  H2O Br  AgBr POAC for C POAC for H POAC for Br nC = n CO2 13.2 nC = 44 nH = 2 × n H2O 3.6 nH = 2 × 18 nBr = nAgBr 37.6 nBr = 188 nC = 0.3 nH = 0.4 nBr = 0.2 C0.3H0.4Br0.2 C3H4Br ] Q.8 Calculate the number of moles of gas present in the container of volume 10 lit at 300 K. If the manometer containing glycerin shows 5m difference in level as shown in diagram. Given: dglycerin = 2.72 gm/ml, dmercury = 13.6 gm/ml. [Sol. hgdg = hHgdHg 5 × 2.72 = 13.6 hHg hHg = 1 m Pgas = 0.76 dg + 1 dg = 1760 mm of Hg 1760 760 ×10 = n × R × 300 n = 0.94 mole] Q.9 100 ml of a gaseous mixture containing CO, CO2 and N2 on complete combustion in just sufficient amount of O2 showed contraction of 20 ml. When the resulting gases were passed through KOH solution it reduces by 50% then calculate the volume of each component present in the original mixture. [Sol. 2CO + O2  2CO2 if volume of CO is x x x/2 x if volume contraction will be x/2 CO2 N2 x/2 = 20 y 60–y x = 40 ml Total VCO = 40 + y  40 + y 100 = 1  y = 10 ] 2 PART-B Q.10 Acontainer containing 1 mol of H2 and 2 mol of Ar was found to exert a total pressure 12 atm. Calculate the partial pressure of each gas if 3 moles of N2 are added and volume and temperature are doubled to their original value.(There is no chemical reaction) [4] [Sol. 12 × V = 3 × R × T PT ×2V = 6 × R × 2T P 12 × 2 = 2 × 2 PT = 24 PH2 = 4 atm PAr = 8 atm P = 12 atm ] Q.11 State whether the following statements are True or False with proper reasoning for the given curve.(Assuming ideal gas) (i) Both n and V must vary. (ii) If n is constant then V must be constant. (iii) If volume is constant then number of moles at point (2) must be greater than at point (1). [4] [Sol. (i) P = nR T V If V is constant then n may vary and if n is constant V may vary. (ii) If V is constant then n must vary (iii) n2R > n1R V V n2 > n1 ] Q.12 A gaseous mixture containing He, CH4 and SO2 was allowed to effuse through a fine aperture then calculate [4] Molar ratio of gases coming out initially. (i) if mixture contains equal moles of each substances. (ii) If mixture contain He, CH4 and SO2 in 1 : 2 : 3 mole ratio. Assuming composition of effused out mixture is same as that of initial mixture in both the above cases. [Sol. (i) rHe rCH nHe nSO = nHe = = 2 CH = = 4 nHe : nCH : nSO = 4 : 2 : 1 (ii) nHe = = 1 nCH 1 nHe = = 4 nSO 3 nHe : nCH : nSO = 4 : 4 : 3 ] Q.13 Two inflated balloons (thin skin) having volume 600 ml & 1500 ml at 300 K are taken as shown in diagram. If maximum volume of inner and outer balloons are 800 ml & 1800 ml respectively then find the balloon which will burst first on gradual heating. [5] [Sol. Case I - Suppose Inner balloon burst first 600 = 800  T = 400 K 300 T2 2 Case II - Suppose outer balloon burst first 1800   600 T ' 1500  600  300  2 300 =   ' 2 T2 = 360 K ] Q.14 100 ml of a gaseous mixture containing a compound of Nitrogen & Oxygen and some NH3 was sparked to cause complete decomposition to give N2, O2 & H2 only. Volume was found to increase by 70 ml. The resulting mixture was again sparked and volume was found to be 80 ml. 30 ml of resulting gaseous mixture is absorbed by alkaline pyrogallol solution. Then calculate the volume of each component and molecular formula of oxide. (There is no reaction between N2 and O2) [5] x y [Sol. NxOy  2 N2 + 2 O2 Px Py P 2 2 2NH3  N2 + 3H2 q 3q q 2 2 2H2 + O2  2H2O (l) 3q Py 2 2  Py  3q     2 4  Px + Py  q 3q 2 2 + 2q = 170  2q  + 2 4 = 90 Px + q + Py  3q  5q  2q + = 2 2 2 4 = 80 3q 90 4 P + q = 100 q = 40 ; P 60 30 x + 30 y = 90 x + y = 3 x =1 Py  30 = 30 2 y = 2 ] Q.15 438 gm of a mixture containing Na2SO4, NH4CN & CH3COONa reacts completely with 5 litres of 1 M HCl & same amount of same sample requires 1.5 litre of 2M NaOH for complete reaction. Calculate number of moles of each substance in original sample. (HCN is a weak acid) Reactions: NH CN + HCl  NH Cl + HCN; 4 4 CH3COONa + HCl  CH3COOH + NaCl; NH4CN + NaOH  NH4OH + NaCN [5] [Sol. Na2SO4 x NH4CN + HCl  NH4Cl + HCN y y CH3COONa + HCl  CH3COOH + NaCl z z y + z = 5 × 1 NH4CN + NaOH  NH4OH + NaCN y y y = 1.5 × 2 = 3 z = 2 Na2SO4 + 3 × 44 + 82 × 2 = 438 nNa SO = 1 ] Q.16 Match the column–I with column–II and column–III and show your calculations [6] Column–I Column-II Column-III (i) 400 gm/lit NaOH (X) 6.25 m NaOH (P) 10 M NaOH if dsolution = 1.0 gm/ml if dsolution = 2 gm/ml (ii) 20% w/w NaOH (Y) 0.167 mole fraction (Q) 5 M NaOH of NaOH if dsolution = 1 gm/ml if dsolution = 1.3 gm/ml [Sol. (i) xB = 10 = 0.167 10 +  1300  400     18  400 g / L M = 40 g mol 1 = 10 M (i)  (Y)  (P) (ii) 100 gm contains 28.57 gm NaOH 0.51000 m = 80 = 6.25 m 20 1000 M = 40 100 = 5 M (ii)  (X)  (Q) ] Q.17 With the help of following reactions Fe2(SO4)3 + KCN  Fe(CN)3 + K2SO4 (1) Fe(CN)3 + KCN  K3Fe(CN)6 (2) K3Fe(CN)6 + FeSO4  Fe3[Fe(CN)6]2 + K2SO4 (3) Calculate (a) Number of moles of Fe3[Fe(CN)6]2 which can be produced by 4 kg Fe2(SO4)3 with 100% efficiency. (b) Calculate the mass of FeSO4 used in part(a). (c) Calculate the total number of moles of KCN which will be used to produce 11.84 kg Fe3[Fe(CN)6]2 if reaction (1), (2) and (3) have efficiency 80%, 60% and 40% respectively. [6] [Sol. Fe2(SO4)3 + 6KCN  2Fe (CN)3 + 3K2SO4 6  2  20 0.4  0.6  0.8 2  20 0.4  0.6 Fe(CN)3 + 3KCN  K3Fe(CN)6 2  20 0.4  0.6 3 2  20 0.4  0.6 2  20 0.4 2K3Fe(CN)6 + 3FeSO4  Fe3[Fe(CN)6]2 + 3K2SO4 2  20 0.4 3 20 0.4 20 4000 (a) No. of moles of Fe2(SO4)3 = 400 =10 nFe [ Fe(CN ) ] = 10 × 2 1 1 × × =10 moles 3 6 2 1 1 2 (b) WFeSO = nFeSO (reacted) × 152 4 4 = 30 × 152 = 4560 gm (c) moles of KCN = 1250 + 500 = 1750 ] Q.18 Metal salt of organic acids for a metal shows behaviour similar to that shown by silver salt using silver salt method, the molecular mass of dibasic acid was determined by silver salt method as 326. For the metal salt, same mass of metal salt was taken and same mass of residue was obtained as in the silver salt experiment. From this information (a) Calculate Msilver salt(Mss) & Formula mass of A2– if acid is H2A. (b) Establish a relationship between Ametal & valnecy of Metal & if atomic weight of metal is 324 then calculate its valency. (c) If 17.10 gm of above metal salt is used & subjected to heating then what will be weight of residue obtained. [6] [Sol. (a) Formula mass of A–2 = 324 Msilver salt = 324 + 216 = 540 (b) Mass of Ag = Mass of M Mass of salt of Ag = Mass of salt of M wg = wg 2  w 108 216 + 324 = 108 = M 2  w  M 2M + 324x 540 2M + 324x 2M + 324x 324x 5 = M 324x = 2 + M M = 3 324 x = 3M M = 108 x 324 x = 108 = 3 (c) Hence, the metal salt has the formula = M2A3 Molar mass of metal salt = 2 × 324 + 324 × 3 = 1620 wt. of residue = 17.10  2  324g 1620 = 6.840 g ]