Chemistry-71.Phase-1-Subjective-Solutions

FACULTY TRAINING TEST PHASE - I CHEMISTRY SOLUTIONS 1. a) Given P1 = 15.6 atm P2 = 12.5 atm T1 = ? T2 = 300° K Since volume of Cylinder remains constant  T1 = = 374.4K b) P = atm, T = 273 + 150 = 423 K Let density be d for N2 P = d = = gm Lit–1 = 1.06 gm Lit–1 2. a) = (As the mixture contains 40% ozone) =  30 min = 32.86 min b) P = 10–10 mm = atm r = 10–3 lit, T = 293 K  PV = nRT  n = =  n = 5.47  10–18 Therefore number of molecules in / ml = 5.47  10–18  6.023  1023 = 3.29  106 molecules 3. a) % ionic character = =  100 = 16.90 b) i) SbH3  AsH3  PH3  NH3 ii) LiF  LiCl  LiBr  LiI iii) H2Te  H2S  H2O iv) I  Br  F  Cl v) Al3+  Mg2+  Na+  F–  O2– vi) NaI  NaBr  NaCl  NaF vii) Mg  Si  Al  Na 4. Let x ml CO, y ml CH4 and z ml N2 be present in mixture Thus, x + y + z = 20 …(i)  Volume of CO = x  Volume of CO2 = x because CO(g) + = CO2(g) CH4(g) + 2O2 = CO2(g) + 2H2O(l)  Volume of CH4 = y  Volume of CO2 = y N2 + O2  No reaction (at normal state) This volume of CO2 is absorbed by KOH  x + y= 14 ml …(ii) Thus (initial volume of CO + CH4 + N2) + Volume of O2 taken – Volume of CO2 formed – Volume of N2 – Volume of O2 left = 13  + 2y = 13 …(iii) ( volume of O2 used = +2y) Solving Equations (i), (ii) and (iii), we get x = 10 ml, y = 4 ml, z = 6 ml 5. XeOF4  Hybridization sp3d2 Shape. Square pyramidal XeOF6  Hybridization sp3d3 Shape pentagonal bipyramidal (b) M.O. distribution of O2 1s2 * 1s22s2*2s22px2 Since two unpaired e– are present in O2 molecules , this is paramagnetic in nature 6. a) + 2H2O  4MnO2 + 3O2 + 4OH– b) 6Cu3P + 124H+ + 11  18Cu+2 + 6H3PO4 + 22Cr+3 + 53H2O 7. Radius of the nucleus is of the order of 10–13 cm and thus uncertainty in position of electron i.e., (x), if its within the nucleus will be 10–13 cm. Now x. u  u = = 5.79  1012 cm/sec. i.e., order of velocity of electron will be 100 times greater than the velocity of light which is impossible. Thus, possibility of electron to exist is nucleus is zero. 8. a) Using normality equation 25  N = 20  0.01  N = = Amount of HCl in 2500 mL of N/125 = = 0.73g Amount of HCl in 700 mL of N/10 = = 2.555 g Amount of HCl consumed = (2.555 – 0.73) = 1.825 g Let amount of CaCO3 in the mixture is 'x' g Amount of MgCO3 = (2.36 –x)g CaCO3 + 2HCl  CaCl2 + H2O+ CO2 100g 73 g xg xg MgCO3 + 2HCl  MgCl2 + H2O + CO2 84 g 73g (2.36 –x)g g But, = 1.825  x = 1.63 g % of CaCO3 = = 69.06 % of MgCO3 = 100 – 69.06 = 30.94 b) Eq. weight of KMnO4 = 1/5  molecular weight of KMnO4 M/50 KMnO4 = N/10 KMnO4 Meq. of KMnO4 = 20  1/10 = 2 Meq. of N2H6SO4 in 10 ml solution = m.eq. of FeCl3 reacting with N2H6SO4 = m.eq. of Fe2+ formed = m.eq. of KMnO4 used Meq. of N2H6SO4 in 10 ml solution = 2 Eq. wgt. of N2H6SO4 = = = 32.5 weight of N2H6SO4 in 10 mL = 32.5  2  10–3 g = 0.065 g  weight of N2H6SO4 in one litre = 6.5 g 9. Let the amount of NaCl in the mixture be 'x' gms. Hence the amount of MCl in the mixture = (1–x) g The equation of the reactions are as follows + AgNO3   + MNO3 (W  atomic wgt. Of M) Since 58.5 g NaCl produces 143.5 g AgCl x g NaCl produces = Similarly, (1–x) g MCl produces = g of AgCl Since the weight of the precipitate is 2.567g, hence + = 2.567 g According to the (ii) part of the question one component is volatile. Since NaCl is not a volatile compound hence MCl may be considered as volatile. This compound forms AgCl according to the equation where it is equal t 1.341 g. Thus, = 1.341  + 1.341 = 2.567  x = 0.5 g Also, = 1.341  W = – 35.5 = 18.0 Hence molecular weight of MCl = 18 + 35.5 = 53.5 10. a) 25cm3 of remaining oxalic acid solution  32cm3 of 0.1N KMnO4  N1  V1 = 0.1  32  N1 N Unreacted oxalic acid = 250 cm3 of N M.eq. of unreacted oxalic acid = 32 M.eq. of total oxalic acid = 1  50 = 50 M.eq. of used oxalic acid = 50 – 32 = 18 MnO2 + H2SO4 + H2C2O4  MnSO4 + 2H2O + 2CO2 Meq. of H2C2O4  m.eq. of MnO2 m.eq. of MnO2 = 18 Wgt. Of MnO2 = 18  43.469  10–3 = 0.7824 g % of MnO2 =  100 = 48.9 M.eq. of MnO2 = m.eq. of O2 = 18 Wgt. Of O2 = m.eq. of O2  its eq. wgt.  10–3 = 18  8  10–3 = 0.144g % of available oxygen =  100 = 9 b) Let the mixture contain 'x' g of K2Cr2O7 Amount of KMnO4 = (0.561 – x) g The reactions involved are i) 2KMnO4 + 8H2SO4 + 10KI  6K2SO4 + 2MnSO4 + 8H2O + 5I2 5I2 + 10Na2SO310NaI + 5Na2S4O6 2KMnO4  10(Na2S2O3.5H2O)  10I  10H ii) K2Cr2O7 + 6KI + 7H2SO4  4H2SO4 + Cr2(SO4)3 + 7H2O + 3I2 3I2 + 6Na2S2O3  6NaI + 3Na2S4O6 K2Cr2O7  6(Na2S2O3.5H2O)  6I  6H From (i) we have that 316 g of KMnO4 will require hypo = 10  248 g (0.561 – x) g of KMnO4 will require hypo = g Similarly from (ii), we have 249 g of K2Cr2O7 will require hypo = 6  248 g x g of K2Cr2O7 will require hypo = g We also know, that 100 mL of 0.15 N hypo = g hypo = 3.720 g hypo  accordingly + = 3.72 7.848 (0.561 – x) + 5.061 x = 3.72 7.848  0.561 – 7.848x + 5.061x = 3.72 2.787x = 0.68 x = 0.244 g Amount of K2Cr2O7 = 0.244  43.5% Amount of KMnO4 = 0.317 g  56.5% 11. a) By Heisenberg's uncertainty principle x,p  h/4 given that, x = p, hence x = p = = 0.726  10–17 Also, x.v  h/4m Thus v = h/4m.x = h/4m = = = 7.98  1012 ms–1 We find that uncertainty in the measurement of position (or) momentum is negligible, but not in velocity. b) Using de Broglie equation  = h/mv v = h/m = = 8.09  107 ms–1 KE = mv =  9.1  10–31  (8.09  107)2 = 2.98  10–15J = eV = 1.86  104 eV