CHEMISTRY-27-08- 11th (J-Batch)

Q.14 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 humidityin both vessels. [Assume constant temperature of 300 K. Aqueous tension of water is 15.2 mmHg. ] [4+4] [Ans. In 246 lit vessel : PT = 0.205 atm and RH = 25% In 24.6 lit vessel : PT = 2.02 atm and RH = 100% ] [Sol. Moles of H2O = 0.1 mol 0.1 0.082 300 Partial pressure of H2O = 246 = 0.01 atm Aqueous tension of water = 15.2 = 0.02 atm 760 All water is in vapour form 0.4  246 moles of N2 in vessel = 24.6 = 4 moles In each vessel there are 2 moles of N2 & 0.05 moles of water vapour. In 246 lit. vessel total pressure 2.05 24.6 P = 246 = 0.205 atm 0.05 24.6 In 24.6 lit vessel, partial pressure of water = 24.6 = 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 = Total pressure = 2.02 atm RH of 2nd vessel = 100% 2  24.6 24.6 = 2 atm RH of 1st vessel = 0.005 × 100 = 25% ] 0.02 0.195 moles of C = 44 moles of O = 1 0.2  0.04  12  0.195 16  9 44  C : H : O = 0.0044 : 0.0044 : 0.0089 = 1 : 1 : 2 Empirical formula = CHO2 Let n be basicity of acid 1 100 × 0.2 × n = 400 × 10  n = 2 Let H2X be acid and Ag2X its silver salt  Ag2X  2Ag 0.5 g 0.355 0.5 2 216 + M  2 0.355 = 108 M = Mol. wt. of acid 1 M + 214 M = 90.2 0.355 = 108 molecular weight n = emipirical formula weight 90.2 = 45  2 Molecular Formula = C2H2O4 ] Q.12 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. [Use: Atomic weight of metal × specific heat = 6.4] [7] [Ans. 0.7 %] [Sol. MBrx + xHCl  MClx + xHBr 6.4 Atomic mass of metal = 0.11 = 58.2 1.8  58.2 + 80x 1 = 58.2 + 35.5x  0.8(58.2) = (80 – 35.5 × 1.8)x  x = 2.89  x = 3 Let M be exact atomic wt. of metal 1.8 M + 80(3) 1 = M + 35.5(3) (a) ppm of Ca(HCO3)2 = 20 106 162 800 × 106 = 4.05 ppm (b) Molality of Na+ = 40 106 800 80106 × 1000 = 5 × 10–5 mol/kg (c) Molarity of Ca2+ = 800103 = 10–4 M ] Q.8 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 dp bulb gas leaked causing pressure decrease as dt = – 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 228 P0 = 3 + 76 = 152 cm Hg At t = 7 min. 114 P = 3 + 76 = 114 cm Hg  152e–kt = 114  1 152 1 –1 –1 –1 k = t ln 114 = 7 × 0.28 = 0.04 min = 0.04 × 60 hr = 2.4 hr ] Q.9 Calculate minimum number of balloons each of volume 82 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] 29 1 [Ans. 25 balloons] 1 [Sol. Density of air at sea level, d0 = 0.082  290 = 0 .82 g/lit  29103 10831  – Density at 831 m = d = d0e–Mgh/RT = Let n be number of balloons 1 0 .82  × e  8.31 290   = 0 .82 × e–0.1 = 0 .9 0 .82 g/lit  103 × g + n × 50g =  1000 + 50 n = 90 n 0 .9 0 .82 × 82 × g × n  n = 25 balloons ] Q.10 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) Molarity of H2O2 = 0.45 0.1 = 4.5 M Vol. strength = 11.2 × 4.5 = 50.4 volumes ] Q.3 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. [2+2] [Ans. 60%, 0.6] [Sol. rmix r 2  = = 0.4 32 = 0.8  M Mavg 2 O2  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.4 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] PV  3 PV [Sol.(a) 300R 1 5  = T(R)   T = 300 ×  5 2 = 750 K = 477°C (b) Let x be fraction escaped PV PV 300R (1 – x) =  x = 2/3 PV 900R PV (c) 300R (1 – 0.5) = TR  T = 327°C ]