Chemistry-8.Half Course Test -2-Solution

HALF COURSE TEST - II (MAILING) CHEMISTRY SOLUTIONS 1. We know that year–1 [2] Let the relative number of atoms C12 at t = 0 be N0, then relative number of atoms after t years = 0.617 N0. So putting these values in the equation [2] [4] t = 4026.76 years 2. The kinetic equation for a second order reaction in which both reactants have same initial concentration. K = [3] Initial concentration of ester and base (a) = 0.01 M, K = 6.36 litre mol–1 min–1 t = 20 minutes. On substituting the values 6.36 = [3] x = [2] Fraction of ester hydrolysis = = 0.56 3. Sulphide ions in alkaline solution react with solid sulphur and to form S2–2 and S3–2 ions as i) S(s) + S2– S22– K1 = 1.7 ii) 2S(s) + S2– S32– K2 = 5.1 iii) and S22– + S(s) S32– K3 = ? equation (iii) = equation (ii) – equation (i) So K3 = = 3 4. Molarity of N2H4 = = 0.01 M N2H4 + H2O N2H5+ + OH– Initially 0.01 M –– –– –– At equilibrium (0.01 – x)m –– xM xM Kb = x = 2 × 10–4 M % of N2H4 reacting with water = = 2.1 5. λ = 13.4 × 10–10 = ∴ v = 5.43 × 105 m/s [2] We know that 2.18 × 106 × = vn 2.18 × 106 × = 5.43 × 105 ⇒ n = 4 [2] First excited state means n = 2 = 109637 λ1 = 4864 [2] Longest wavelength means transition from 2 → 1 = 109737 λ2 = 1216 [2] 6. C3H8 + 5O2 ⎯→ 3CO2 (g)+ 4H2O(l) CH4 + 2O2 ⎯→ CO2 (g)+ 2H2O(l) CO + O2 ⎯→ CO2 (g) [2] Let a,b and c ml be the volumes of C3H8,CH4 and CO respectively. than a + b + c = 100 where a = 36.5 [2] Volume of CO2 formed = 3a + b + c = 3 × 36.5 + (100 – 36.5) = 173 ml [2] \7. Let the moles of Na2C2O4 be ‘a’ and that of KHC2O4.H2C2O4 be ‘b’ and the volumes of KMnO4 and NaOH consumed be V litre. equivalent of KMnO4 = equivalent of Na2C2O4 + equivalent of KH2O4.H2C2O4 0.1 × V = ----------------- (i) [3] equivalent of NaOH = equivalent of KHC2O4.H2C2O4 0.1V = ----------------- (ii) [3] ⇒ from (i) & (ii) a : b = 1 : 6 [1] 8. i) (a) sp3d2 [2] (b) sp3d [2] (c) sp3d [2] (d) sp3d [2] ii) (a) NH3 < PH3 < AsH3 < SbH3 [2] (b) BeCO3 < MgCO3 < CaCO3 < BaCO3 [2] 9. 2 P (g) ⎯→ 4Q(g) + R (g) + S (l) at initial P0 0 0 0 at time t (P0-x) 2x Ps at time ∞ 0 2P0 Ps [2] given P0 –x + 2x + + Ps = 317 and 2P0+ + PS = 617 putting Ps = 32.5 we get P0 = 233.8 x = 33.8 [4] Now K = = K = 5.2 × 10-3 min-1 [2] T1/2 = = 133.3 min [2] 10. AgCl (s) Ag+ (aq) + Cl- (aq) K1 = KSP Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2]+ (aq) K2 = Kf –––––––––––––––––––––––––––––––––––––––––––––––––––––– AgCl (s) + 2NH3 (aq) [Ag(NH3)2]+ (aq) + Cl- (aq) K = [2] 0.1 0 0 at initial (0.1-2x) x x at equilibrium K’ = K1 × K2 = [4] ⇒ = 0.0536, ⇒ x = 4.9 × 10-3 M hence the solubility of AgCl (s) in 0.100 M NH3 (aq) is 4.9 × 10-3 mole/litre [2] 11. a) Butane 2,3 dione exist in the trans from, due to flipping which is highly stable hence exclusively exist in keto form. But in case of cyclopentanone no such kind of flipping is possible hence form enol in which it is stable through H – bonding. b) [3] 12. A O3, H2O [1] B LiAlH4 [1] C. Conc. H2SO4, H+ [1] D NaNH2 in Liq. NH3 [1] E CH3I [1] F H2SO4, Hg+2 [1] 13. a) [4] b) 6