Chemistry-6.Half Course Test-1-Solutions

HALF COURSE TEST - I (MAILING) CHEMISTRY SOLUTIONS 1. No. of gm equivalents of NaOH required for back titration = = no. of gm equivalents of H2SO4 remained unreacted ∴ no. of gm equivalents of H2SO4 reacted = = 17.76 × 10–3 [4] = no. of gm equivalents of M1M2CO3 = no. of gm equivalents of pure M1M2C4H4O6, 2H2O Let wt. of the pure sample was x gm ∴ moles = ∴ no. of gm equivalents = ∴ = 17.76× 10–3 [4] or, x = 2.1845 gm ∴ % purity = = 20.355% [2] 2. = 765 mm at 35°C So, at 0°C the total pressure will be = 678.1mm [2] = (678.1 – 645) = 33.1mm So = 645 mm at 0°C = 345 mm at 25°C [4] ∴ at 0°C = mm = 316mm ∴ =(645 – 316) = 329 Since p = mole fraction × total pressure ∴ Mole fraction of H2O = = 4.9% Mole fraction of N2 = = 46.6% Mole fraction of CO2 = = 48.5% [4] 3. Radius of the Bohr orbit in hydrogen atom is = (1)2 = 5.3 × 10–11 m Uncertainty is 2% i.e. 5.3 × 10–11 = 10.6× 10–13m [4] Now we know ΔpΔx ≥ Or (m Δv) Δx ≥ Or Δv ≥ × ≥ [4] Δv ≥ 0.547 × 108m sec–1 So the uncertainty in the velocity of the electron is 5.47 × 107 m sec–1 [2] 4. Now, At the time of earth formation, For X193, λ1 = For X192, λ2 = [4] If ‘t’ is the age of earth ∴ λ1 – λ2 = = log or, = ∴ t = 32.56 × 107 years [6] 5. CO(g) + H2O(g) CO2(g) + H2(g) Moles at equilibrium 0.2 0.3 0.4 0.5 ∴ Kc = = 3.333 [4] [Δn = 0, so no effect of volume] Now when CO2 was added to the vessel the reaction proceeds into backward direction. Let ‘a’ moles of CO2 was introduced to get the desired conversion. CO2(g) + H2(g) CO(g) + H2O(g) 0.4 + a 0.5 0.2 0.3 at new equilibrium (0.4 + a –0.2) (0.5 – 0.2) (0.2 + 0.2) (0.3 + 0.2) = (0.2 + a) 0.3 0.4 0.5 ∴ [4] ∴ a = 2.022 moles [2] 6. a) All the compounds contain active methylene group (–CH2 – flanked between two carbonyl group). Now higher the stability of conjugate base, higher will be the acidity. Here the conjugate base stabilization depends on the terminal groups. In 1st compound if we replace one terminal H-atom by –CH3 group, which has +I effect, conjugate base will be destabilized. And if the terminal H-atom is replaced by –OEt group, which has +M effect, conjugate base will be least stabilized. [4] So the order of acidity is b) We know higher the stability of conjugate acid higher is the basicity. Now if we consider the stability of conjugate acids, the compound I is most basic. Because all the resonating figures of it’s conjugate acid are identical, having most stability. [4] The conjugate acids of compound II and III are respectively and out of which 1st one is more stable due to +M effect of –NH2 comparing +I effect of –CH3 group in the later one. 7. a) Among the halides of lithium LiF, LiCl, LiBr, LiI the covalency is maximum in LiI according to the Fajan’s rule and LiF has maximum ionic character. So the melting point of LiF is maximum. [3] b) Bond orders of both and are 2.5 B.O. of = = 2.5 B.O. of = = 2.5 But in , one of the N-atom being positively charged attracts the bonding electron cloud of the other N-atom in higher extent comparing in . So the bond length in is lower than the bond length in . And bond energy of > . [4] 8. a) (i) sp3 (ii) sp3 [1½ × 2] b) (i) Hex 2-ene 4-yne [2 × 2] (ii) 1, 3-propanedioic acid c) [3] [3] 9. a) This is a meso compound which posses plane of symmetry. The upper part of the molecule is the mirror image of the lower part of the molecule. [2] The number of stereoisomers of a compound is 2n where n = no. of asymmetric carbon. Here the meso compound and its mirror image are actually same. So number of optical isomers is 2n-1= 3. [1½] Out of these three stereoisomers meso is optically inactive and remaining two are optically active. [2] b) X = O2N – CH2 – CH2 – CH2Br [1½] Here anti Markownikov’s product is obtained although there is no peroxide. Actually reaction passes through carbocation intermediate. Here primary carbocation is more stable than secondary carbocation due to strong –I effect of –NO2 group. –I effect of –NO2 destabilises the 2° carbocation. [2½] c) [2] 10. A = CH2 = CH – CH3 B = HCHO C = CH3CHO D = CH3CH3 E = CH3CH2Br [5 × 1½] 11. (a) Half reactions are ⎯⎯→ ⎯⎯→MnO2 Balancing the atoms 1st, then the charges and the e–s for both the half reactions 1st step ⎯⎯→ +4H+ ⎯⎯→ + 2H2O 2nd step ⎯⎯→ + e– + 4H+ + 2e–⎯⎯→MnO2 + 2H2O 3rd step [ ⎯⎯→ + e– ] × 2 adding + 4H+ ⎯⎯→ 2 + MnO2 + 2H2O [3] b) Half reactions are As2S3 ⎯⎯→ H3AsO4 + S ⎯⎯→ NO Balancing 1st the atoms, then the charges and the e–s for both the half reactions 1st step As2S3 ⎯⎯→ 2H3AsO4 + 3S ⎯⎯→ NO + 2H2O 2nd step As2S3 + 8H2O ⎯⎯→ 2H3AsO4 + 3S + 10H+ + 10e– 3rd step [As2S3 + 8H2O ⎯⎯→ 2H3AsO4 + 3S + 10H+ + 10e– ] × 3 [ ] × 10 adding 3As2S3 + 4H2O + + 10H+ ⎯⎯→ 6H3AsO4 + 9S + 10NO [3] 6