CHEMISTRY MARKING SCHEME Bhubaneswar 2015 Set 2 - Code No. 56/2/B. Ques. Value points Marks. ½ +½ (Any two of these) 2. CH 3 CH CH 2 CH 2 Br

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1 CHEMISTRY MARKING SCHEME Bhubaneswar 205 Set 2 - Code No. 56/2/B Ques. Value points Marks. HOCl, HOClO, HOClO 2, HOClO 3 + (Any two of these) 2. CH 3 CH CH 2 CH 2 Br CH 3 3. Negative charge 4. XY Phenylpropan-2-ol 6. Formula: w zi t wvalance96500 timetakeninsec Mol Mass Current in Amp Substituting the values in the formula we get:.7 g C mol timetakeninsec 58.5 g mol 5amp timetakeninsec t=772 s ( Or by any other correct method) 7. (i) Potassium hexacyanidoferrate (III) [Co(NH 3 ) 5 NO 2 ] (i) Due to comparable energies of 5f, 6d and 7s orbitals. Because 5f electrons have poorer shielding effect than 4f electrons. 9. (i) Positive deviation, lowering of temperature or absorption of heat. By applying an external pressure greater than the osmotic pressure on the solution or P > π Reverse osmosis is used in desalination of hard water / sea water. 0. (i) H 2 / Pd-BaSO 4 NaOH/CaO,,, 0. i) C 6 H 5 CO C 6 H 5 < CH 3 COCH 3 < CH 3 CHO ii) Cl CH 2 COOH < Cl 2 CH COOH < CCl 3 COOH

2 . (i) Distillation Collector/ enhancing the non-wettability of mineral particles. As S is positive / G is more negative 2. (i) Stoichiometric Defect Frenkel Defect Due to small size of Ag + ion 3. (i) CH 3 CH(OH) CN C 6 H 5 COOH CH 3 CH 2 NH 2 4. t 2g 3 e g Hybridization dsp 2, Shape Square planar or diagram (Marks of (i) part is merged into and part ) 5. (i) Due to the stability of benzyl carbocation/resonance/diagram Because 2-Bromobutane has a chiral centre. Due to I effect of halogen. 6. NaNO2 HCl H2OH (i) C 6 H 5 NH 2 C 6 H 5 N 2 Cl C 6 H 5 OH o o 0 5 C Or Hydrolysis HBr KOH CH 3 CH = CH 2 CH Organic peroxide 3 CH 2 CH 2 Br Aq CH 3 CH 2 CH 2 OH (Or any correct method)

3 6. (i) CH 3 CH 2 CH 2 OH CH 3 CHO + H 2 Cu/573K Dehydrogenation / Oxidation C 2 H 5 Cl + NaOCH 3 C 2 H 5 -O-CH 3 + NaCl 7. (i) Maltose 8. Sugar Present in DNA is Deoxyribose whereas in RNA it is Ribose Thymine is present in DNA whereas in RNA Uracil is present (Any one) Beri-Beri = E 0 cell V log [ 0.000] 2 [0.00] = E 0 cell V log 0 - = E 0 cell V (-) = E 0 cell V E 0 cell = E 0 cell =2.650 V 9. (i) Glyptal: and HO-CH 2 - CH 2 -OH (ethylene glycol) Teflon: Monomer:,,2,2-Tetrafluoroethene

4 ,,2,2-Tetrafluoroethene Nylon-6 Monomer: Caprolactum (Note : half mark for structure/s and half mark for name/s) 20. (i) Because of higher oxidation state of Mn in Mn 2 O 7. Due to almost similar atomic size / comparable size. 2MnO 2 + 4KOH + O 2 2K 2 MnO 4 + 2H 2 O 2. (i) Solution is homogeneous colloid is heterogeneous In solution the size of particles (solute) is less than nm whereas in colloids the range of size of particles is 000 nm (0 9 to 0 6 m)(any one point) In homogeneous catalysis the reactant and catalyst are in the same phase whereas in heterogeneous catalysis they are in different phase. In O/W emulsion oil is the dispersed phase while in W/O water is dispersed in oil The O/W type emulsion can be diluted with water whereas the W/O emulsion can t be diluted with water. (Any one point) p p w2 M Formula 0 p M 2 w mm mm 5.0 g 8 g / mol mm M g 5.0 g 8.0 g / mol mm M 2 95 g mm M 2 = 60.0 g/mol 23. (i) Concern for students health, Application of knowledge of chemistry to daily life, empathy, caring or any other (Any two) Through posters, nukkad natak in community, social media, play in assembly or any other (Any two) Tranquilizers are drugs used for treatment of stress or mild and severe mental disorders. Eg: equanil (or any other suitable example) (iv) Aspartame is unstable at cooking temperature. 24. a) (i) The +3 Oxidation state of Bi is more stable than Sb(III). Because the electronegativity of Cl is greater than that of I. Due to decrease in electronegativity and increase in the atomic size.,,

5 + (b) i) Due to formation of fumes of HCl or equation PCl 5 + H 2 O POCl 3 + 2HCl ii) Rhombic sulphur or -Sulphur iii) Due of loss of Chlorine. The yellow colour is due to dissolved Cl 2. On standing the Cl 2 is consumed in reacting with water to form colourless products: Cl 2 + H 2 O HOCl + HCl 2HOCl 2HCl + O 2 iv) 4H 3 PO 3 3H 3 PO 4 + PH 3 Oxidation state of P is +3 Oxidation state of P is +5 Oxidation state of P is 3 v) 2F 2 + 2H 2 O 4HF + O 2 x 5

6 b) (CH 3 ) 3 N < CH 3 NH 2 < (CH 3 ) 2 NH c) Dye Test: On treating with benzene diazonium Chloride at low temperature C 6 H 5 -NH 2 will form coloured dye while CH 3 -NH 2 will not form. (or any other correct distinguishing test) (a) CH3COOCH 3 Formula: k log t CH COOCH M k log 20s 0.2M k = 0.03 s M k2 log 40s 0.M k 2 = 0.03 s Since constant values of rate constants are obtained by applying st Order integrated rate law, the reaction is pseudo first order reaction. total changein concentration (b) Av rate total changeintime [ CH3COOCH 3] final [ CH3COOCH 3] initial Av rate Time( f ) Time( i) 0.0 M 0.20 M Av rate 40 Sec 20 Sec Av rate = M sec or 5.0 x 0 3 mol L sec a) i) Collision frequency: No of collisions taking place per second per unit volume. ii) Rate Constant: It is the rate of reaction when the concentration of reactants

7 is unity i.e. M. It is temperature dependent k Ea T T log k 2.303R TT 2 k2 Ea T2 T log k 2.303R TT 2 b) 2 2 Ea 50 log Ea Ea Ea = J Ea = 3.29 kj/mol