Cbse 12 class question paper 2020 scanned copy of notes, Cheat Sheet of Family Law

Download Cbse 12 class question paper 2020 scanned copy of notes and more Cheat Sheet Family Law in PDF only on Docsity! Marking scheme – 2019 CHEMISTRY (043)/ CLASS XII 56/1/1 Q.No Value Points Marks SECTION A 1 AgCl , Due to large difference in their size/ Due to small size of Ag+ ion. ½ , ½ 2 (CH3)3N < C2H5NH2 < C2H5OH 1 3 Due to large surface area these are easily assimilated or adsorbed. 1 OR 3 Emulsion – both dispersed phase and dispersion medium are liquid Gel- Dispersed phase is liquid while dispersion medium is solid 1 4 Nucleophiles having two nucleophilic centres. CN- /SCN- / NO2 - (any one) ½ , ½ 5 Glucose has aldehydic group while fructose has ketonic group/ Glucose is aldose while fructose is ketose. 1 OR 5 Glucose and Galactose 1 SECTION B 6 i) ii) 1 1 OR 6 i) H2O < H2S < H2Se < H2Te ii) HF> HCl > HBr > HI 1 1 7 For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. (i) (ii) (iii) The components have nearly same intermolecular force of attraction (any two) 1 ½, ½ 8 i) Rate = k [H2O2] [ I -] ii) order = 2 iii) Step 1 1 ½ ½ 9 A = K2MnO4 / MnO4 2- , B= KMnO4 / MnO4 -- , C= IO3 - or KIO3 , D= I2 ½ ×4 10. Bis(ethan-1,2-diamine)dichloridoplatinum (II) Cis Trans 1 ½ , ½ OR 10. i) [Co(NH3)6]2(SO4)3 ii)K3[Cr(ox)3] 1 1 11 i) [CoF6] 3- ii)[Co(en)3] 3+ iii) [Co(en)3] 3+ iv) [CoF6] 3- ½ ×4 CBSE Class 12 Chemistry Question Paper Solution 2019 12 i) A= B= ii) A= B= ½ ×4 SECTION C 13 t= 𝑅 0 ─ 𝑅 𝑡 𝑘 = [0.1−0.064] 4𝑋 10−3 = 9 s 1 1 1 14 i) Adsorption of toxic gases ii) Negative charge ; iii) Increases with increase in temperature/ First increases then decreases 1 ½ , ½ 1 15 d= 𝑧𝑚 𝑎3 𝑁 ; m=Mass of element , N=number of atoms 𝑁 = 108 X 4 10.8X27X10-24 = 1.48 X 1024 atoms Or M= 𝑎3 × 𝑁𝑎 ×𝑑 𝑍 = 27 × 10−24 ×6.022 × 1023 ×10.8 4 = 43.88 g mol -1 43.88 g mol -1 contains 6.02 × 1023 atoms So , 108 g contains = 6.02 × 1023 × 108 43.88 = 1.48 × 1024 atoms 1 1 1 ½ 1 ½ 1 16 ΔTf = Kf m Kf = ΔTf X M2x w1 w2 x1000 = 2x 342 x 96 4x1000 = 16.4 K ΔTf = Kf m’ = Kfw2 x1000 M2x w1 = 16.4 x 5 x 1000 95x180 = 4.8 K ΔTf = Tf o- Tf 4.8 = 273.15 - Tf Tf = 268.35 K ½ 1 1 ½ 2.8 𝑚𝑍𝑛 = 56 2 X 2 65.3 mZn = 3.265 g b) i)A- strong electrolyte , B-Weak electrolyte ii)Λ0m for weak electrolytes cannot be obtained by extrapolation while Λ0m for strong electrolytes can be obtained as intercept. 1 1 1 26 a) i) ii) CH3CH2OH PCC,Heat CH3-CHO i)CH3MgBr ii)H+ CH3CH(OH)-CH3 (or any other correct method) b) c) Due to involvement of lone pair of oxygen in delocalisation makes the benzene ring electron rich. 1 1 ½ ½ 1 1 OR 26 a) i) o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding while p-nitrophenol is less volatile due to intermolecular hydrogen bonding. ii) Due to the formation of stable intermediate tertiary carbocation / CH3O - being a strong base favours elimination reaction. b) i) ii) (Award 1 mark if attempted in any way) c) Add neutral FeCl3 to both the compounds, phenol will give violet colouration while ethanol does not. 1 1 1 1 1 27 a) i) In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons like O2 . ii) Due to greater interelectronic repulsion iii) Because decomposition of ozone into oxygen results in the liberation of heat (H is negative) and an increase in entropy (S is positive), resulting in large negative Gibbs energy change (G) for its conversion into oxygen. b) i) NO gas/ Nitric oxide ii) NO2 gas / Nitrogen dioxide 1 1 1 1,1 OR 27 a) i) 1 ii) b) i) Due to small size and low bond dissociation enthalpy ii) As the size increases, electronegativity decreases / non-metallic character decreases c) 1 1 1 1