Class 12

class 12 chemistry question paper 2020 with solution

56/1/1 – Set
MARKING SCHEME
Sr. SECONDARY SCHOOL EXAMINATION, 2020
Subject: CHEMISTRY

Expected Answer / Value Points

SECTION – A

1 -Racemic Mixture
2- Polarimeter
3- Pent-2-ene / CH₃CH=CHCH₂CH₃
4- Antiseptic
5-CH₃I + C₆H₅OH
6- A
7- Zn
8-No
9- CH₂=CH-Cl
10- Branched hydrocarbon part 1
11- B
12- D
13- C
14- C
15- A
16- iii
17- ii
18- i
19- ii
20- i

SECTION – B

Ans.21-(a) The drugs which are used to control stress / anxiety / tension / mild or severe mental diseases
(b) The drugs which are used to kill or to prevent the growth of micro-organism, applied externally on living tissues.

OR

Ans.21-Soap molecules form micelle around the oil droplet or dirt in such a way that hydrophobic part interacts with the oil droplet and hydrophilic part projects out. Micelles can be washed away on rinsing with water. Thus soap helps in emulsification and washing away of oil and fats.

Ans.23(a) 1^st order
(b) No, due to exponential relation / the curve never touches the x-axis.

Ans.24 –

Ans.25-(a) K₂[Zn(OH)₄]
(b) [Pt(NH₃)₆]Cl₄

Ans.26 (a) (CH₃)₃C-OH / tertiary butyl alcohol is formed.
b) C₆H₅COCH₃ / acetophenone is formed

Ans.27-(a) C₆H₅OH + HCHO , Phenol + formaldehyde
(b) CH₂= C(Cl) – CH=CH₂, Chloroprene

SECTION – C

Ans.28-(a) (A) → CH₃CONH₂ (B) → CH₃NH₂
(b) (A)→ C₆H₅NH₂ (B) → C₆H₅N₂Cl
(c) (A) → C₆H₅CN (B) → C₆H₅COOH

OR

Ans.28-(a) (i) Add Ice cold (NaNO₂ + HCl) followed by phenol or β-Naphthol to both the compounds.
Aniline forms orange red dye while ethylamine doesn’t.
(ii) Add CHCl₃ and KOH (alc.) to both the compounds.
Aniline gives foul smelling isocyanide while N-Methylaniline doesn’t.
(or any other suitable chemical test)
b) Butanol > Butanmine > Butane

Ans.29-(a) Because the – CHO group in glucose is involved in hemiacetal formation and thus is not free / due to cyclic structure of glucose -CHO group is not free.
(b) Because the hydrogen bonds are formed between specific pairs of bases.
(c) Starch is a polymer of D – glucose while cellulose is a polymer of E – glucose

Ans.30-(a) Because sulphur readily gets oxidized itself to more stable +6 state.
(b) Because of absence of d-orbital in Fluorine.
(c) Because size increases from Helium to Radon. / dispersion or van der Waal forces increase from Helium to Radon

OR

Ans.30 – (a) MnO₂+ 4HCl → MnCl₂ + Cl₂ + 2H₂O
(b) XeF₆ + KF → K⁺[XeF₇]^–
(c) 4I^–(aq.) + 4H⁺(aq.) + O₂(g) → 2I₂(s) + 2H₂O(l)

Ans.31-(a) NaCN act as a depressant.
(b) SiO₂ act as a flux. / used to remove FeO as slag
(c) I₂ is used to convert Ti into volatile compound (TiI₄)

Ans.32-

Ans.33-(a) Decreases.
(b) Increases
(c) Increases

Ans.34 – △T_f = K_f m
1.5=(3.9xW_Bx1000)/176×75
Mass of ascorbic acid = 5.08g

SECTION – D

Ans.35 – (a) E⁰ cell = E^o_c – E^o_A
= 0.34 – (-0.76)
= 1.10V

△G^o = -nFE^o
= -2 x 1.10 x 96500
= -212300 J/mol or -212.3 kJ/mol

(b) (i) Pollution free
(ii) High efficiency.

OR

Ans.35-(a) (i) Silver wire at 30^o C because as temperature decreases, resistance decreases so conduction increases.
(ii) 0.1 M CH₃COOH, because on dilution degree of ionization increases hence conduction increases.
(iii)KCl solution at 50^o C, because at high temperature mobility of ions increases and hence conductance increases

Ans.36-

(a) (i) Cu⁺¹(3d¹⁰) compounds are white because of absence of unpaired
electrons while Cu⁺² (3d⁹) compounds are coloured due to unpaired e- / shows d- d transition.
(ii) chromate (CrO₄^2-) changes to dichromate (Cr₂O₇^2-) ion in acidic medium.
(iii)due to completely filled d-orbitals in their ground state as well as in oxidized state.

(b) Co = [Ar]4s² 3d⁷ , Co⁺² =[Ar] 3d⁷
μ=√n(n+2)
=√3(3+2)
=√15 =3.92B.M

OR

Ans.36-

(b) (i) Sc⁺³, because of absence of unpaired electron.
(ii) Cr, because of presence of strong intermetallic bonding than Cu

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