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Chemistry
Practice Examination for G.C.E A/L Students- 2020
Conducted by Tamil students of Faculty of Engineering
University of Moratuwa
Serial- 03
2020
Marking scheme
CHO
N2O4(g)
jhf;fp
NO2(g)
tpisT
G G
B
A
C .
.
.
CH3−CH=CH−CH−C≡C−CHO
│ CH3
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Answer for MCQ
Q Ans Q Ans Q Ans Q Ans Q Ans
01. 5 02. 4 03. 4 04. 4 05. 4
06. 5 07. 3 08. 3 09. 2 10. 1
Marks Distribution
Part-I=10×2=20 Marks
Part-II=250 Marks
Part-II (Structured essay question) =100 Marks
Part-II (Essay question) =150 Marks
Total; = Part -I + Part –II
= 20 + 𝟐𝟓𝟎 /𝟓𝟎
= 70 Marks = 𝟕𝟎/ 𝟕𝟎 × 𝟏𝟎𝟎
= 100 Marks
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Part-II
Structured essay
(a)
(b)
1.
(05)
2. 4-methyl-2-hexene (04)
3. 4-methyl-3-hexanal \ 4-methylhexan-3-ol (04)
4.
(05)
5. Nucleophilic substitution reaction. (03)
6. Positional isomers ;. (04)
CH3 – CH2 – C – CH3
CH3
OH
C→ H– C – C – CH3
CH3 H
CH3 OH
A→
CH3
CH3 – CH2 – CH – CH2OH B→ (i)
CH3 – CH2 – CH–O –CH3
CH3
E→
CH3
CH3 – CH – CH2CH2OH D→
(7×5=35 Marks)
CH3 – CH – CH2CH2OH
CH3
CH3 – CH – CH=CH2
CH3
OH
CH3 – CH – CHCH3
CH3
CH3 – CH = CH–CH3
CH3
OH
CH3 – C – CH2CH3
CH3
(3)
(4)
(3)
(4)
(3)
(3)
(4)
(ii)
(24 Marks)
CH3
CH3 – CH2 – CH – CH=CHCH3
CH3
CH3 – CH2 – CH – CH2CHBrCH3
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7.
(05)
(05)
8. G, the one which gives immediate turbidity,
C, the one which gives turbidity after some time (06)
Part -II
CH3
CH3 – CH2 – CH = CHCH2CH3 F→
CH3
Br
CH3 – CH2 – CH – CH2CH2CH3 G→
(Total=100 Marks)
PCl5 (g) PCl3(g) + Cl2(g)
initial 0.5𝑚𝑜𝑙 Equilibrium mol; 0.5(1 − 𝛼)𝑚𝑜𝑙 0.5𝛼 0.5𝛼 𝛼-degree of dissociation
𝑃𝑉 = 𝑛𝑅𝑇 (system show ideal behavior)
𝑃 𝛼 𝑛𝑇 (𝑉, 𝑅 constants )
𝑃1 𝛼 0.5 × 𝑇1
𝑃2 𝛼 0.5(1 + 𝛼) × 𝑇2
𝑃1
𝑃2=
𝑇1
(1 + 𝛼)𝑇2
𝑃1𝑇2𝛼 = 𝑃2𝑇1 − 𝑃1𝑇2
𝛼 =𝑇1𝑃2 − 𝑇2𝑃1
𝑇2𝑃1
……….(10)
………(5)
……….(5)
graph I→wrong →concentrations must be[PCl3]=0.6M, [Cl2]=0.6M
Graph II→ wrong →when [Cl2]increases [PCl3] decreases
Graph III→ wrong →[PCl3]has changed by 0.5M but
[PCl5]is not changed by0.5M.
Graph IV→right
……….(5)
……….(5)
……….(5)
……….(5)
5) (a) (i)
(ii)
(40Gs;spfs;)
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KC =[ Y(g)][Z(g))]
[X(g)]=
0.4moldm−3 × 0.4moldm−3
0.2moldm−3
= 0.8moldm−3
PV = nRT ,y; P ∝ 1
V (n, R, T constants)
P → P
2 Mf P → 2V MFk;
C = n
V ⟹ concentration becomes
1
2 times
[X(g)] = 0.2moldm−3, [Y(g)] = [Z(g)] = 0.4moldm−3
QC =[ Y(g)][Z(g))]
[X(g)]=
0.2moldm−3 × 0.2moldm−3
0.1moldm−3
= 0.4moldm−3
KC=0.8moldm-3, QC=0.4moldm-3
here KC > QC
/ reaction takes place in forward direction
……….(3)
……….(2)
……….(5)
……….(5)
……….(3)
……….(2)
b) (i)
(ii)
(iii)
(iv) X (g) Y(g) + Z(g)
Initial concentration 0.1 0.2 0.2 moldm-3
final concentration 𝑥 - - moldm-3
equilibrium concentration 0.1−𝑥 0.2+𝑥 0.2+𝑥 moldm-3
T constant so, KC = 0.8moldm-3
(0.2 + 𝑥)moldm−3 × (0.2 + 𝑥)moldm−3
(0.1 − 𝑥)moldm−3= 0.8moldm−3
(0.2 + 𝑥)2 = (0.1 − 𝑥)0.8
𝑥2 + 1.2𝑥 − 0.04 = 0
𝑥 =−1.2 ± ξ1.44 − 0.16
2
𝑥 = −0.6 ± ξ0.1 × 2
𝑥 = 0.632 − 0.6 = 0.032
/at equlibrium
[X(g)]=0.68moldm-3
[Y(g)]= [Z(g)]=0.232moldm-3
……….(5)
……….(5)
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KP=KC(RT) Δn
Δn=2-1=1
Kp=0.8moldm-3 x 8.314Jmol-1x400K
=2.66X 103 Pa
P→ 2P ⟹ concentration doubles
𝑄𝑐 =0.8moldm−3 × 0.8moldm−3
0.4moldm−3= 1.6moldm−3
QC>KC /reaction takes place in backward direaction
……….(3) ……….(2)
(v)
(vi) X (g) Y(g) + Z(g)
Initial concentration 0.4 0.8 0.8 moldm-3
final concentration 𝑦 𝑦 moldm-3
Equilibrium concentration 0.4+𝑦 0.8−𝑦 0.8−𝑦 moldm-3
𝐾𝑐 =(0.8 − 𝑦)moldm−3 × (0.8 − y)moldm−3
(0.4 + y)moldm−3= 0.8moldm−3
(0.8 − y)2 = (0.4 + y)0.8
𝑦2 − 2.4𝑦 + 0.32 = 0
𝑦 =2.4 ± ξ5.76 − 1.28
2
𝑦 = +1.2 ±1
2ξ4.48
𝑦 = 0.142
/at equilibrium
[X(g)]=0.542moldm-3
[Y(g)]= [Z(g)]=0.658moldm-3
……….(5)
……….(5)
……….(5)
……….(3)
……….(2) (55
Gs;spfs;)
(vii)
……….(5)
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……….(2)
……….(2)
……….(3)
……….(2)
……….(2)
……….(2)
……….(2)
……….(2)
……….(2)
(c)
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……….(2)
……….(2)
……….(2)
……….(2)
……….(5)
……….(5)
(35
Gs;spfs;)
……….(5)
(d)
(i)
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therefore point c
……….(5)
……….(2)
……….(3)
……….(5) (20
Gs;spfs;)
nkhj;jk;-150 Gs;spfs;
(ii)
(iii)
(iv)
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