HS 1 / 3Your Target is to secure Good Rank in Pre-Medical 2012
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25A. 1 2 1 2 2 3 3 3 2 3 2 1 4 2 2 3 1 1 2 2 3 2 2 1 3Q 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50A. 3 4 1 2 3 1 1 2 2 1 2 2 2 2 3 1 2 2 3 3 4 4 4 1 4Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75A. 3 2 3 3 1 4 1 3 3 1 3 2 4 1 2 3 3 2 3 2 1 4 4 1 3Q. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100A. 4 1 3 1 4 1 2 1 3 4 4 1 4 2 2 1 3 2 4 4 3 4 2 1 3Q. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125A. 3 2 4 2 2 1 1 2 4 2 2 3 2 2 3 2 4 3 4 1 2 2 2 4 1Q. 126 127 128 129 130 131 132 133 134135 136 137 138 139140 141 142 143 144 145 146 147 148 149 150A. 3 4 2 2 1 1 1 1 2 2 1 2 1 1 2 2 2 4 3 1 4 1 3 3 2Q. 151 152 153 154 155 156 157 158 159160 161 162 163 164165 166 167 168 169 170 171 172 173 174 175A. 1 2 1 1 3 1 1 1 1 4 3 4 4 4 3 2 1 4 1 3 2 2 1 4 4Q. 176 177 178 179 180 181 182 183 184185 186 187 188 189190 191 192 193 194 195 196 197 198 199 200A. 2 4 4 2 2 1 3 1 4 2 4 3 4 3 1 1 2 3 3 3 3 1 1 2 2
HINT – SHEET
DATE : 20 - 05 - 2012ALLEN SPECIAL AIIMS # 01
TARGET : PRE-MEDICAL 2012
LEADER, ACHIEVER & ENTHUSIAST COURSE
MAJOR TEST
43. Reactivity µ Stability of carbocation intermediate
45. Moles of sodium in 2.3mg of sodium
= -´ 32.3 10
23 = 1 × 10–4 mol
Energy required to convert 1 mole of sodium
vapour into ions = 495 KJ
\ Energy required to convert 1 × 10–4 mole of
sodium vapours into ions = 495 × 10–4 KJ
= 49.5 J
46. Only solvent molecules can pass through SPM
so only dilution is possible
47.
CH OH2
HÅ HBr Aqu. KOHD
Br OH
OCH2
PCC
Ph P=CH3 2
48. Oxides of metals lies in ECS below H are
unstable
AgNO3 D¾¾® Ag + NO2 +
12
O2
49. NH3 + H2O � NH4+ + OH–
Kb = [ ]
[ ][ ][ ]
[ ]+ - - -é ù ´ ´ë û =
3 34
3 3
NH OH 1.5 10 1.5 10NH NH
Equilibrium [NH3] = ( )-
-
´´
23
2
1.5 101.8 10
= 0.125 M
Initial concentration of [NH3] = 1.5 × 10–3 + 0.125
= 0.1265 M
» 0.13 M
51. O OHC–O Na
O O O
O OO
H O3Å NaOH Kolbe
OHC–O Na
Electrolysis
HC |||HC
52. Pb3O4 is a mixed oxide (2PbO×PbO2)
53. (NH4)2S(s) � 2NH3(g) + H2S(g)
2P P
(NH4)2 Se(s) � 2NH3(g) + H2Se(g)
2q q
20–05–2012
2/3 HSYour Target is to secure Good Rank in Pre-Medical 2012
MAJOR TEST : AIIMS
TARGET : PRE-MEDICAL 2012
( ) ( )1 3 2
2
p NH H SK P P= = (2p + 2q)2. p = 9 × 10–3
( ) ( )2 3 2
2
p NH H SeK P P= = (2p + 2q)2. q = 4.5 × 10–3
\ pq = 2 Þ p = 2q
\ 1pK = (2p + q)2. p = 9 × 10–3
\ p = 0.1 atm and q = 0.12
= 0.05 atm.
Total pressure = 2p + 2q + p + q
= 3p + 3q
= 3 × 0.1 + 3 × 0.5
= 0.45 atm
54. R–CH=O + H N–NH–C H2 6 5 ¾®
R–CH=N–NH–C6H5
weight = (R + 119)
Q 28
100R 119
´´
= 20.9
R = 15 means CH3–
\ R–CH=O CH3–CHO
55. CH3–CH2–CºCH 2
3 2
(i) NaNH(ii)CH CH Br¾¾¾¾¾®
CH3–CH2–CºC–CH2–CH3
Na + NH3 (liquid)
CH3–CH2–CH=CH–CH2–CH3
Br2 in CCl4
Br
Br
3,4-Dibromohexane
57.[ ]-d A
dt = K[A]n
On doubling [A], rate is doubled, hence n = 1
\ Kf = [ ] -- ´
=2d A / dt 1 10
[A] 0.1 = 0.1 mol L–1s–1
KC = f
b
KK
\ Kb = =f
c
K 0.1K 10 = 1 × 10–2 L mol–1s–1
61. CaCO3(s) � CaO(s) + CO2(g)
Vapour density (D) = M2
Actual density (d) = PMRT
d PM 2 2PD RT M RT
´= =
´
At equilibrium P and T remain constant
63. Glucose with C6H5NHNH2 form glucosazone
and structure of glucosazone is
CH=N–NH
C=N–NH
(CHOH)3
CH OH2
64. Bond angle in the hydrides of VA & VI A gp.
third period downwards is ; 90°.
65. Density crystal decreases due to schottky
defects, effective no. of atoms per unit cell
= 4 æ ö-ç ÷è ø
0.251
100 = 3.99
d = ( )-
´
´ ´ ´323 7
50 3.99
6 10 0.50 10
= ´
´ ´50 3.99
6 125 0.1
d = 2.66 g/cc
67. Salol is the product of reaction between phenol
and salicylic acid
Salol ® Phenyl salicylate
HS 3/3
20–05–2012
Your Target is to secure Good Rank in Pre-Medical 2012
MAJOR TEST : AIIMS
PRE-MEDICAL : LEADER, ACHIEVER & ENTHUSIAST COURSE
69. Au+ + e ® Au E° = 1.69 V ... (1)°D 1G = –1 × F × 1.69
Au+3 + 3e ® Au E° = 1.40 V ... (2)°D 2G = –3 × F × 1.40
Au+ ® Au+3 + 2e– E° = ? ... (3)°D 3G = –2 × F × E°
From (1) and (2)° ° °D - D = D1 2 3G G G
Þ –1 × F × 1.69 + 3 × F × 1.40 = –2 × F × E°
Þ E° = - ´1.69 3 1.40
2
= -1.69 4.202
E° = –1.255 Volt
71. In the period L to R IP normally increases
73. DHionisation = –106.68 – (– 2 × 55.84)
= 5 KJ/mol
74. Structure in option no.(2) contains 9 carbon
atoms so not the answer of related question.
77. ° ° °D = D - D200 200 200G H T S
Þ °D 200H = 20 – 4 = 16 KJ/mol
° °D = D + D -2 1T T P 2 1H H C (T T )
° °D = D - D400 200 PH H C (400 – 200)
= 16 + ´20 200
1000
= 16 + 4
= 20 KJ/mol78.
3 moles
2 moles
1 mole
— mole 23
3 R–CH = CH + BH (R–CH –CH ) B2 3 2 2 3
79. Red–P
P P
P P—P P—P P—
n
92. Only estrogen is steroid, rest are peptide
buesa ls dsoy bLVªkstu gh LVhjkWbM gSA96. Amount of water decreases in blood leads to
increased concentration but decrease in
volume
ikuh dh deh ls IykTek lkanzrk c<+rh gS vkSj vk;ru de gksrk
gSA100. This enzyme attacks cellulose.
;g ,atkbe lSywykst dks ipkrk gSA119. More diameter & Myelination.
T;knk O;kl vkSj ekbfyu
142.
Anthracene Phenanthrene
144. In aromatic compounds percentage of carbon
is high so burns with sootyflame
146. Alkenyl halide not show substitution at normal
condition [Due to double bond character in
C–X bond by resonance]
148.O O
O OCr
O
It has O2– and O22– units
155. Benzyl carbocation more stable than ethyl
carbocation,
156. H – P – O K– +
O
O K– +
K2HPO3, H – O – P – O K– +
O
O K– +
K2HPO4
174. Lacteals absorb fatty acids but do not increase
surface area. These are lymphatic ducts in a
villus
ySDVhy] olh; vEyksa dks vo'kksf"kr djrh gS ij lrg {ks=
ugha c<+krhA buesa jä dh txg fyEQ mifLFkr gksrk gSA
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