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

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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

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Your Target is to secure Good Rank in Pre-Medical 2012

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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