C1. 1. Answer (2)
Let the % of B-10 = X then % of B-11 = (100 – X)
Average atomic mass 10.2
100 11 ) X 100 ( X
10 + − =
= 10X + 1100 – 11X = 1020
X = 80
% of B–10 = 80 2. Answer (3)
Average atomic mass = 35.5 4
37 1 35
3× + × =
3. Answer (2)
Since X–, Y2– and Z3– are isoelectonic therefore
Number of electrons in increasing order will be X > Y > Z X–, Y2– and Z3– all have same no. of neutrons.
Therefore atomic no. increasing order will be Z < Y < X
C2. 1. Answer (2)
Let the moles of water = 1 mole Moles of urea will also be 1.
Mass percentage of water 100 60 18
18 ×
= +
78
=1800
= 23.07 2. Answer (1)
1 kg water has 11.11 moles of solute
mole fraction of solute 0.167
55 . 55 11 . 11
11 . 11 18
11 1000 . 11
11 .
11 =
= + +
= 3. Answer (3)
106 gm water contains = 300 gm CaCO3
∴ Molarity mol/L
1000 100 /
= 300
M 003 . 1000 0
3 =
= C3. 1. Answer (1)
= π Δ
⋅
Δ 4
p h
x (Δx = Δp)
= π
Δ 4
x h
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Maximum no. of electrons = 2n2
= 2 × 42 = 32
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-For the visible region transition belongs to Balmar series. -For the Balmar series the electron should jumps from higher level to 2nd energy level.
C5. 1. Answer (3)
Mn2+ – 1s2, 2s2, 2p6, 3s2, 3p6, 3d5
3d5 Total spin =
2 5 2 5×1 = 2. Answer (4)
Spin quantum no. is not derived from Schrondiger wave equation.
3. Answer (3)
Maximum number of electrons in any subshell having same value of spin quantum number is (2l + 1) C6. 1. Answer (2)
For the maximum wavelength energy should be minimum. In the option (2) there is no shell change.
Energy emission is minimum in this transition. So the wavelength will be maximum.
2. Answer (2)
Energy electron for H-atom depends only on the value of n not the value of l. Therefore 4s > 3d = 3p = 3s
3. Answer (1)
Radial nodes = (n – l – 1) For 3s ; n = 3, l = 0 Radial nodes = 3 – 0 – 1 = 2 For 2p ; n = 2, l = 1 Radial nodes = 2 – 1 – 1 = 0 C7. 1. Answer (1)
For the first excited state n = 2
Energy = 2
2
n 6 Z . 13 ×
−
eV
4 6 . 13 2
6 1 .
13 × 2 = −
−
=
= –3.4 eV 2. Answer (4)
1 529n . 0 r
2 H =
= 0.529 × 1 Å
= 0.529 Å
4 529 2 . 0 r
2 Be3+ = ×
= 0.529 Å 3. Answer (4)
1s electronic level allow the H-atom to absorb a photon but not to emit a photon. If it emit a photon it will drop into nucleus, that is not possible.
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-C8. 1. Answer (1)
N(7) – 1s2, 2s2, 2p3 – Half filled (more stable) O(8) – 1s2, 2s2, 2p4 – Less stable
∴ Electron Affinity of Nitrogen is less than Oxygen.
2. Answer (4)
Electron Affinity of Br is less than Chlorine.
3. Answer (1)
Na(g) ⎯→ Na+ (g) + e ΔH = ionisation energy …(1) Na+(g) + e ⎯→ Na ΔH = Electron affinity …(2)
Since (1) and (2) process are opposite therefore ionisation energy of Na is equal to electron affinity of Na+. C9. 1. Answer (4)
In He+ ions the electrons are tightly held up by the nucleus therefore its ionisation energy is more than He.
2. Answer (1)
Be (4) ⎯→ 1s2, 2s2 – full filled (more stable) B (5) ⎯→ 1s2, 2s2, 2p1 – Less stable
∴ Be has more first I.E. than I.E. of B.
3. Answer (1)
Due to half filled P-orbital nitrogen electronic configuration is more stable. Therefore its ionisation energy is more.
C10. 1. Answer (2)
Since IE of element B is less therefore it is most reactive metal amongst given elements.
2. Answer (4)
Element D has high IE but less than IE of A. Therefore (D) is non metal.
3. Answer (1)
The first ionisation potential is highest for element A therefore A is noble gas.
C11. 1. Answer (2)
B — F F
N H H H F
..
N
F F F
..
μ = 0 Addition Subtraction
∴ BF3 < NF3 < NH3. 2. Answer (3)
Cl Cl
Cl μ = 0
3. Answer (2)
% ionic character = 16 (ΔEN) + 3.5 (ΔEN)2 = 16(0.2) + 3.5(0.2)2 = 3.34
AakashIIT-JEE CO molecule
Total no. of electrons = 14.
Arrangement – σ1s2σ*1s2σ2s2σ* 2s2
In CO+ ion, one electron is released from antibonding molecular orbital.
∴ B.O. = 2
Percentage ionic character = 100 16.87% 10
There is no unpaired electrons
∴ paramagnetism is not shown.
2. Answer (2)
Bond order of N2 is 3.
Bond order of O2 is 2.
Bond order of F2 and Cl2 are 1.
∴ Bond length ∝ Bond1order
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-3. Answer (1)
In NO molecule there are total no. of 15 electrons.
Arrangement - σ1s2σ* 1s2σ2s2σ*2s2σ2px2
∴ Unpaired electrons = 1.
C15. 1. Answer (3)
Initial moles 1 0
At equilibrium 1 – x 2x
Total no. of moles at equilibrium = 1 – x + 2x = 1 + x = 1 + 0.14 = 1.14
AakashIIT-JEE Coming out would be 2.17 x
68
Partial pressure of hydrogen Ptotal
16
Dalton’s law of partial pressure is not valid for reacting gases
fumes
Volume percent of H2 100
mixture Pressure 2
× 750 100
150× =
= C17. 1. Answer (1)
3O2(g) 2O3(g) Initial volume 1 litre 0 After reaction 1 – 3x 2x
Total volume of gaseous mixture = 1 – 3x + 2x = 0.8
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-2. Answer (2)
mL 402 oil turpentine mL
1003 O2) O
O
( + ⎯⎯⎯⎯⎯⎯→ Therefore volume of O3 is 60 mL
2O3 3O2
2 mL 3 mL
60 mL 90 mL
Total volume now = 40 + 90
= 130 mL
Change (increase) in volume = 130 – 100
= 30 mL 3. Answer (1)
Let the volume of NH3 be x mL Volume of H2 = (50 – x) mL
2 2
3 H
2 N 3 2 NH ⎯⎯→1 +
Since 40 mL of O2 is added and sparked it must have reacted with H2 to form liquid water. Moreover since 6 mL contraction occurs with alkaline pyrogallol, 34 mL is the volume of O2 is used up.
Total volume of H2 is 68 (Q 2H2 + O2→ 2H2O)
( )
x 682 x 3
50− + = x = 36
% NH3 = 72 C18. 1. Answer (2)
PV = K(constant) 2. Answer (4)
PV = nRT
n is not directly proportional to T
∴ Its graph will not be straight line.
3. Answer (2)
M(average) = 20
y x
y 28 x
16 =
+ +
16x + 28y = 20x + 20y 4x = 8y
Maverage =
y x
x 28 y 16
+ +
3 24 72 y
3 ) y 2 ( 28 y
16 + = =
= C19. 1. Answer (4)
At point A near low pressure region volume is very high thus (V – b) ~ V
RT V V
p a
2⎟ =
⎠
⎜ ⎞
⎝⎛ +
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At point B at high pressure volume is low
⎟⎠
Comparing to equation of state
⎟⎠
19(a). Answer (1, 3, 4) (IIT-JEE 2008)
Because V is very large, so in Van der Waal’s equation ⎟
⎠ neglected and equation becomes PV = RT. Coefficients depends on the identity of the gas but are independent of the temperature. Real gas exert lower pressure than the same gas behaving ideally due to intermolecular force of attraction.
C20. 1. Answer (3)
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-2. Answer (2)
Amount of heat required to lower the temperature of 500 g of water from 20º to 0ºC
= 4.2 × 500 × 20 = 42000 J
No. of moles of ice needed to melt to absorb this heat = 7 10 6
42000
3 =
×
Since each ice cube contains one mole water so at least 7 ice cube are required.
3. Answer (3)
Enthalpy change ΔH = nCpΔT, since process is isothermal ΔT = 0
= 0 C21. 1. Answer (4)
For spontaneous process ΔG should be negative.
∴ ΔH < 0 and ΔS > 0 2. Answer (2)
Entropy change in isothermal process.
ΔS
1 2
V logV nR 303 .
=2
1 log10 314 . 8 2 303 .
2 × ×
=
= 38.29 JK–1 mol–1 3. Answer (2)
ΔS = Tb
ΔH
K 75 400
1000 Tb = 30× =
C21(a). Answer (4) (IIT-JEE 2008)
ΔG = ΔG° + RT ln Q at equilibrium
ΔG = 0 Q = Keq
∴ ΔG° = –RT ln Keq
C22. 1. Answer (4)
Work done in AB process = –PΔV
W1= 3 (30 – 10) = –60 atm×litre Work done in B → C process = 0
Work done in C → A process
W2 = –2.303 nRT log
1 2
V V
= 30
log10 ) 30 1 ( 303 .
2 ×
−
= +2.303 × 30 × 0.5
= 34.54 atm × litre Net work done in the process
= –60 + 34.54 = –25.45 atm × litre
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-At point A → PV = nRT
375K
1 08 . 0
10
T 3 =
×
= ×
Temperature at point C = 375K
08 . 0 1
30 nR
T pV =
= ×
= 3. Answer (2)
Since process is cyclic therefore ΔH and ΔU will be zero because in cyclic process state functions will be zero.
C23. 1. Answer (2)
Since heat of ionisation for HCN is more than CH3COOH. It means HCN is weaker acid than acetic acid.
Therefore Ka(HCN) will be less than Ka(CH3COOH)
∴ pKa(HCN)>pKa(CH3COOH)
2. Answer (3)
Na+(aq) + OH– (aq) → NaOH (aq) Heat of ionisation of strong base = 0
0 ) OH ( H ) Na ( H ) NaOH ( H
H=Δ f −Δ f −Δ f =
Δ + −
⇒ −470.7−ΔHf(Na+)−(−228.8)=0
kJ 9 . 241 ) Na (
Hf =−
Δ +
3. Answer (2) ΔQ = nSΔT
Molar heat capacity = T n
Q Δ Δ
Since in ice water equilibrium there is change in temperature is zero.
∴ Molar heat capacity = ∞ C24. 1. Answer (1)
For acidic buffer
pH = pKa + log [[acidsalt]]
= 4.74 + log
1 . 0
500 1000 59
2 ⎥⎦⎤
⎢⎣⎡ ×
= 4.74 + log 59 40
= 4.74 – 0.1739
= 4.57
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-2. Answer (3) pH = pKa + log
] acid [
] salt [ [salt] = [acid]
pH = pKa
= 4.74.
3. Answer (3)
pH = 4.74 + log [[acidsalt]] 6.74 = 4.74 + log [acid] ] salt [
] acid [
] salt
[ =
1 100
% of acid in the mixture = 101
1 × 100 = 1 %
∴ % dissociation of acid = 99%
C25. 1. Answer (3)
Keq =
r f
K K
Kf = Keq × Kr
= 1.16 ×10–3 × 57
= 66.12 × 10–3 Kf = 6.612 × 10–2 2. Answer (1)
Kf > Kr. 3. Answer (4)
Since reaction is endothermic on increasing the temperature equilibrium constant increases ⎟
⎠
⎜ ⎞
⎝
⎛ =
r c Kf
K K
but Kf increases more than Kr. C26. 1. Answer (1)
CH3COOH CH3COO– + H+
1 – α α α + 0.01
Ka = 1.8 10–5 1
) 01 . 0
( = ×
α
− + α α
on solving we get pH = 1.937.
2. Answer (4) [OH–] = 2 × 10–7 pOH = – log (2×10–7)
= 7 – log 2 = 6.7 pH = 14 – 6.7 = 7.3.
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-Kw = [H+] [H–] = 10–14
Ka = [H[H][OHO] ] 5510.55
14 2
− −
+ =
= 0.018 × 10–14
= 18 × 10–17. C27. 1. Answer (4)
CH3COONa → CH3COO– + Na+
CH3COO– + H2O CH3COOH + OH– (basic).
2. Answer (3)
H2S H+ + HS solution is acidic [H+] > 10–7 m NaCl → solution is neutral [H+] = 10–7 NaNO2 → solution is basic [H+] < 10–7 H2SO4 → solution is strong acidic
0.01 MNaNO2 < 0.01 MNaCl < 0.01 MH2S < 0.01 MH2SO4 3. Answer (2)
CH COOH + NaOH3 CH COONa + H O3 2
10 4 0 0
6 0 4 4
m.moles (initial) After reaction
buffer solution
pH = pKa + log [[acidsalt]]
pH – pKa = log 6 4
= log ⎟
⎠
⎜ ⎞
⎝
⎛ 3 2 .
C28. 1. Answer (3)
There is no effect at equilibrium at constant volume.
2. Answer (4)
NaNO3(s) NaNO2(s) + 2 1 O2(g)
∴ since no. of gaseous moles decrese in backward direction thererfore on increasing the presure reverse reaction favour.
3. Answer (2)
C2H4(g) + H2(g) C2H6(g), ΔH = – 136.8
Since reaction is exothermic therefore decrease in temperature favour forward reaction. Since no. of moles decreases in forward direction therefore on increasing the pressure forward reaction favours.
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-Specific conductivity is directly proportional to the concentration.
2. Answer (3)
K = ×⎜⎝⎛ ⎟⎠⎞ a R
1 l
1.2 = ⎟
⎠
⎜ ⎞
⎝
×⎛ a 55
1 l
m 1
66 55 2 . a 1
= −
×
⎟=
⎠
⎜ ⎞
⎝
⎛ l
or 66 × 10–2 cm–1. 3. Answer (4)
On doubling the edge length, volume of cube becomes 8 times and the concentration of solution decreases 8 times. Specific conductance is directly proportional to concentration. Hence, it decreases 8 times.
C31. 1. Answer (3)
Λ = 0.001028
10 95 . 4 1000 N
K
1000× = × × –5
= 48.15
degree of dissociation = ⎟=⎜⎝⎛ ⎟⎠⎞
⎠
⎜ ⎞
⎝
⎛ ΛΛ
∞ 390.5 15 . 48
= 0.123.
2. Answer (3)
Λ∞m = 119 + 160 = 279
degree of dissociation α = ΛΛ∞
m
0.1 = 279
Λ
Λ = 27.9.
3. Answer (1) Λm =
M K 1000×
K = 1000
1 . 0 9 . 27 ×
= 2.79 × 10–3.
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-C32. 1. Answer (4)
Q Anion is bigger Q it will constitute lattice 294
. 21 0 . 2
65 . 0 r
r = =
−
+ (i.e. range of coordination number = 4)
2. Answer (1) 414 . r 0 r =
−
+ (for maximum packing efficiency)
∴ 193.23pm
414 . 0 r−= 80 =
3. Answer (2)
oh voids in fcc is situated at all edge centres and body centre and td voids are present at each body diagonal.
Nearest oh and td void pair is oh void at body centre and td void at edge centre.
∴ a
4 distance= 3
oh void
td void
body diagonal √3a
td void
√3 4 a
C33. 1. Answer (2)
2 4 4 1 4 12 1 2 2 1 8
8 1B C
A × × + × × +
⇒ AB4C6 2. Answer (1)
Tetrad axis will remove 2 atoms at face centre and 1 of body centre (which is not present in this compound) 2 face centre atoms can be B or C.
If B is removed, formula of left compound = AB3C6 If C is removed, formula of left compound = AB4C5 3. Answer (1)
A at corner
B at 2 face centre and C at rest 4 face centres i.e. the basic lattice is ccp
∴ No. of oh voids = 4 and td voids = 8
Out of 4 oh voids 3 are filled (atoms are present at edge centre) and 4 td voids are filled
∴ fraction occupied = 0.58 127 =
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-Density of CsCl
Z = 1 (1 formula unit is present) For bcc structure
3 pm ) r r ( a 2
) r r ( 2 a 3
− +
− +
= + +
=
Volume = 3 30 3
3
cm 10 )
3 (
)]
r r ( 2
[ ++ − × −
∴ 30
A 3
3 CsCl
10 N )]
r r ( 2 [
) 3 ( M
− −
++ × ×
= × ρ
2. Answer (1)
48 . 2 5
) 8 . 2 ( ) 2 (
4 3
bcc 3
fcc = × =
ρ ρ
3. Answer (2)
V = (100)3 × 10–30 = 10–24 cm3 1 cm3 = 10 g
∴ 10–24 cm3 = 10–23 g 10–23 g = 2 atoms (Bcc)
∴ 100 g = 2 × 100 × 1023 g = 2×1025 atoms C35. 1. Answer (3)
body diagonal plane of fcc.
2. Answer (1)
Rectangular plane (fcc)
3. Answer (3)
body diagonal plane (Bcc)
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O2– replaced = 6 (face centred)
Charge replaced = 6 × 2 = 12 (negative units)
∴ Y to be doped = 4
On increasing temperature, K increases but Ea and A remains same.
2. Answer (3)
By solving, Ea = 64 kJ 3. Answer (4)
According to Arrhenious equation,
RT
n ratio decreases while in K-electron capture, α-emission and positron emission, p n ratio increases.
2. Answer (4)
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-Tritium (1T3) has highest p
n ratio i.e. 2.
38(a). Answer (2) (IIT-JEE 2008)
Z (At no.)
no. of neutrons 45o
Elements with higher atomic number are more stable if they have slight excess of neutron as this increase the attractive force and also reduces repulsion between protons.
C39. 1. Answer (3) Rate ∝ [A]x 2 ∝ (4)x (4)1/2∝ (4)x
x = 2 1
2. Answer (1)
In the 1st experiment, Rate ∝ [A]x
(3)3 = 27 ∝ (3)x x = 3
In the IInd experiment, Rate∝ [A]x [B]y 8 ∝ (2)3 (2)y y = 0 3. Answer (4)
The order w.r.t. B is zero.
C40. 1. Answer (3)
Lesser is the half life, more is the radioactivity.
2. Answer (4)
) x a ( log a t 303 . K 2
= −
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-20 log100 t 303 . 2 3 . 69
693 .
0 =
t = 161 minutes 3. Answer (1)
60 log100 10
303 .
K1= 2 …(1)
10 log100 t 303 .
K2 =2 …(2)
K1 = K2
10 log100 t 303 . 2 60 log100 10
303 .
2 =
t = 45 minutes C41. 1. Answer (3)
xbenzene = 5
1 = 0.2. xtoluene = 5 4 = 0.8
According to Roult’s law
PT = Pbenzeneo ×xbenzene+Ptolueneo ×xtoluene PT = 700 × 0.2 + 600 × 0.8
PT = 140 + 480 = 620 mm 2. Answer (2)
Suppose, x g benzene and x g toluene are mixed nbenzene =
78
x , ntoluene = 92
x
Total moles =
92 78
x 170 92
x 78
x
= × +
xbenzene =
170 92 92 78
x 170
78 x
=
×
xtoluene =
170 78 170 1− 92 =
PT = Pbenzeneo ×xbenzene+Ptolueneo ×xtoluene
PT =
170 600 78 170
700× 92 + ×
PT = 378.82 + 275.29 PT = 654.11 mm
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-Mole fraction of benzene in vapour phase = 0.225 620
140 P
P
T
benzene = =
C42. 1. Answer (2)
m = 0.1
500 1000 60
3 1000
500 60
3
=
×
=
In beaker ‘A’
ΔTb = i × Kb × m 0.17 = i × 1.7 × 0.1 i = 1
It shows that acetic acid remains normal molecule in acetone.
In beaker B ΔTb = i × Kb × m 0.13 = i × 2.6 × 0.1 i = 0.5
It shows that acetic acid is 100% dimerised in benzene.
2. Answer (1)
In acetone, acetic acid remains the normal molecule. Hence, molecular weight of acetic acid will be 60.
3. Answer (2)
In benzene, acetic acid dimerises 100%. Hence molecular weight of acetic acid will be 120.
C43. 1. Answer (3)
K3[Fe(CN)6] 3K+ + [Fe(CN)6]3–
1 0 0 before dissociation
(1 – α) 3α α after dissociation
1 3 i 1+ α
= (Q α = 0.6)
8 . 1 2
6 . 0 3 i 1+ × =
=
2. Answer (4)
The ratio of effective molarity of 0.5 M AlCl3, 2 M urea and 0.2 M K4[Fe(CN)6] is 2 : 2 : 1. Hence, the ratio of osmotic pressure is also 2 : 2 : 1.
3. Answer (2)
2CH3COOH (CH3COOH)2
1 0 before association
(1 – α)
2
α after association
Here, α = degree of dimerisation 8
. 1 0
5 . 0 i=1− α =
AakashIIT-JEE -4 . 5 0 . 0
2 . 0 =
= α
Percentage of dimerisation = 40%.
C44. 1. Answer (3)
Fe(OH)3 sol is positively charged.
2. Answer (1)
Number of milligram of lyophilic colloid required to protect 10 ml gold sol in 1 ml 10% NaCl solution is equal to gold number of lyophilic colloid.
3. Answer (3)
As2S3 is negatively charged sol.
C45. 1. Answer (3)
According to Freundlich adsorption isotherm, P
nlog K 1 m log
logx = +
Comparing the above equation with y = mx + c then we get, intercept = log K and slope = n 1.
2. Answer (2)
According to Langmuir adsorption isotherm,
aP ) 1 ( a b x m = +
Comparing the above equation with y = mx + c then we get, slope = a
1, intercept = a b.
3. Answer (1)
According to Langmuir adsorption isotherm,
bP 1
aP m
x
= +
At very low pressure, bP is negligible in comparison to 1. Then we get mx =aP.