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So the given graph will be a straight line with slope equal to
1434
1910414.2
10626.6106.1
he
×=×
×=
−
−
25. Answer (1)
⎟⎠⎞
⎜⎝⎛
∞−=
λ 221
111096781
λ = 9.1176 × 10–6 cm= 911.76 Å
26. Answer (4)
Energy of infra radiation is less than the energy of ultraviolet radiation of the given transitions energy emitted intransition n = 5 → n = 4 is less than the energy emitted in transition n = 4 → n = 3.
27. Answer (3)
Light source is radiating energy at the rate 20 Js–1
The lines in the Balmer series are emitted when the electrons jumps from n = 3, 4, 5...... orbits to the secondallowed orbit. Since the difference in energy between the third allowed state and the ground state is 12.09 eV.The electrons will not be excited to the third allowed state and hence no line in the Balmer series will be emitted.
47. Answer (2)
Absorption line in the spectra arise when energy is absorbed. i.e. electron shifts from lower to higher orbit out of (1)and (2), (2) will have lowest frequency as this falls in the Paschen series.
48. Answer (1)
Ni(28) – 1s2, 2s2, 2p6, 3s2, 3p6, 3d8, 4s2
Total no. orbitals = 15
49. Answer (1)
Radius in the third orbit = 9r (∴ rn ∝ n2)
n3λ = 2πr3
3λ = 2π × 9r
λ = 6πr
50. Answer (1)
Orbital angular momentum = π
+2h)1(ll
For s-orbital l = 0
∴ Orbital angular momentum = 0.
51. Answer (1)
Since it is feasible to remove only one electron from the element therefore element belong group 1.
52. Answer (4)
Inert gases has most stable electronic configuration therefore has least electron affinity.
53. Answer (3)
1s2, 2s2, 2p6, 3s1, after removing the first electron it occupy the noble gas configuration therefore it is not feasibleto remove 2nd electron.
54. Answer (3)
BaO2 can exist in form of Ba2+ O22–.
55. Answer (4)
This is because in transition element the effect of increasing nuclear charge almost compensated by extrascreening effect provided by increasing number of d-electrons.
56. Answer (3)
1 mole sodium = 23 gm sodium.
Q 23 gm sodium requires 495 kJ energy for ionisation.
Because intermolecular force of attraction in NH3 is high.
84. Answer (3)
Since in adiabatic process there is no exchange of heat between system and surrounding therefore in expansiontemperature falls down and pressure will be less than the pressure in isothermal process.
85. Answer (2)
Solubility of gases in liquids increases on increasing the pressure.
ΔE = 0, ΔH = 0 and at constant temperature, PV = K(constant) – Boyle’s law
When temperature is constant PV is constant
ΔH = ΔE + Δ(PV) = 0.
100. Answer (3)
Heat of formation of a compound is defined as the change in enthalpy when one mole of the compound has beenformed from its constituent elements.
101. Answer (1)
3O2(g) → 2O3(g) is endothermic
0E3EO42O3 <−
2O O43E
3<
This in equality valid only 3O2O EE > .
102. Answer (2)
2HgO(s) → 2Hg(l) + O2(g)
As the reactant from its solid state is converting to liquid and gas phase heat is required for this decompositionΔH > 0 further more entropy increases ΔS > 0.
103. Answer (3)
For a diatomic gas Cp = R27
, Cv = R25
Only 75
= 0.71 of energy supplied increases the temperature of gas.
The rest is used to do work against external pressure
0.71 × 60 = 42.6 kcal.
104. Answer (1)
122
111 VTVT −γ−γ =
∴ 11
2
1
1
2 2VV
TT −γ
−γ
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
Since γ is more for the gas X. The temperature will also be more for it.
105. Answer (1)
For an adiabatic process
PVγ = constant
log P = –γ log V + constant
Thus slope of log P versus log V graph is –γ. The value of γ is maximum for helium monoatomic gas. Thuscurve C should respond to helium.
The solution contains = 200 × 0.1 = 20 m.mol of NaOH
1 m.mol of CO2 reacts with 2 m.mol of NaOH
2NaOH + CO2 → Na2CO3 + H2O
The resulting solution contains 18 m mol of NaOH and 1 m.mol of Na2CO3. On titration upto phenolphthaleinend point, the NaOH will use 18 m. mol of acid and Na2CO3 will use 1 m.mol of acid, Hence
Normality = N 095.0200
)118(=
+.
112. Answer (3)
CH3COONa + HCl → CH3COOH + NaCl
initially moles 0.1 0.2 0 0
after reaction moles 0 0.1 0.1 0.1
Since CH3COOH is weak acid therefore we assume that H+ ions from CH3COOH is so less than can beneglected
∴ HCl → H+ + Cl–
0.1 0.1 0.1
[H+] = 0.1 M
pH = – log[H+] = – log 0.1
= 1
113. Answer (1)
For CaCO3 (s) CaO(s) + CO2(g)
Kp = 2COP = 0.0095 atm
Since atmospheric pressure is 1 atm so percentage of CO2 in air = %95.0100PP
total
CO2 =× .
Thus to prevent the decomposition of CaCO3 at 100°C the % of CO2 in air must be greater than 0.95%.
Since Qp < Kp. So this pressure is not sufficient to maintain the system in equilibrium therefore total pressurein the chamber would be equal to 30 atm.
If some amount of CO has been added into the vessel at constant volume then the second equilibrium will movefor backward direction. As a result the equilibrium concentration of Cl2 will be less. So the equilibrium constantof the first reaction will also be disturbed and reaction quotient will be less than the equilibrium constant.Therefore to attain the new equilibrium first reaction will move to forward direction and the conc of PCl5 presentat new equilibrium will be less
Higher the reduction potential greater is tendency for reduction. The electrode with higher reduction potential(Pb2+/Pb) acts as a cathode while other electrode (Fe/Fe2+) with lower reduction potential acts as anode
At anode e2FeFe 2 +⎯→⎯ +
At cathode Pbe2Pb2 ⎯→⎯++
Net reaction PbFeFePb 22 +⎯→⎯+ ++
ºanode
ºcathode
ºcell EEE −= = –0.13 V – (–0.44) = + 0.31V
Since the standard emf to the cell is positive the reaction is spontaneous. Hence more of Pb and Fe2+ areformed.
135. Answer (3)
Na2S2O3
Na O — S — O Na+ – +—
S–1 –1
–2
–2O
+2
+1
+1
+2
Since total charge in the sulphurs are –2 and +6 each
∴ Oxidation no. of sulphur in hypo are –2 and +6
136. Answer (2)
Equivalents of Cr deposited = equivalent of O2 evolved
As the Cu2+ ions lost from the solution are compensated by copper anode therefore concentration of the solutionremain same
At anode e2Cu)s(Cu 2 +⎯→⎯ +
At cathode )s(Cue2Cu2 ⎯→⎯++
139. Answer (2)
As the surface area of contact of an electrode with electrolyte increases. Conductance of electrolyte increasethereby time rate of electrolysis increases.
140. Answer (2)
Half cell reactions of the given electrodes are
First electrode ++ ⎯→⎯+ 222 AmOeAmO
Second electrode OH2Ame2H4AmO 242
2 +⎯→⎯++ +++
Third electrode ++ ⎯→⎯+ 24 Ame2Am
Thus it is evident that half cell reaction of only second electrode involves H+ ions so its reduction potentialwill change with varying pH value.
141. Answer (1)
Anode e2H2H2 +⎯→⎯ +
10–6 M
Cathode 2He2H2 ⎯→⎯++
Ecell = 0.118 = 26cathode
2
)10(]H[log
2059.0
−
+
On solving we get
[H+]cathode = 10–4 M
142. Answer (2)
422
2 OZnMne2ZnMnO2 ⎯→⎯++ +
187E
2MnO =
t = days35.5124360087102
965008Ei
96500w3 =
×××××
=×
×−
143. Answer (4)
In pure state sulphuric acid makes cyclic ring type of structure in the absence of water so it cannot give offH2 gas to react with metals.
The metal which has high reduction potential reduce first∴ Sequence of deposition of metal Mg < Cu < Hg, but Mg will not be deposited because H+ preferentially
3 face centre + 1 corner atom forms tetrahedral void in fcc.
170. Answer (3)
The centre atom is surrounded by six atom and one atom lies over this therefore C.N. = 7.
171. Answer (4)
All have same number of formula unit (i.e. Z = 4).
172. Answer (3)
Triclinic is most unsymmetrical crystal system a ≠ b ≠ c and α ≠ β ≠ γ = 90°.
173. Answer (1)
Total volume of sphere = 33 r3
16r344 π=π×
For fcc unit cell, a = r22
Volume of cube = 333 r216)r22(a ==
Packing fraction = 23r216
r3
16
cubeofvolumesphereofvolume
3
3π
=π
=
174. Answer (4)
d = 3
38233AV
Cm/g75.411023.66204
)10(10023.6624
aNMZ
=×
=××
×=
××
−
In Frenkel defect the density remains unaltered.
175. Answer (3)
When one face plane is removed then four corners ions and one face centre ion of B are removed (i.e. effectiveone ion of B) and 4 ions of A are removed from edge centres (effective one ion)
∴ New formula A+ B–
effective atoms 3 3
Since equal number of cations and anions are missing therefore defect is Schottky detect.
176. Answer (1)
r+ + r– = 2a
a = 2.3 × 2 = 4.6 Å
d = 243233AV 10)6.4(10023.6
784aN
MZ−×××
×=
××
= 3)6.4(023.67804×
× = 5.32 gm/Cm3.
177. Answer (4)
Total number of effective atoms in unit cell = 4 + 4 = 8.
Comparing the equation (i) with Ps = 210 – 120XA then we get
oBP = 210 and o
AP = 90
203. Answer (1)
ΔTf = Kf × m
1.86 = 1.86 × m
m = 1
It shows that 1 mole of urea dissolved in 1000 g of water.
nurea = 1, nwater = 5.5518
1000=
xurea = 5.561
5.5511
nnn
waterurea
urea =+
=+
204. Answer (2)
Freezing will start at –1.86°C not 0°C because ΔTf = 1.86 and as the value of ΔTf increases then the molality alsoincreases due to the freezing of water. Glucose doesn’t freeze.
205. Answer (3)
ΔTf = Kf × m …(i)
ΔTb = Kb × m …(ii)
By adding (i) and (ii)
ΔTf + ΔTb = Kf × m + Kb × m = m(Kf + Kb)
(Q ΔTf + ΔTb = 2.38)
2.38 = m(1.86 + 0.52)
m = 1 for non electrolyte solute.
Hence, answer is (3).
206. Answer (4)
For NaCl, m = 0.1
For Ba(NO3)2, m = 0.1
ΔTf for NaCl = i × Kf × m = 2 × 1.86 × 0.1 = 0.372
ΔTf for Ba(NO3)2 = i × Kf × m = 3 × 1.86 × 0.1 = 0.558
Total depression in freezing point = ΔTf for NaCl + ΔTf for Ba(NO3)2 = 0.372 + 0.558 = 0.93
Hence, freezing point of solution = 0 – 0.93 = –0.93°C.
The potential required to stop electro-osmosis is known as Dorn potential.
216. Answer (2)
K is highly reactive towards water.
217. Answer (1)
Gold number is equal to the number of milligram of substance required to prevent the coagulation of 10 ml gold solbefore adding 1 ml of 10% NaCl solution.
218. Answer (2)
van der Waal’s adsorption occurs at low temperature and high pressure.
219. Answer (1)
Higher is the value of van der Waal constant ‘a’ more is the adsorption on charcoal.
220. Answer (3)
Fe(OH)3 is positively charged sol due to adsorption of Fe3+ ions.
Section - B : Multiple Choice Questions1. Answer (2, 4)
The substances having same composition of atoms and similar crystal structure are isomorphous.
5. Answer (2, 3)Those substance can be oxidised and reduced, in which central element is neither in lowest nor in highestoxidation state.For Cl, range of oxidation number is from –1 to 7.In HCl, Cl is present in lowest oxidation stateIn HClO4, Cl is present in highest oxidation state.
There are three possible values of spin quantum number it means an orbital can accommodate 3 electrons.So 1s – 1s3, first period would have 3 vertical columns.
20. Answer (1, 3, 4)
Kinetic energy of the ejected electron depends on the frequency of incident radiation not on the intensity.
Intense and weak beam are having more or less number of photons.
21. Answer (1, 2)
rn = n2r1
)nln(4nlnrrln
AAln 4
21
2n
1
n ==⎟⎟⎠
⎞⎜⎜⎝
⎛
π
π=⎟⎟
⎠
⎞⎜⎜⎝
⎛
ln
ln(n)
AA
n
1
22. Answer (1, 2, 4)
Since only six different wavelengths are emitted therefore highest excited state is n = 4 therefore (1) is correct.
In the emitted radiation two wavelength are shorter than λ0 it means that initially atoms were in excited statetherefore (2) is also correct.
therefore hybridisation of each nitrogen is sp and sp2 respectively.
In gaseous state N2O5 exists as N O N
O
O
O
O
therefore hybridisation of each nitrogen is sp2.
N2O5 is called anhydride of nitric acid because in reaction with H2O, N2O5 forms nitric acid
H2O + N2O5 ⎯⎯→ 2 HNO3
48. Answer (1, 2, 3, 4)
KK (σ2s2) (σ*2s2) (π 2px2 = π 2py
2)
Four electrons are present in 2π molecular orbitals that’s why double bond contains both π bonds.
49. Answer (1, 3, 4)
(1)
Cl
P
Cl
BrBr
Cl
dipole moment 0μ ≠
(2)
Cl
P
Br
BrBr
Cl
dipole moment = 0μ
(3)O O
CH3 CH3
dipole moment is not zerodue to following structure
(4) NH
HO
Hdipole moment is not zero
50. Answer (1, 3, 4)
Order of acidic strength H3PO2 > H3PO3 > H3PO4, hybridisation of phosphorus in all acids are not sp3. InH3PO3, H3PO2 there is P—H bond present, therefore these are reducing in nature.
51. Answer (2, 4)
Under the critical condition gases does not follow ideal behaviour for Z is not equal to 1 at absolute zerotemperature the kinetic energy of gas molecules will be zero.
∴ It is zero is cyclic process internal energy of ideal gas depends only on temperature
∴ In isothermal process ΔE = 0.
63. Answer (1, 2, 4)
In isothermal process T = constant
P1V1 = P2V2 (Boyle’s law at constant T)
ΔU = 0
ΔH1 = ΔH2
64. Answer (2, 4)
Standard heat of formation of all elements in their standard states is zero
ΔHf(O) ≠ 0 and
ΔHf (diamond) ≠ 0 because these are not standard state.
65. Answer (1, 3, 4)
State function depends only on initial and final position therefore Enthalpy, Entropy, Gibbs free energy arestate function.
66. Answer (2, 3)
During the streching of rubber band the long flexible macromolecules get uncoiled the uncoiled arrangementhas more specific geometry and more order thus entropy decreases.
67. Answer (1, 3, 4)
Work done in reversible process is more than work done in irreversible process at equilibrium ΔG is zero.
Pressure favours the forward reaction. The temperature at which atmospheric pressure is equal to vapourpressure is called boiling point if pressure boiling point will be increased .
71. Answer (1, 3)
On increasing the ammonia the partial pressure of NH3 increases where as increasing the temperature favoursthe dissociation of NH4HS therefore more NH3 will be formed.
72. Answer (1, 2, 4)
NaCN Na+ + CN–
CN– + H2O HCN + OH– (basic)
CH3COONa CH3COO– + Na+
CH3COO– + H2O CH3COOH + OH– (basic)
Na2CO3 2Na+ + CO32–
CO32– + 2H2O H2CO3 + 2OH– (basic)
73. Answer (1, 2)
N2(g) + 3H2(g) 2NH3(g) ΔH = negative
Since the no. of moles of gases decreases in product sides therefore on increasing the pressure forwardreaction favours. Catalyst increases the rate of reaction, therefore NH3 formation will be fast.
74. Answer (1, 2, 3)
Electron deficient species are called lewis acid
therefore BF3, Ag+ are electron deficient
SnCl4 can expand its octet due to vacant d-orbital therefore behaves as a lewis acid.
75. Answer (2, 3)
The aqueous solution of NH4Cl and CuSO4 are acidic.
CH3COOH is weak acid its concentration of H+ ions is less than 10–6 M therefore pH > 6
CH3COOH + NaOH → CH3COONa + H2O
initial m. mol 2 6 0 0
after reaction 0 4 2 2
Since solution is basic therefore pH > 7.
The aqueous solution of CH3COONH4 is generally neutral
∴ pH > 6.
77. Answer (3, 4)
NO3– is a conjugate base of HNO3 which is strong acid
HSO4– is a conjugate base of H2SO4 which is strong acid
78. Answer (1, 2, 4)
Due to common ion effect solubility of AgCl will be less than water in NaCl, AgNO3 and CaCl2 solution.
79. Answer (1, 2)
Hln H+ + In–
Ka = ]HIn[
]In][H[ −+
[H+] = ]base[]acid[Ka
for 75% red [H+] = 55 103257510 −− ×=⎥⎦
⎤⎢⎣⎡
pH = 4.52
for 75% blue [H+] = 55 1031
752510 −− ×=⎟
⎠⎞
⎜⎝⎛ = 5.47
80. Answer (2, 3)
On increasing the temperature ionic product of water increases so pH and pOH decreases but water willremain neutral.
81. Answer (1, 2, 4)
⎥⎦
⎤⎢⎣
⎡−
Δ=
211
2T1
T1
R303.2H
KKlog
Since in the option (3)
ΔH = 0 because (Eaf = Eab)
∴ Therefore only this reaction is independent of temperature and (K2 = K1) on the other hand there is notΔH = 0 therefore equilibrium constant depends on temperature.
t = t1 the reaction attains the equilibrium therefore the amount of NH3 remins constant after t1 time
2x = 2
x = 1 mole
Moles of N2 = 2, moles of H2 = 6, moles of NH3 = 2
W(N2) + W(H2) + W(NH3) = 2 × 28 + 2 × 6 + 2 × 17 = 102 g at t = 2t1
Molar ratio of N2 and H2 same at two time i.e., 3t1 and
2t1 because initial molar ratio is 1 : 3.
⎟⎠⎞
⎜⎝⎛=
−−
=3
142)x39(
28)x3()H(W)N(W
2
2 remain same at 3tt 1= as well as
2tt 1= .
83. Answer (1, 2, 4)
Upto phenolphthalein NaOH is fully neutralised and Na2CO3 will be converted to NaHCO3. In next step NaHCO3coming from Na2CO3 neutralised by HCl using methyl orange indicator. So y ml should be less than x mlthat required for phenolphthalein end point.
84. Answer (2, 3)
Due to smaller size of Li+ is more solvated than Na+ ion therefore conductivity is less than Na+ ion.
85. Answer (1, 3)
Pb(s) (Pb2+)A(aq)
(Pb2+)c(aq) Pb(s)
for this cell E°cell = 0
Ecell = C
2A
2
]Pb[]Pb[log
2059.0
+
+
−
Ksp = [Pb2+][SO42–] = s2
s = )PbSO(K 4sp
Ksp = [Pb2+][I–]2 = 4s3
s = 31
2sp )PbI(4
K⎥⎦
⎤⎢⎣
⎡
Ecell = 21
4sp
31
2sp
]PbSO(K[
4)PbI(K
log2059.0
⎥⎦
⎤⎢⎣
⎡
86. Answer (2, 3)
Ecell = E°cell – ]Cu[]Cd[log
2059.0
2
2
+
+
on increasing the concentration of [Cu2+] and decreasing the concentration of [Cd2+] Ecell will be more positive.
Molar conductance of an electrolyte depends upon its degree of dissociation with increase in dilution the molarconductance increases due to increase in dissociation specific conductance decreases upon dilution becausenumber of current carrying ions per unit volume of solution decreases.
89. Answer (1, 2)
1
2n
5.2
6422n
0
21n
3
2
22 aIN2OSNaIOSNa2−
=
+
==
++⎯→⎯+
90. Answer (1, 3)
Ecell = 23
25
)10()10(log
2059.00
−
−
− = positive
0E2H/H
=° +
Cr ⎯→⎯ Cr3+(aq) + 3e
Cu2+(aq) + 2e ⎯→⎯ Cu
2Cr + 3Cu2+ ⎯→⎯ 2Cr3+ + 3Cu
Ecell = E°cell – 32
23
)Cu()Cr(log
6059.0
+
+
= E°cell – 3
2
)2.0()1.0(
6059.0
= E°cell + 2log36059.0
91. Answer (2, 3)
The value of the constant A for a given solvent and temperature depends on the type of electrolyte i.e., chargeon cation and anion produced on the dissociation of the electrolyte in the solution.
If a given metal ion has negative reduction potential H+ will be reduced by metal. Similarly if reduction potentialis positive metal will be reduced by H2. If metal ion with negative potential is coupled H-electrode the hydrogenhalf cell should function as cathode thus metal electrode will be negative half cell (anode). In aqueous solutioncontaining Zn2+, Na+ and Mg2+ the H+ will get preferentially reduced while in aqueous solution of Cu2+, Ag+,Au3+ these ions will be discharged ahead of H+.
94. Answer (1, 4)
As cell proceeds Ecell tend to zero to attain equilibrium state reaction quotient also increases to reach thestate of equilibrium.
95. Answer (1, 2, 3)
Mole of Fe3+ = 0.1 × 1 = 0.1
Mole of electron = mole149.096500
43600=
×
Fe3+ + e → Fe2+
0.100 mol electron required to reduce all the Fe3+ to Fe2+ 0.049 mol electron to reduce the Fe2+ to Fe
Hence, the formula of compound is A1/2 BC4 or AB2 C8. On placing body diagonal plane, 2 corner atoms,2 edge atoms, 1 body atoms are removed but 2 face atoms may or may not be removed.
Possibility I : Suppose, 2 face atoms are removed.
Number of A atoms = 41
812–
21
=×
Number of B atoms = 1 – 1 = 0
Number of C atoms = ⎟⎠⎞
⎜⎝⎛ +× 12
41–4 =
212
Hence, formula of compound is 25
41 CA or AC10
Possibility II : Suppose 2 face atoms are not removed.
Number of A atoms = 412
81–
21
=×
Number of B atoms = 1 – 0 = 1
Number of C atoms = 21212
41–4 =⎟
⎠⎞
⎜⎝⎛ +×
Hence, formula of compound is 212
41 BCA or AB4 C10
100. Answer (1, 3, 4)
In spinel structure (MgAl2O4), O2– ions occupy ccp lattice, two Al3+ occupy 50% octahedral voids and one Mg2+
occupies 12.5% tetrahedral voids.
101. Answer (1, 2, 4)
Frenkel defect is shown by those ionic solids in which the difference in the size of cation and anion is verylarge.
Ist Combination : Two corner Cl– ions are removed.
Number of Na+ ions = 4
Number of Cl– ions = 812–4 × = 3.75
Hence, formula per unit cell is Na4Cl3.75.
IInd Combination : Two face Cl– ions are removed
Number of Na+ ions = 4
Number of Cl– ions = 221–4 × = 3
Hence, formula per unit cell is Na4Cl3IIIrd Combination : One corner and one face Cl– ions are removed.
Number of Na+ ions = 4
Number of Cl– ions = 1211
81–4 ×+× = 3.375
Hence, formula per unit cell is Na4Cl3.375.
103. Answer (1, 3)
86.095.169.1
rr
==−
+ i.e. coordinate no. = 8
∴ bcc structure
∴ a3)rr(2 =+ −+
∴732.1
)95.169.1(2a +=
104. Answer (1, 4)
Body diagonal touches corner & body centre.
105. Answer (1, 4)
The given plane represents face plane in fcc.
106. Answer (3, 4)
Fluoride structure (CaF2) has cation constituting ccp whereas anions are present in all tetrahedral voids whereasfor antifluorite structure anion constitute lattice of cation are present at tetrahedral voids.
107. Answer (1, 3)
Octahedral voids form at the edge centre as well as the body centre at fcc.
Zn2+ is present in alternate tetrahedral voids therefore its C.N is 4. In rock salt structure there is 4Na+ and4Cl– ions present in a unit cell.
110. Answer (1, 3)
Doping of group 14 elements with group 15 elements produces excess of electrons and doping of group 14elements with group 13 elements produces holes in the crystals.
111. Answer (1, 2, 3)
Statement 4 is incorrect because K = A e–Ea/RT.
112. Answer (1, 3)
Statement 4 is incorrect because of B decreases then C increases hence there must be a –ve sign.
113. Answer (3, 4)
Maxwell and Ostwald theories are exclusively related to chemical kinetics.
114. Answer (2, 3)
A has units of rate constant of reaction not rate of reaction.
115. Answer (1, 2, 3, 4)
Rate of reaction depends on nature of reactants, temperature, nature of catalyst and surface area of reactants.
116. Answer (1, 2, 4)
303.2t·k
]R[]R[log 0 =
Comparing with y = mx + C
then we get (1) and (4) t½ does not depend on concentration for first order reaction.
117. Answer (1, 2, 4)
1n½ )a(1t −∝ (n = order of reaction) and the unit of frequency factor ‘A’ is equal to the unit of ‘k’’
118. Answer (1, 3, 4)
From the rate expression, overall order of reaction is two & first order w.r.t. [I–]
119. Answer (1, 2, 4)
The alkaline hydrolysis of ethyl acetate is second order while others are first order reaction.
120. Answer (1, 2)
A is called pre exponential factor.
121. Answer (1, 2, 4)
In α-decay, positron emission & k-electron capture, n/p ratio increases while in β-decay, n/p ratio decreases.
122. Answer (1, 2, 3)
In (1), (2) and (3) options, effective molarity are same.
123. Answer (1, 2, 3, 4)
In all options, effective molarity are same i.e. 2.2 M. So, boiling point of solutions is same.
124. Answer (2, 3, 4)
Effective molarity of Ba3(PO4)2 is more than MgSO4. So, the osmotic pressure of Ba3(PO4)2 is more. SPMallow the movement of solvent only not the solute. So, no ppt. of BaSO4 is formed in right side.
When CuSO4 is dissolved in NH4OH then association takes place instead of dissociation.
CuSO4 + 4NH4OH → [Cu(NH3)4]SO4
Hence, freezing point of solution is raised and boiling point of solution is lowered.
126. Answer (1, 3)
Acetone and chloroform shows negative deviation from Raoult’s law while ethanol and water shows positivedeviation from Raoult’s law.
127. Answer (3, 4)
Statement 1 and 2 are incorrect because sol particles neither move toward anode nor cathode.
128. Answer (2, 3)
An anion caused the precipitation of a positively charged sol and vice versa. The higher the valency of theeffective ion, greater is the penetrating power.
129. Answer (2, 3)
Size of suspension particles are > 10–5 cm in diameter.
Size of colloidal particles are 10–7 –10–5 cm in diameter.
Size of true particle < 10–7 cm in diameter.
130. Answer (1, 2)
When dispersion medium is gas, the colloidal system is called aerosol, smoke, dust are example of aerosolsof solids whereas fog, clouds are example of aerosol of liquids.
131. Answer (1, 3, 4)
Starch, gum and protein in water are examples of lyophilic sols. Sulphur in water is an example of lyophobicsol.
Sulphur in water is an example of ryophobic sol.
132. Answer (2, 4)
Chemisorption is specific in nature and it is shown by the gases which can react with adsorbent. Chemisorptionis unimolecular not multimolecular and favourable at high temperature not at low temperature.
133. Answer (2, 4)
Peptization is the preparation method of colloids, electrophoresis is the property of colloids. While ultrafilterationand electrodialysis are the purification method of colloids.
134. Answer (2, 3)
In homogeneous catalysis, the physical state of reactants and catalyst are same
N2(g) + 3H2(g) Fe(s)
2NH3 (Haber’s process)
2SO2(g) + O2(g) Pt(s)
2SO3 (Contact process).
135. Answer (1, 3)
Scattering of light can’t done by water and CaCl2 is more effective coagulant than NaCl because As2S3 isnegatively charged sol.
Section - C : Linked ComprehensionC1. 1. Answer (2)
Let the % of B-10 = Xthen % of B-11 = (100 – X)
Average atomic mass 2.10100
11)X100(X10=
−+=
10X + 1100 – 11X = 1020X = 80
% of B–10 = 802. Answer (3)
Average atomic mass = 5.354
371353=
×+×
3. Answer (2)Since X–, Y2– and Z3– are isoelectonic thereforeNumber of electrons in increasing order will be X > Y > ZX–, Y2– and Z3– all have same no. of neutrons.Therefore atomic no. increasing order will be
Z < Y < XC2. 1. Answer (2)
Let the moles of water = 1 moleMoles of urea will also be 1.
3. Answer (4)For the visible region transition belongs to Balmar series. For the Balmar series the electron should jumpsfrom higher level to 2nd energy level.
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 = 3s3. Answer (1)
Radial nodes = (n – l – 1)For 3s ; n = 3, l = 0Radial nodes = 3 – 0 – 1 = 2For 2p ; n = 2, l = 1Radial nodes = 2 – 1 – 1 = 0
C7. 1. Answer (1)For the first excited state n = 2
Energy = 2
2
nZ6.13 ×−
eV4
6.13216.13 2
−=×−=
= –3.4 eV2. Answer (4)
1n529.0r
2
H =
= 0.529 × 1 Å= 0.529 Å
42529.0r
2
Be3 ×=+
= 0.529 Å3. Answer (4)
1s electronic level allow the H-atom to absorb a photon but not to emit a photon. If it emit a photon itwill drop into nucleus, that is not possible.
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 thanHe.
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 energyis 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.
3. Answer (1)Let the volume of NH3 be x mLVolume of H2 = (50 – x) mL
223 H23N
21NH +⎯→⎯
Since 40 mL of O2 is added and sparked it must have reacted with H2 to form liquid water. Moreover since6 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)
( ) 68x23x50 =+−
x = 36% NH3 = 72
C18. 1. Answer (2)PV = K(constant)
2. Answer (4)PV = nRTn is not directly proportional to T∴ Its graph will not be straight line.
3. Answer (2)
M(average) = 20yx
y28x16=
++
16x + 28y = 20x + 20y 4x = 8y
Maverage = yxx28y16
++
24372
y3)y2(28y16
==+
=
C19. 1. Answer (4)At point A near low pressure region volume is very high thus
Because V is very large, so in Van der Waal’s equation ⎟⎠⎞
⎜⎝⎛ + 2V
aP (V – b) = RT, 2Va
and b are
neglected and equation becomes PV = RT. Coefficients depends on the identity of the gas but areindependent of the temperature. Real gas exert lower pressure than the same gas behaving ideallydue to intermolecular force of attraction.
There is no effect at equilibrium at constant volume.
2. Answer (4)
NaNO3(s) NaNO2(s) + 21 O2(g)
∴ since no. of gaseous moles decrese in backward direction thererfore on increasing the presure reversereaction 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 molesdecreases in forward direction therefore on increasing the pressure forward reaction favours.
Specific conductivity is directly proportional to the concentration.
2. Answer (3)
K = ⎟⎠⎞⎜
⎝⎛× aR
1 l
1.2 = ⎟⎠⎞⎜
⎝⎛× a55
1 l
1m66552.1a
−=×=⎟⎠⎞
⎜⎝⎛ 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 decreases8 times. Specific conductance is directly proportional to concentration. Hence, it decreases 8 times.
Elements with higher atomic number are more stable if they have slight excess of neutron as thisincrease 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 = 21
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.
Electrons are filled in the orbital from lower energy level to higher energy level 3d orbital has more energythan 4s. Therefore 4s orbital will be filled first.
15. Answer (4)
Fixed circular path around nucleus is not possible because it violates the Heisenberg’s uncertainty principle.
16. Answer (3)
Since the wavelength for the electron and proton is same but mass of electron is less than proton so itsvelocity will be more than proton.
It is clear from this expression mVh
=λ
17. Answer (4)
de Broglie wavelength mVh
=λ . It is applicable to moving particle.
18. Answer (2)
Zn(30) → 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2
Zn2+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d10
Since Zn2+ has no unpaired electrons therefore it is diamagnetic.
19. Answer (4)y
x
dx2 – y2
In 22 yxd − the electron density lies along the x and y axis.
20. Answer (1)
The value of l depends on the n.
For n = 3 l = 0, 1, 2 (because value of l ranging from 0 to (n – 1)
Value of m depends on l and ranging from –l to +l.
21. Answer (3)
Energy of entron in nth orbit for H-atom or H-like ions 2
2
nZ6.13−=
It is clear the energy of electrons depends on principal quantum number.
In other atom the energy depends on n as well as value of l.
H2O is liquid due to hydrogen bonding but no H-bonding present in H2S.
Oxygen is more electronegative than sulphur.
42. Answer (2)
Intermolecular force of attraction of inert is very low. Since inert gases have highly stable electronic configurationtherefore ionisation energy is quite high.
43. Answer (3)
PV = nRT
P ∝ T (when n and V are kept constant)
44. Answer (2)
Heat absorbed during isothermal expansion of an ideal gas against vaccum is zero. Volume of the gasmolecules is negligible compared to the volume occupied by gas.
45. Answer (4)
Gases molecules have different types of speed, (i.e. Vrms, Vmp, Vav) molecules of ideal gas neither attractnor repel to each other.
46. Answer (2)
Molar heat capacity at constant pressure is more than that of molar heat capacity at constant volume. The
average kinetic energy of ideal gas depends only on temperature (i.e. 23
RT)
47. Answer (1)
Due to great intermolecular force among the NH3 molecules the value at ‘a’ is more for NH3 than that of N2.
48. Answer (4)
Compresibility of non ideal gas is not equal to 1 i.e. (Z ≠ 1) it may be greater than 1 or less than 1. Dueto intermolecular force of attractions pressure of non ideal gas is lower than expected.
49. Answer (1)
Due to the free valancies the transition metals adsorb the gases.
50. Answer (3)
Due to stronger inter molecular force of attraction SO2 is easily liquified. Since critical temperature is directlyproportional to the value of ‘a’
∴ Critical temperature of SO2 will be more than H2.
51. Answer (3)
⎟⎠
⎞⎜⎝
⎛2V
a represents the inter molecular force of attraction.
It can not be neglected because in real gas the molecules are closer to each other.
52. Answer (4)
nRT)nbV(V
anp 2
2=−⎟
⎟⎠
⎞⎜⎜⎝
⎛+
Unit of V = unit of nb
Unit of b = litre mol–1
‘a’ represents inter molecular force of attraction
At constant pressure gases follows the Charle’s law
V ∝ T (at constant P)
54. Answer (1)
ΔE is a state function because it depends only on initial and final position only.
55. Answer (1)
In cyclic process the change in enthalpy (ΔH), change in entropy (ΔS) and change in free energy (ΔG) arezero because these are state function.
56. Answer (1)
Heat of neutralisation of HF with NaOH is more than 13.7 KCal because heat of hydration of F– ion is veryhigh due to smaller size of F ion.
57. Answer (2)
For spontaneous process ΔG must be negative.
ΔG = ΔH – TΔS
58. Answer (1)
Enthalpy of formation of H2O(l) is more than H2O(g) because some extra heat evolved when the water vapouris condensed.
59. Answer (4)
Pressure and temperature are intensive properties whereas volume is extensive property. Extensive propertydepends on the mass of substance.
60. Answer (2)
ΔG = ΔH – TΔS
If ΔS is negative then at high temperature TΔS will be more than ΔH therefore ΔG will be positive and reactionwill be non spontaneous at low temperature. ΔH will be more than TΔS therefore ΔG will be negative andreaction will be spontaneous.
61. Answer (3)
ΔG = ΔH – TΔS
Since ΔS = negative ∴ TΔS is positive.
If ΔG = (–) then ΔH must be negative.
62. Answer (2)
Solubility of HgI2 is more in KI solution due to complex formation
2KI + HgI2 → K2[HgI4]
due to bigger size I– ion is highly polarizable.
63. Answer (1)
Kc = [CO2] on increasing the volume the concentration of CO2 decreases. So to maintain the concentrationof CO2 equilibrium shifts in forward direction.
CH3COOH CH3COO– + H+ on addition of CH3COONa the concentration of CH3COO– increases andequilibrium shifts in backward direction so pH will increases because [H+] decreases.
65. Answer (1)
For neutral solution [H+] = [OH–] ∴ pH = pOH
pH + pOH = pKw
∴ pH = 2
pKw
66. Answer (1)
CuO + H2 Cu + H2O at 75ºC the water in form of liquid ∴ ΔH = –1
Kp = Kc(RT)–1 ∴ Kc > kp
But at 175º water in forms of gaseous ∴ ΔH = 0
Kp = Kc
67. Answer (4)
The aqueous solution of salt of strong base with weak acid is basic ∴ pH > 7
due to cationic hydrolysis the solution becomes acidic.
∴ pH < 7
68. Answer (3)
Solubility of salt increases on dilution the solubility product (Ksp) depends only on temperature.
69. Answer (3)
43POH −+ + 42POHH 1Ka
−42POH −+ + 2
4HPOH Ka2
Ka1 > Ka2 ∴ H3PO4 is stronger acid then −42POH
pKa1 < pKa2
70. Answer (1)
Catalyst speeds up the rate of forward reaction as well as backward reaction.
71. Answer (2)
AT the equilibrium rate of forward reaction is equal to the rate of backward reaction so concentration ofreactants as well as product does not change with time.
NH4OH −+ + OHNH4 on addition of NH4Cl the concentration of +4NH increases and equilibrium will be
shifted in backward direction therefore [OH–] decreases as a result pOH increases and pH will decreases.
75. Answer (1)
Blood is a basic buffer solution of )HCOCOH( 332−+ .
76. Answer (2)
+4NH
H — N — H
H
H
+form of charge on nitrogen is +1.
77. Answer (3)
CrO
O
O
O–1
O
–1
–1 –1–2
+1
+1 +1
+2
+1 oxidation number of chromium is +6.
78. Answer (1)
23
3n42 COFeOFeC +⎯→⎯ +
=
232
)5n(4
)3n(42 COFeMnMnOOFeC ++⎯→⎯+ ++
=
−
=
Equivalents of FeC2O4 = equivalent −4MnO
1 × 3 = 0.6 × 5
79. Answer (2)
OH2S3SH2SO 222 +⎯→⎯+
Bleaching action of SO2 is due to reduction.
80. Answer (2)
In electrochemical cell the chemical energy is converted to electrical energy for this process the cell reactionshould be spontaneous.
For spontaneous process ΔG must be negavite.
81. Answer (3)
Reduction potential of Na is very less than water therefore water reduced first and H2 gas evolved at cathode.Liberation or deposition depends on the reduction potential.
Inverse and normal spinel structure have same packing efficiency.
93. Answer (1)
Ions come closer in crystals suffering from frenkel defects.
94. Answer (2)
ccp has ABC ABC..... pattern of stacking.
95. Answer (1)
Distance between 2 nearest spheres in fcc = a22
Distance between 2 nearest spheres in bcc = a23
96. Answer (2)
Ferromagnetic substances are those paramagnetic substance which persists their magnetic moment (i.e. spinalignment) even in absence of magnetic field. They turn to paramagnetic substance when heated above curiestemperature.
97. Answer (3)
Common salt has F electrons, responsible for colour.
98. Answer (4)
10–3 mole of SrCl2 will develop 10–3 NA cationic vacancies.
99. Answer (2)
Each tetrahedral void is at ¼th distance from corner at body diagnol.
100. Answer (1)
Co-ordination number in bcc is 8.
101. Answer (1)
For LiCl, Cl– will constitute lattice.
102. Answer (1)
ZnS :- Molar ratio = 1 : 1
∴ Simple coordination number ratio = 1 : 1
Q Simple coordination number of Zn2+ = 4 (present in tetrahedral void)
∴ Simple coordination number of S2– = 4
103. Answer (4)
The value of Arrhenious constant is not affected by temperature.
104. Answer (1)
The end product of (4n + 2) series is Pb20682 .
105. Answer (1)
For second order, t1/2 is inversely proportional to concentration.
During β-decay, neutron is converted into proton and the atomic number increases by one.
108. Answer (3)
Alkaline hydrolysis of ester is known as saponification and it is second order reaction.
109. Answer (3)
Rate constant of any order reaction is directly proportional to temperature.
110. Answer (3)
The effective molarity of 1 M CuSO4, 0.5 M AlCl3 and 2 M urea solution are same and elevation in B.P. is acolligative property. Boiling point is not a colligative property.
111. Answer (1)
In 1 M aqueous solution, solvent is less than 1000 g while in 1 m aqueous solution, solvent is 1000 g.
Hence, 1 M is more concentrated than 1m.
112. Answer (4)
When mercuric iodide is added in KI solution then association takes place and freezing point is raised.
HgI2 + 2KI → K2HgI4 potassium mercuric iodide
In aqueous solution, HgI2 dissociates as
HgI2 → Hg2+ + 2I–
113. Answer (3)
Molality and mole fraction is independent of temperature and molality is the number of gm moles of solutedissolved in 1 kg of solvent.
114. Answer (1)
The effective molarity of KCl is twice the sugar because KCl gives two ions. So, the vapour pressure of KClis less than sugar.
115. Answer (3)
In presence of more volatile solute, vapour pressure of solution is developed by solute and solvent both.
116. Answer (4)
In benzene, acetic acid dimerises. So, the van’t Hoff factor is less than 1.
117. Answer (4)
Micelles is formed at above CMC and at above Kraft temperature.
118. Answer (1)
The extent of adsorption of CO2 is much more higher than H2 because critical temperature and van der Waal’sforce of attraction of CO2 is much higher than H2.
119. Answer (4)
Solution of starch in water is a lyophilic sol it is reversible sol.
120. Answer (2)
For coagulation, higher is the charge on oppositely charged ions, greater will its coagulating power.
So the reaction forwarded in forward direction by decreasing temperature, decreasing pressure andaddition of inert gas at constant pressure
(C) N2O4 (g) 2NO2(g) ΔH = 57.2 kJ
Δn = 1
Reaction is endothermic therefore by increasing temperature, decreasing the pressure and addition of inertgas at constant pressure equilibrium shift in forward direction
(D) N2 (g) + O2 (g) 2NO(g) ΔH = +180.5 kJ
Reaction is endothermic reaction shift in forward direction by increasing temperature. No effect of pressurebecause there is no change in no. of gaseous moles
(A) Since CaCO3 is solid therefore its concentration remains constant. No effect of addition of CaCO3.
Since reaction is endothermic therefore high temperature favours forward direction reaction.
Since Δng= 1 therefor low pressure favour the forward direction reaction and addition of inert gas atconstant pressure favours the forward direction because Δng > 0.
(B) On addition of amount of reactant favours the forward direction and low temperature favours the forwarddirection because reaction is exothermic.
Since (Δng = –2) therefore high pressure favours forward direction and addition of inert gas at constantpressure favours backward reaction.
(C) Addition of amount of reactants favours forward direction there is no effect of pressure and addition of inertgases of constant pressure because Δng = 0.
(D) Addition of reactant favours forward direction and Δng = 1 therefore addition of inert gas at constantpressure favours forward direction.
49. Answer - A(q, r, s), B(p, q, s), C(p, q), D(p, q)(A) In Wurtzite structure, S2 form hcp and Zn2+ are present in Half of tetrahedral voids.(B) In Zinc Blend structure, S–2 form cubical closed packed structure and Zn+2 occupy half of tetrahedral
voids.(C) In antiflourite strucrure, O–2 show c.c.p. like packing and Na+ is placed on all tetrahydral void.(D) In Rock salt structure Cl– occupy f.c.p. and Na+ occupy octahedral voids.
14. pπ – pπ back bonding in BF3 gives some double bond character which is absent in BF4–.
15. (i) O Cl
FO
FF
Trigonal bipyramidalwith hybridisation of Clsp d3
(ii) Xe
OF
Square pyramidalwith hybridisation of Xesp d3 2
F
F
F
(iii) I
O
Square pyramidalwith hybridisation of Isp d3 2
Cl
Cl
Cl
Cl
⎡⎢⎣
⎤⎥⎦
–
(iv) IClCl
⎡ ⎤⎢ ⎥⎣ ⎦
+
V-shapedwith hybridisation of I sp3
16. (i) NO2 is paramagnetic.(ii) NO2
+ is linear with bond angle 180° NO2– is bent with bond angle slightly less than 120°.
(iii) The NO2+ ion has the shortest and strongest bond. The NO2
– ion has the longest and the weakest bondsof the three.
17. The molecule BrF5 has bromine atom in centre surrounded by six electron pair, five of which are used to formbonds to fluorine atoms. Figure shows the structure of BrF5 molecule. It is better to imagine six electrons pairaround the bromine atom directed towards the corner of octahedron with five of these corners occupied byfluorine atoms. As we know that the non-bonded pair of electrons occupied more space than bonding electronpairs so geometry of BrF5 depart from that of a regular octahedron.
19. Effective molar mass of the gas mixture mol/g43.131
273082.06.0p
dRT=
××==
Let the mole fraction of methane in the gaseous mixture be x
x(16) + (1 – x)4 = 13.43
x = 0.78
Therefor, partial pressure of methane in the mixture = 0.78 × 1 = 0.78 atm
Partial pressure of helium in the mixture (1 – 0.78) × 1 = 0.22 atm.
20. Let the formula of nitrogen hydride be NxHy and the initial volume of nitrogen hydride by aml
NxHy ⇒ )g(H2y)g(N
2x
22 +
initial a 0 0
02
ax2
ay
Thus a22
ay2ax
=+
x + y = 4 …(i)
when O2 is added
)l(OH)g(O21)g(H 222 →+
Since the gaseous mixture was needed to pass through alkaline pyrogallol solution. The oxygen was left inexcess and hydrogen would have been consumed completely. After passing through pyrogallol solution volumeleft would be corresponding to nitrogen only
Total moles of gases in two vessel after the reaction = R600V5.6
R600V5.1
R600V5
=+
When the two vessel are kept in water bath at different temperatures diffusion will take place till the pressureof the gases in two vessel becomes equal. Let the moles of gases left in one of vessel maintained at 27°C
be x. Thus moles of gases in another vessel maintained at 52°C would be ⎟⎟⎠
⎞⎜⎜⎝
⎛− x
R600V5.6
.
V325Rx
R600V5.6
V300Rx ×
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
××
x = RV1065.5 3−×
P = atm7.1~695.1V300R
RV1065.5 3 =
××× − .
22. For one mole of a real gas
RT)bV(VaP 2 =−⎟
⎠⎞
⎜⎝⎛ +
If volume correction is ignored
RTvVaP 2 =⎟
⎠
⎞⎜⎝
⎛ +
at STP V = 22.4 L
∴ Pr < Pi (= 1 atm)
If pressure correction is ignored
P(V – b) = RT
at STP Pr = Pi = 1 atm
∴ Vr > Vi (22.4 L)
At low pressure volume is so high that b <<< V
∴ RTVVaP 2 =⎟
⎠⎞
⎜⎝⎛ +
RTVaPV =+
VaRTPV −=
RTVa1
RTPV
−=
∴ Z < 1.
∴ At high temperature motion of molecules is so fast that P >>> 2V
(iv) Repulsive forces dominate since the actual density is less than the density if it were ideal.
24. Work done for irreversible process
W = –P(V2 – V1) = –ΔnRT = 37320181
××⎟⎠⎞
⎜⎝⎛ −− = –41.44 cal/gm
Now q = 540 cal/gm
ΔE = Δq + W = 540 – 41.44 = 498.56 cal/gm
For isothermal process
ΔG = 2.303 nRT2
1
VVlog = 2.303 × 4 × 8.314 × 400 10
1log = –30635.4 J.
25. Since T is constant ΔU = 0
W = 2.303nRT1
2
PPlog = 2.303 × 3 × 8.314 × ⎟
⎠⎞
⎜⎝⎛
15log400 = 1.61 × 104 J
also q = –W = –1.61 × 104 J
of course work done on the gas W is positive for compression. The heat q is negative because heat mustflow from the gas to surrounding of constant temperature maintain the temperature of the gas at 400 K whenit is compressed.
26. The vapour pressure of mercury (in atm) is equal to Kp for the reaction
Hg(l) Hg(g); Kp = PHg
Note that Hg(l) is omitted from equilibrium constant expression because it is pure liquid. Because the standardstate for elemental mercury is the pure liquid ΔGf° = 0 for Hg(l) and ΔG° for the vapourisation reaction is simplyequal to ΔGf° for 1 mol of Hg(g)
log Kp = 58.5298314.8303.2
1085.31RT303.2
G 3−=
×××−
=°Δ−
Kp = antilog(–5.58) = 2.63 × 10–6
Since Kp is defined in units of atmosphere therefore the vapour pressure of mercury at 25°C is 2.63 × 10–6 atm.
36. The degree of dissociation of acetic acid is given by
013.07.390
2.5c ==λλ
=α∞
CH3COOH CH3COO– + H+
initially C 0 0
At equilibrium C(1 – α) Cα Cα
Dissociation constant for acetic acid
Ka = Cα2
= 0.1 × (0.013)2
= 1.69 × 10–5 M.
37. For calculating the minimum weight of NaOH which is supposed to be added to cathode compartment we arerequired to know the [H+] present in this compartment. The reaction occuring in the cell is
Zn(s) + 2H+ → Zn2+ + H2(g)
∴ Ecell = Zn/ZnH/H 2
2EE ++ °−° – 2
2H
]H[
]Zn[Plog
2059.0 2
+
+
0.701 = 2]H[1.01log
2059.0)76.0(
+
×−−−
[H+] = 0.0316 M
Since sufficient NaOH is to be added to cathode compartment to consume entire H+ the equivalent of HCl mustbe equal to the equivalent of NaOH. Let the weight of NaOH added be x gm
10316.0140x
×=×
x = 1.264 g.
After the addition of sufficient NaOH the solution becomes neutral and we have [H+] = 10–7 M
x may be taken as approx. 0.095 due to the less amount of Hg(CN)2 & K+ is also present due to completedissociation.
So, total molarity = −+
−+ ++ n2n)CN(HgnCNK
= 0.1892 + (0.1892 – 0.095 n) + 0.095
m = f
fkTΔ
0.1892 + (0.1892 – 0.095 n) + 0.095 = 86.153.0
0.4734 – 0.095 n = 0.2849
n ≈ 2
So, the complex becomes −24)CN(Hg
53. There is formation of an electrical double layer of opposite charges on the surface of colloidal particles.
[AgI] I– K+
[AgI] Ag+ NO3–
[As2S3] S2– 2H+
solid
++++
––––
Fixed layer Diffused layer
Potential difference across this electrical double layer is called zeta potential or electrokinetic potential, Z andis given by
D4Z πημ
=
Where η is called coefficient of viscosity, D is dielectric constant of the medium and μ is the velocity of thecolloidal particles when an electric field is applied.
54. Whenever a mixture of gases is allowed to come in contact with a particular adsorbent under the sameconditions, the more strongly adsorbable adsorbate is adsorbed to a greater extent irrespective of its amountpresent.
e.g. air besides moisture contains a number of gases such as N2, O2 etc., yet moisture is adsorbed morestrongly on silica than the other gases of the air.
55. Gold number is the weight of protective colloid in milligrams which prevents the coagulation of 10 cm3 of goldsol.