PHYSICAL CHEMISTY
1. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is JEE-2010
(A ) K
T
(B)
K
T
(C ) K
T
(D ) K
T
1. A K = Ae–Ea/RT Rate constant increases with increase in temperature exponentially.
2. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is JEE-2010 (A) Br2(g) (B) Cl2(g) (C) H2O(g) (D) CH4(g) 2. B The standard molar enthalpy of formation of elements in standard state is zero. For chlorine standard state is gas.
3. The bond energy (in kcal mol–1) of a C–C single bond is approximately JEE-2010 (A) 1 (B) 10 (C) 100 (D) 1000
3. C
4. Among the following, the intensive property is (properties are) JEE-2010 (A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity
4. A, B Molar conductivity and EMF are physical quantities independent of amount of matter in the system.
Paragraph for Questions 5 to 6 The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is : JEE-2010 M(s)|M+ (aq; 0.05 molar) || M+ (aq; 1 molar) | M(s) For the above electrolytic cell the magnitude of the cell potential |Ecell| = 70 mV.
5. For the above cell (A) Ecell < 0; ∆G > 0 (B) Ecell > 0; ∆G < 0 (C) Ecell < 0 ; ∆G° > 0 (D) Ecell > 0; ∆G° < 0
6. If the 0.05molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be (A) 35 mV (B) 70 mV (C) 140 mV (D) 700 mV
5. B E = E° – 10 2.303 RT log Q nF Here temp is not given
E = 0 – 10 2.303 RT .05 log nF 1 So E = 70 mV. ∆G = – nFE Here Ecell is +ve then ∆G is negative.
6. C When E° = 0
10 11
2 10 2
log QE E log Q =
() 2
2
70 log.05 E log .05 = E2 = 140 mV.
7. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is JEE-2010
7. 3
Average titre value =
25.2 25.25 25 3 + +
= 25.15
The minimum digits after decimal in data decides the extent of significant digit after decimal in “average value”. Hence answer is 25.1 with total no. of significant digits = 3.
8. The concentration of R in the reaction R → P was measured as a function of time and the following data is obtained : JEE-2010 [R] (molar) 1.0 0.75 0.40 0.10 t(min.) 0.0 0.05 0.12 0.18 The order of the reaction is 8. 0 t(min.) [R]0 – [R]t [R]0 – [R]t/∆t 0.05 0.25 5 0.12 0.60 5 0.18 0.9 5
since ,
[ ] [ ] 0 t R R t − ∆
ratio is constant. This indicates that the given reaction is of zero order.
For zero order reactions, [ ] [ ] () 0 t 1 K R R t = − ∆ 9. The number of neutrons emitted when 235 92 U undergoes controlled nuclear fission to 142 54 Xe and 90 38Sr is JEE-2010 9. 3 1422 35 90 1 549 2 38 0 U Xe Sr 3 n → + + No. of neutrons emitted = 3
10. Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5 is JEE-2010 10. 0
FF
FF
Br
F
Because LP–BP repulsions are stronger than BP–BP repulsions, there will be a distortion in the geometry & bond angles deviate from regular 90°.
11. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is JEE-2010
(A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic 11. A Electronic configuration of ‘B2’ is 2 2 2 2 2 0 1s 1s 2s 2s 2p 2py x * *σ σ σ σ π = π Number of unpaired electrons = 0 ⇒ diamagnetic
Bond order = bo abo n n 4 2
1
2 2 − − = = 12. The species having pyramidal shape is JEE-2010
(A) SO3 (B) BrF3 (C) 2 3S iO − (D) OSF2 12. D
S O F F Central atom ‘S’ has 3 σ bond pairs and 1lp of electrons. 13. The packing efficiency of the two–dimensional square unit cell shown below is JEE-2010
L (A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54% 13. D
L Number of atoms per unit cell
=
1 4 1 2 4 × + =
2 L = 4a. ⇒ a =
L 2 2
Packing fraction =
2
2
L
2 .
2 2 L ×π
= 0.7854 8 π = % efficiency = 78.54%.
14. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is wS and that along the dotted line path is wd, then the integer closest to the ratio wd/ws is JEE-2010
0.0 0.5
0.50 .0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
a
b
V (lit.)
P (atm.)
14 2 Wd = 4(2–0.5) + 1(3–2) + 0.5(5.5–3) = 6 + 1 + 1.25 = 8.25 L–atm
Ws = 2.303 nRT 2 1 V log V
nRT = PV = 4 x 0.5
= 2.303 x 2 x
5.5
log
0.5
= 4.79. L–atm.
d
s
W 8.25
1.72
W 4.79 = = . = 2.
15. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y x 10x. The value of x is JEE-2010
15. 7 ‘Ag’ crystallizes in ‘CCP’ manner and n = 4
3
0
n A
d
N .a × = 10.5 = 23 3 4 108 6 10 a × × × 8a 4 10 cm −≅ × Volume of the solid given = 10–12 x 104 x 4 x 10–8 = 4 x 10–16 cc.
Number of atoms per CC = 23 22 10.5 6 10 5.8 10 108 × × = × ∴ no of ‘Ag’ atom = 4 x 10–16 x 5.8 x 1022 = 2.74 x 107
Paragraph for questions 16 to 18
The hydrogen–like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. JEE-2010
16. The state S1 is (A) 1s (B) 2s (C) 2p (D) 3s
17. Energy of the state S1 in units of the hydrogen atom ground state energy is (A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50
18. The orbital angular momentum quantum number of the state S2 is (A) 0 (B) 1 (C) 2 (D) 3
16. B No. of radial nodes = 1 ⇒ – 1 = 1 n – ⇒ n – l = 2 Orbital n - l 1s 1 2s 2 2p 1 3s 3
17. C
Energy of state S1 =
( )1 2 2 E H Z n ×
=
( )1E H
9
4
×
= 2.25 E1(H)
18. B Energy of state S2 = E1(H)
⇒
( ) ()1 12 E H 9 E H n × = ⇒ n = 3 Now, no. of radial nodes = 1 ⇒ n – l – 1 = 1 ⇒ 2 – l = 1 ⇒ l = 1
19. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is JEE-2009 (A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05 19. B Sol : The abundance of 54Fe 5% = 56Fe 90% = 57Fe 5% = .
∴ The atomic mass of Fe is =
( ) ( ) ( )5 54 90 56 57 5 100 + + × 20. The term that corrects for the attractive forces present in a real gas in the van der Waals equation is JEE-2009
(A) nb (B)
2
2 an V
(C)
2
2 an V − (D) – nb
20. B 21. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 sol is JEE-2009 (A) Na2SO4 (B) CaCl2 (C) Al2(SO4)3 (D) NH4Cl 21. C Sol: Sb2S3 is an negatively charged sol, so among the given cations Al3+ has highest floculating power.
22. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is JEE-2009
(A) 4.0 x 10–4 (B) 4.0 x 10–5 (C) 5.0 x 10–4 (D) 4.0 x 10–6 22. A Sol : Henry’s law is H P K X =
2
2 2 2
N N H O N H P n n K = × Where
2Nn = no. of moles of nitrogen, N2 dissolved.
2 2N N airP X P=
2H On = no. of moles of solvent
= 0.8 × 5
2NP = partial pressure of gas dissolved.
2NP 4 = atm 2 N HK = Henry’s law constant for gas dissolved
2N 5 4 n 10 1 10 = × × 2 NX = mole fraction of Nitrogen, N2 in air
2
4 Nn 4 10− = × moles .
23. The correct statement(s) regarding defects in solids is(are) JEE-2009 (A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion (B) Frenkel defect is a dislocation defect (C) Trapping of an electron in the lattice leads to the formation of F–center (D) Schottky defects have no effect on the physical properties of solids 23. B, C
24. Match each of the diatomic molecules in Column I with its property/properties in Column II. JEE-2009 Column – I Column - II (A) B2 (p) Paramagnetic (B) N2 (q) Undergoes oxidation (C) 2O − (r) Undergoes reduction (D) O2 (s) Bond order ≥ 2 (t) Mixing of ‘s’ and ‘p’ orbitals
24. A→ p, q, r, t ; B → q, r, s, t; C → p, q, r ; D → p, q, r, s
25. For a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow
the equation log k = –(2000)
1
6.0.
T
+ The pre–exponential factor A and the activation energy Ea ,
respectively, are JEE-2009 (A) 1.0 × 106 s–1 and 9.2 kJ mol–1 (B) 6.0 s–1 and 16.6 kJ mol–1 (C) 1.0 × 106 s–1 and 16.6 kJ mol–1 (D) 1.0 × 106 s–1 and 38.3 kJ mol–1 25. D
Sol. log K = log A – a E 2.303RT aE 2000 2.303R = log A = 6
26. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is JEE-2009 (A) 0 (B) 2.84 (C) 4.90 (D) 5.92 26. A Sol. CO is strong field ligand
Cr(CO)6 : Cr is in zero oxidation state. All 6 electrons present in Cr will be paired. So no unpaired electrons
27. For the reduction of 3 NO− ion in an aqueous solution, E° is + 0.96 V. Values of E° for some metal ions are given below. JEE-2009 V2+ (aq) + 2e– → V E° = –1.19 V Fe3+ (aq) + 3e– → Fe E° = –0.04 V Au3+ (aq) + 3e– → AuE° = + 1.40 V Hg2+ (aq) + 2e– → HgE° = + 0.86 V
The pair(s) of metals that is(are) oxidized by 3 NO−in aqueous solution is (are) (A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V 27. A, B, D Sol. 2 0 V / V E 1.19V + = +
3
0 Fe / Fe E 0.04V + = +
3
0 Au / Au E 1.4V + = −
2
0 Hg/Hg E 0.86V + = − The cell constructed with Au will have negative E°. (+ 0.96 + (–1.4) = –0.44) with remaining three metals it will be positive.
28. Among the following, the state function(s) is (are) JEE-2009 (A) Internal energy (B) Irreversible expansion work (C) Reversible expansion work (D) Molar enthalpy 28. A, D
29. The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10–4. The pH of a 0.01 M solution of its sodium salt is JEE-2009 29. 8 Sol : C6H5COO– + H2O C6H5COOH + OH–
Initial conc . 0.01 0 0 Eq. conc. 0.01 (1 – ∝) 0.01 ∝ 0.01 ∝
For conjugate acid – base pair, Kw = Ka × Kb
⇒
14
10
b 4 a Kw 10 K 10 K 10 −
−
−= = =
()b 0.01 0.01
K
0.01 1 ∝× ∝
=
− ∝
⇒
101 0 0.01 − = ∝
⇒
10 81 0 10 0.01 − −∝ = = 6O H 0.01 10 − − = ∝ = POH = 6 PH = 8 30. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is JEE-2009 30. 4
Sol : 1 2
y
3RT 2RT 40 M =
⇒
1 2
y
3T 2T 40 M =
⇒
y
2 60
30
M ×
=
⇒ My = 4 Mol. Wt. of gas y = 4.
31. The total number of α and β particles emitted in the nuclear reaction 238 214 92 82 U Pb → is JEE-2009 31. 8
Sol :
214x ,y2 38 829 2 U Pb α β → 92 – 2x + y = 82 ⇒ 2x – y = 10 ..(1) 238 – 4x = 214 ..(2) ⇒ 4x = 24 ∴ x = 6 from equation (1) , y = 2 Total. No of α & β particles = 6 + 2 = 8
32. The coordination number of Al in the crystalline state of AlCl3 is . JEE-2009 32. 6 Sol : The crystalline solid has a layer lattice with six co–ordination Al. Coordination No. of Al = 6.
33. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K–1, the numerical value for the enthalpy of combustion of the gas in kJ mol–1 is JEE-2009 33. 9 Sol : q = Heat capacity × ∆T = 2.5 × 0.45 = 1.125 / g
for 1 mole,
28 q 1.125 9 3.5 = × = Heat of combustion = – 9 kJ mol–1
34. Under the same reaction conditions, initial concentration of 1.386 mol dm-3 of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio 1 0 k k of the rate constants for first order (k1) and zero order (k0) of the reactions is JEE-2008 (A) 0.5 mol-1 dm3 (B) 1.0 mol dm-3 (C) 1.5 mol dm-3 (D) 2.0 mol-1 dm3 34. (A)
1
1 2
0.693
K
t = 0
0 1 2
a
K
2t = , 0(1/ 2) 1 0 1 2 2tK 0.693 K t a = ×
or, 1
0
K 0.693 40 K 40 1.386 = × or , _1 3 1 0 K 0.693 0.5 mol dm K 1.386 = =
35. 2.5 mL of
2
M
5
weak monoacidic base (Kb = 1 x 10-12 at 25oC) is titrated with 2 M 15
HCl in water at 25oC. The concentration of H+ at equivalence point is (Kw = 1 x 10-14 at 25oC) JEE-2008 (A) 3.7 x 10-13 M (B) 3.2 x 10-7 M (C) 3.2 x 10-2 M (D) 2.7 x 10-2 M 35. (C)
M1 V1 = M2 V2
2 5
2
2.5
15 × = VHCl ⇒ VHCl = 7.5 ml
Meq of base = 1
[ Base] =
1
M
10 [H+] = w
b
K
C
K
×
[H+] =
_14
_12
10
0.1
10
×
⇒ [H+] = _ 2 3.2 10 M ×
36. A gas described by van der Waals equation JEE-2008 (A) behaves similar to an ideal gas in the limit of large molar volumes (B) behaves similar to an ideal gas in the limit of large large pressures (C) is characterized by van der Waals coefficients that are dependent on the identify of the gas but are independent of the temperature (D) has the pressure that is lower than the pressure exerted by the same gas behaving ideally
36 (A), (C) and (D)
• 2 m a P V +
(Vm – b) = RT
If Vm is large PVm = RT • ‘a’ and ‘b’ are characteristics of vanderwaal’s gas and independent of temperature. • Due to attractive forces among the gaseous molecules, the pressure exerted is lower than the pressure exerted by the ideal gas. 37. STATEMENT – 1: The plot of atomic number (y–axis) versus number of neutrons (x-axis) for stable nuclei shows a curvature towards x-axis from the line of 45o slope as the atomic number is increased. JEE-2008 and STATEMENT – 2: Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides. (A) Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1 (B) Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct Explanation for Statement–1 (C) Statement–1 is True, Statement–2 is False (D) Statement–1 is False, Statement–2 is True
37. (A)
10 200 30 40 50 60 70 80 90 100
20
40
60
80
100
120
Belt containing stable nuclei
NUMBEROFPROTONS(P)
NUMBER OF NEUTRONS (N)
N
1
P
=
38. STATEMENT – 1: For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. JEE-2008 and STATEMENT – 2: At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. (A) Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1 (B) Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct Explanation for Statement–1 (C) Statement–1 is True, Statement–2 is False (D) Statement–1 is False, Statement–2 is True 38. (D) ∆G = ∆G0 + RT lnQ At equilibrium , ∆G = 0 (∆G0 ≠ 0) ∆G0 = – RT lnK If ∆G is – ve, reaction will be spontaneous.
Paragraph for Question Nos. 39 to41 Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Applications of colligative properties are very useful in day–to–day life. One of its examples is the use of ethylene glycol and water mixture as anti–freezing liquid in the radiator of automobiles. JEE-2008 A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given : Freezing point depression constant of water (Kfwater) = 1.86 K kg mol–1 Freezing point depression constant of ethanol (Kfethanol) = 2.0 K kg mol–1 Boiling point elevation constant of water (Kbwater) = 0.52 K kg mol–1 Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol–1 Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol–1 Molecular weight of ethanol = 46 g mol–1
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non–volatile and non–dissociative.
39. The freezing point of the solution M is (A) 268.7 K (B) 268.5 K (C) 234.2 K (D) 150.9 K 39. (D)
Molality of the solution
0.1 1000 m 2.4 0.9 46 × = = × ∆Tf = Kf × m = 2 × 2.4 = 4.8
Tf = 155.7 – 4.8 = 150.9 K
40. The vapour pressure of the solution M is (A) 39.3 mm Hg (B) 36.0 mm Hg (C) 29.5 mm Hg (D) 28.8 mm Hg 40. (B) P = xA p0A = (0.9 × 40) = 36 mm of Hg
41. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is (A) 380.4 K (B) 376.2 K (C) 375.5 K (D) 354.7 K 41. (B) Molality of the solution in which mole fraction of H2O 0.9, is, 0.1 m 1000 6.17m 0.9 18 = × = ×
∆Tb = Kb × m = 0.52 × 6.17 = 3.2
∴ Tb = 373 + 3.2 = 376.2 K
42. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol–1) JEE-2008 (A) 9.65 x 104 (B) 19.3 x 104 sec (C) 28.95 x 104 sec (D) 38.6 x 104 sec 42. (B) 0.01 mole of H2 = 0.02 equivalents of H2 ≡ 0.02 F Q i t = ⇒ 10 x 10–3 = 0.02 96,500 t ×
⇒
_ 2 0.02 96,500
t
10 ×
=
⇒ t = 19.3 x 104 sec
43. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is JEE-2008 (A) CH3 (CH2)15 N+(CH3)3 Br– (B) CH3(CH2)11OSO3– Na+ (C) CH3(CH2)6 COO– Na+ (D) CH3(CH2)11 N+(CH3)3 Br– 43. (B) Detergents have more tendency to form micelles. Due to higher molar mass and greater polarity, CH3 (CH2)11 OSO3– Na+ will form micelles at lowest molar concentration.
44. Solubility product constants (Ksp) of salts of types MX, MX2 and M3X at temperature “T” are 4.0 x 10–8, 3.2 x 10–14 and 2.7 x 10–15 respectively. Solubilities (mol dm–3) of the salts at temperature “T” are in the order JEE-2008 (A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2 44. (D) _ MX M X + + Ksp = s2 = 4 x 10–8 ⇒ s = 2 x 10–4
_2
2M X M 2X + + Ksp = 4s3 = 3.2 x 10–14 ⇒ s = 2 x 10–5
_3
3M X 3M X + + Ksp = 27s4 = 2.7 x 10–15 ⇒ s= 1 x 10–4 so, MX > M3X > MX2 45. STATEMENT–1 : JEE-2008 There is a natural asymmetry between converting work to heat and converting heat to work. and STATEMENT–2 : No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. (A) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is a correct explanation for STATEMENT–1 (B) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is NOT a correct explanation for STATEMENT–1 (C) STATEMENT–1 is True, STATEMENT–2 is False (D) STATEMENT–1 is False, STATEMENT–2 is True 45. (A)
Paragraph for Question Nos. 46 to 48
In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space–filling model of this structure, called hexagonal close–packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ‘r’. JEE-2008
46. The number of atoms in this HCP unit cell is (A) 4 (B) 6 (C) 12 (D) 17
46. (B) Effective number of atoms per unit cell of HCP = 6
47. The volume of this HCP unit cell is
(A) 3 24 2r (B) 3 16 2r (C) 3 12 2r (D) 3 64 r 3 3 47. (A) The volume of HCP unit cell = 3 24 2 r
48. The empty space in this HCP unit cell is (A) 74% (B) 47.6% (C) 32% (D) 26%
48 (D) % packing fraction = 74% % of empty space = 26%.
49. Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 x 4 matrix given in the ORS. Column I Column II (A) Oribital angular momentum of the electron in a hydrogen-like atomic orbital JEE-2008 (p) Principal quantum number (B) A hydrogen-like one-electron wave function obeying Pauli principle (q) Azimuthal quantum number (C) Shape, size and orientation of hydrogen–like atomic orbitals (r) Magnetic quantum number (D) Probability density of electron at the nucleus in hydrogen-like atom (s) Electron spin quantum number 49. (A) – q (B) – p, q, r (C) – p, q, r (D) – p, q, r
50. When 20g of naphthoic acid (C11H8O2) is dissolved in 50g of benzene (Kf = 1.72 K kg mol–1), a freezing point depression of 2K is observed. The van’t Hoff factor (i) is JEE-2008 (A) 0.5 (B) 1 (C) 2 (D) 3
50. (A) ∆Tf = i Kf m
2 = i x 1.72 x
20 172
x
1000 50
∴ i = 0.5
51. The value of log10K for a reaction A B is JEE-2008 (Given: ∆rH o 298K = –54.07 kJ mol–1, ∆rS o 298K = 10 JK–1 mol–1 and R = 8.314 JK–1 mol–1; 2.303 x 8.314 x 298 = 5705) (A) 5 (B) 10 (C) 95 (D) 100 51. (B) ∆Go = ∆Ho – T∆So = –2.303 RT log10 K
or, –54.07 –
10
298 x
1000
= –
5705 1000
log10 K
∴ log10 K = 10 52. The species having bond order different from that in CO is JEE-2008 (A) NO– (B) NO+ (C) CN– (D) N2
52. (A) Except NO–, all other species have 14 electrons as there are in CO, so they have same bond order of 3. NO– has 16 e–s and has a bond order of 2
53. The percentage of p-character in the orbitals forming P–P bonds in P4 is JEE-2008 (A) 25 (B) 33 (C) 50 (D) 75
53. (D) Since the hybridisation of P in P4 is sp3, the % p-character is 75 P
P
P
P
54. STATMENT-1: Micelles are formed by surfactant molecules above the critical micellar concentration CMC. JEE-2008 because STATEMENT-2: The conductivity of a solution having surfactant molecules decreases sharply at the CMC. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement–1 (B) Statement-1 is True, Statement–2 is True; Statement-2 is NOT a correct explanation for statement–1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
55. (B) No. of ions per unit volume decreases due to aggregation of ions at CMC, hence conductivity decreases. Paragraph for Question Nos. 56 to 58
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 x 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. JEE-2008
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 Faraday = 96500 coulombs).
56. The total number of moles of chlorine gas evolved is (A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0 56. (B) At anode: 2Cl– →Cl2↑ + 2e– At cathode: 2H2O + 2e– → H2↑ + 2OH– The solution contains 2 moles of Cl– ions, which on electrolysis produce 1 mole of Cl2 gas.
57. If the cathode is a Hg electrode, the maximum weight(g) of amalgam formed from this solution is (A) 200 (B) 225 (C) 400 (D) 446 57. (D) If Hg is used as cathode then following reactions occur At anode: 2Cl– → Cl2↑ + 2e– At Cathode: Na+ + Hg + e– → Na-Hg (Sodium amalgam) Since 2 moles of Na+ ions are there in solution, 2 moles of sodium amalgam will be formed.
∴ weight of amalgam = 223 x 2 = 446 g
58. The total charge (coulombs) required for complete electrolysis is (A) 24125 (B) 48250 (C) 96500 (D) 193000
58. (D) At anode: 2Cl– → Cl2↑ + 2e– At Cathode: 2H2O + 2e– → H2↑ + 2OH– 2 moles of e–s are involved in electrolysis ∴ total charge required = 96500 x 2 = 193000 Coulombs.
59. Match gases under specified conditions listed in Column I with their properties / laws in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 x 4 matrix given in the ORS. JEE-2008
Column I Column II (A) hydrogen gas (P = 200 atm, T = 273 K) (p) compressibility factor ≠ 1 (B) hydrogen gas (P ≈ 0 atm, T = 273 K) (q) attractive forces are dominant (C) CO2 (P = 1 atm, T = 273 K) (r) PV = nRT (D) real gas with very large molar volume (s) P(V – nb) = nRT 59. A- p, s ; B- r; C- p, q ; D- p, s
(A) H2 is the lightest gas and has very weak attractive forces, hence ‘a’ is negligible. It always shows positive deviation. (B) H2 at very low pressure i.e. p ≈ 0, shows ideal behaviour i.e. PV = nRT (C) For CO2, the following graph is valid
z
p
At low pressure, z < 1 & at high pressure, z > 1. According to conditions given at very low temperature, attractive forces are dominant.
(D) 2 a P (V b) RT v + − = at large molar volume v – b ≠ v; but 2 a P v +
≈ P
hence, the above Vanderwaal’s equation becomes P(v – b) = RT & hence z ≠1
1. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is JEE-2010
(A ) K
T
(B)
K
T
(C ) K
T
(D ) K
T
1. A K = Ae–Ea/RT Rate constant increases with increase in temperature exponentially.
2. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is JEE-2010 (A) Br2(g) (B) Cl2(g) (C) H2O(g) (D) CH4(g) 2. B The standard molar enthalpy of formation of elements in standard state is zero. For chlorine standard state is gas.
3. The bond energy (in kcal mol–1) of a C–C single bond is approximately JEE-2010 (A) 1 (B) 10 (C) 100 (D) 1000
3. C
4. Among the following, the intensive property is (properties are) JEE-2010 (A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity
4. A, B Molar conductivity and EMF are physical quantities independent of amount of matter in the system.
Paragraph for Questions 5 to 6 The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is : JEE-2010 M(s)|M+ (aq; 0.05 molar) || M+ (aq; 1 molar) | M(s) For the above electrolytic cell the magnitude of the cell potential |Ecell| = 70 mV.
5. For the above cell (A) Ecell < 0; ∆G > 0 (B) Ecell > 0; ∆G < 0 (C) Ecell < 0 ; ∆G° > 0 (D) Ecell > 0; ∆G° < 0
6. If the 0.05molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be (A) 35 mV (B) 70 mV (C) 140 mV (D) 700 mV
5. B E = E° – 10 2.303 RT log Q nF Here temp is not given
E = 0 – 10 2.303 RT .05 log nF 1 So E = 70 mV. ∆G = – nFE Here Ecell is +ve then ∆G is negative.
6. C When E° = 0
10 11
2 10 2
log QE E log Q =
() 2
2
70 log.05 E log .05 = E2 = 140 mV.
7. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is JEE-2010
7. 3
Average titre value =
25.2 25.25 25 3 + +
= 25.15
The minimum digits after decimal in data decides the extent of significant digit after decimal in “average value”. Hence answer is 25.1 with total no. of significant digits = 3.
8. The concentration of R in the reaction R → P was measured as a function of time and the following data is obtained : JEE-2010 [R] (molar) 1.0 0.75 0.40 0.10 t(min.) 0.0 0.05 0.12 0.18 The order of the reaction is 8. 0 t(min.) [R]0 – [R]t [R]0 – [R]t/∆t 0.05 0.25 5 0.12 0.60 5 0.18 0.9 5
since ,
[ ] [ ] 0 t R R t − ∆
ratio is constant. This indicates that the given reaction is of zero order.
For zero order reactions, [ ] [ ] () 0 t 1 K R R t = − ∆ 9. The number of neutrons emitted when 235 92 U undergoes controlled nuclear fission to 142 54 Xe and 90 38Sr is JEE-2010 9. 3 1422 35 90 1 549 2 38 0 U Xe Sr 3 n → + + No. of neutrons emitted = 3
10. Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5 is JEE-2010 10. 0
FF
FF
Br
F
Because LP–BP repulsions are stronger than BP–BP repulsions, there will be a distortion in the geometry & bond angles deviate from regular 90°.
11. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is JEE-2010
(A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic 11. A Electronic configuration of ‘B2’ is 2 2 2 2 2 0 1s 1s 2s 2s 2p 2py x * *σ σ σ σ π = π Number of unpaired electrons = 0 ⇒ diamagnetic
Bond order = bo abo n n 4 2
1
2 2 − − = = 12. The species having pyramidal shape is JEE-2010
(A) SO3 (B) BrF3 (C) 2 3S iO − (D) OSF2 12. D
S O F F Central atom ‘S’ has 3 σ bond pairs and 1lp of electrons. 13. The packing efficiency of the two–dimensional square unit cell shown below is JEE-2010
L (A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54% 13. D
L Number of atoms per unit cell
=
1 4 1 2 4 × + =
2 L = 4a. ⇒ a =
L 2 2
Packing fraction =
2
2
L
2 .
2 2 L ×π
= 0.7854 8 π = % efficiency = 78.54%.
14. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is wS and that along the dotted line path is wd, then the integer closest to the ratio wd/ws is JEE-2010
0.0 0.5
0.50 .0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
a
b
V (lit.)
P (atm.)
14 2 Wd = 4(2–0.5) + 1(3–2) + 0.5(5.5–3) = 6 + 1 + 1.25 = 8.25 L–atm
Ws = 2.303 nRT 2 1 V log V
nRT = PV = 4 x 0.5
= 2.303 x 2 x
5.5
log
0.5
= 4.79. L–atm.
d
s
W 8.25
1.72
W 4.79 = = . = 2.
15. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y x 10x. The value of x is JEE-2010
15. 7 ‘Ag’ crystallizes in ‘CCP’ manner and n = 4
3
0
n A
d
N .a × = 10.5 = 23 3 4 108 6 10 a × × × 8a 4 10 cm −≅ × Volume of the solid given = 10–12 x 104 x 4 x 10–8 = 4 x 10–16 cc.
Number of atoms per CC = 23 22 10.5 6 10 5.8 10 108 × × = × ∴ no of ‘Ag’ atom = 4 x 10–16 x 5.8 x 1022 = 2.74 x 107
Paragraph for questions 16 to 18
The hydrogen–like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. JEE-2010
16. The state S1 is (A) 1s (B) 2s (C) 2p (D) 3s
17. Energy of the state S1 in units of the hydrogen atom ground state energy is (A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50
18. The orbital angular momentum quantum number of the state S2 is (A) 0 (B) 1 (C) 2 (D) 3
16. B No. of radial nodes = 1 ⇒ – 1 = 1 n – ⇒ n – l = 2 Orbital n - l 1s 1 2s 2 2p 1 3s 3
17. C
Energy of state S1 =
( )1 2 2 E H Z n ×
=
( )1E H
9
4
×
= 2.25 E1(H)
18. B Energy of state S2 = E1(H)
⇒
( ) ()1 12 E H 9 E H n × = ⇒ n = 3 Now, no. of radial nodes = 1 ⇒ n – l – 1 = 1 ⇒ 2 – l = 1 ⇒ l = 1
19. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is JEE-2009 (A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05 19. B Sol : The abundance of 54Fe 5% = 56Fe 90% = 57Fe 5% = .
∴ The atomic mass of Fe is =
( ) ( ) ( )5 54 90 56 57 5 100 + + × 20. The term that corrects for the attractive forces present in a real gas in the van der Waals equation is JEE-2009
(A) nb (B)
2
2 an V
(C)
2
2 an V − (D) – nb
20. B 21. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 sol is JEE-2009 (A) Na2SO4 (B) CaCl2 (C) Al2(SO4)3 (D) NH4Cl 21. C Sol: Sb2S3 is an negatively charged sol, so among the given cations Al3+ has highest floculating power.
22. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is JEE-2009
(A) 4.0 x 10–4 (B) 4.0 x 10–5 (C) 5.0 x 10–4 (D) 4.0 x 10–6 22. A Sol : Henry’s law is H P K X =
2
2 2 2
N N H O N H P n n K = × Where
2Nn = no. of moles of nitrogen, N2 dissolved.
2 2N N airP X P=
2H On = no. of moles of solvent
= 0.8 × 5
2NP = partial pressure of gas dissolved.
2NP 4 = atm 2 N HK = Henry’s law constant for gas dissolved
2N 5 4 n 10 1 10 = × × 2 NX = mole fraction of Nitrogen, N2 in air
2
4 Nn 4 10− = × moles .
23. The correct statement(s) regarding defects in solids is(are) JEE-2009 (A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion (B) Frenkel defect is a dislocation defect (C) Trapping of an electron in the lattice leads to the formation of F–center (D) Schottky defects have no effect on the physical properties of solids 23. B, C
24. Match each of the diatomic molecules in Column I with its property/properties in Column II. JEE-2009 Column – I Column - II (A) B2 (p) Paramagnetic (B) N2 (q) Undergoes oxidation (C) 2O − (r) Undergoes reduction (D) O2 (s) Bond order ≥ 2 (t) Mixing of ‘s’ and ‘p’ orbitals
24. A→ p, q, r, t ; B → q, r, s, t; C → p, q, r ; D → p, q, r, s
25. For a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow
the equation log k = –(2000)
1
6.0.
T
+ The pre–exponential factor A and the activation energy Ea ,
respectively, are JEE-2009 (A) 1.0 × 106 s–1 and 9.2 kJ mol–1 (B) 6.0 s–1 and 16.6 kJ mol–1 (C) 1.0 × 106 s–1 and 16.6 kJ mol–1 (D) 1.0 × 106 s–1 and 38.3 kJ mol–1 25. D
Sol. log K = log A – a E 2.303RT aE 2000 2.303R = log A = 6
26. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is JEE-2009 (A) 0 (B) 2.84 (C) 4.90 (D) 5.92 26. A Sol. CO is strong field ligand
Cr(CO)6 : Cr is in zero oxidation state. All 6 electrons present in Cr will be paired. So no unpaired electrons
27. For the reduction of 3 NO− ion in an aqueous solution, E° is + 0.96 V. Values of E° for some metal ions are given below. JEE-2009 V2+ (aq) + 2e– → V E° = –1.19 V Fe3+ (aq) + 3e– → Fe E° = –0.04 V Au3+ (aq) + 3e– → AuE° = + 1.40 V Hg2+ (aq) + 2e– → HgE° = + 0.86 V
The pair(s) of metals that is(are) oxidized by 3 NO−in aqueous solution is (are) (A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V 27. A, B, D Sol. 2 0 V / V E 1.19V + = +
3
0 Fe / Fe E 0.04V + = +
3
0 Au / Au E 1.4V + = −
2
0 Hg/Hg E 0.86V + = − The cell constructed with Au will have negative E°. (+ 0.96 + (–1.4) = –0.44) with remaining three metals it will be positive.
28. Among the following, the state function(s) is (are) JEE-2009 (A) Internal energy (B) Irreversible expansion work (C) Reversible expansion work (D) Molar enthalpy 28. A, D
29. The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10–4. The pH of a 0.01 M solution of its sodium salt is JEE-2009 29. 8 Sol : C6H5COO– + H2O C6H5COOH + OH–
Initial conc . 0.01 0 0 Eq. conc. 0.01 (1 – ∝) 0.01 ∝ 0.01 ∝
For conjugate acid – base pair, Kw = Ka × Kb
⇒
14
10
b 4 a Kw 10 K 10 K 10 −
−
−= = =
()b 0.01 0.01
K
0.01 1 ∝× ∝
=
− ∝
⇒
101 0 0.01 − = ∝
⇒
10 81 0 10 0.01 − −∝ = = 6O H 0.01 10 − − = ∝ = POH = 6 PH = 8 30. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is JEE-2009 30. 4
Sol : 1 2
y
3RT 2RT 40 M =
⇒
1 2
y
3T 2T 40 M =
⇒
y
2 60
30
M ×
=
⇒ My = 4 Mol. Wt. of gas y = 4.
31. The total number of α and β particles emitted in the nuclear reaction 238 214 92 82 U Pb → is JEE-2009 31. 8
Sol :
214x ,y2 38 829 2 U Pb α β → 92 – 2x + y = 82 ⇒ 2x – y = 10 ..(1) 238 – 4x = 214 ..(2) ⇒ 4x = 24 ∴ x = 6 from equation (1) , y = 2 Total. No of α & β particles = 6 + 2 = 8
32. The coordination number of Al in the crystalline state of AlCl3 is . JEE-2009 32. 6 Sol : The crystalline solid has a layer lattice with six co–ordination Al. Coordination No. of Al = 6.
33. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K–1, the numerical value for the enthalpy of combustion of the gas in kJ mol–1 is JEE-2009 33. 9 Sol : q = Heat capacity × ∆T = 2.5 × 0.45 = 1.125 / g
for 1 mole,
28 q 1.125 9 3.5 = × = Heat of combustion = – 9 kJ mol–1
34. Under the same reaction conditions, initial concentration of 1.386 mol dm-3 of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio 1 0 k k of the rate constants for first order (k1) and zero order (k0) of the reactions is JEE-2008 (A) 0.5 mol-1 dm3 (B) 1.0 mol dm-3 (C) 1.5 mol dm-3 (D) 2.0 mol-1 dm3 34. (A)
1
1 2
0.693
K
t = 0
0 1 2
a
K
2t = , 0(1/ 2) 1 0 1 2 2tK 0.693 K t a = ×
or, 1
0
K 0.693 40 K 40 1.386 = × or , _1 3 1 0 K 0.693 0.5 mol dm K 1.386 = =
35. 2.5 mL of
2
M
5
weak monoacidic base (Kb = 1 x 10-12 at 25oC) is titrated with 2 M 15
HCl in water at 25oC. The concentration of H+ at equivalence point is (Kw = 1 x 10-14 at 25oC) JEE-2008 (A) 3.7 x 10-13 M (B) 3.2 x 10-7 M (C) 3.2 x 10-2 M (D) 2.7 x 10-2 M 35. (C)
M1 V1 = M2 V2
2 5
2
2.5
15 × = VHCl ⇒ VHCl = 7.5 ml
Meq of base = 1
[ Base] =
1
M
10 [H+] = w
b
K
C
K
×
[H+] =
_14
_12
10
0.1
10
×
⇒ [H+] = _ 2 3.2 10 M ×
36. A gas described by van der Waals equation JEE-2008 (A) behaves similar to an ideal gas in the limit of large molar volumes (B) behaves similar to an ideal gas in the limit of large large pressures (C) is characterized by van der Waals coefficients that are dependent on the identify of the gas but are independent of the temperature (D) has the pressure that is lower than the pressure exerted by the same gas behaving ideally
36 (A), (C) and (D)
• 2 m a P V +
(Vm – b) = RT
If Vm is large PVm = RT • ‘a’ and ‘b’ are characteristics of vanderwaal’s gas and independent of temperature. • Due to attractive forces among the gaseous molecules, the pressure exerted is lower than the pressure exerted by the ideal gas. 37. STATEMENT – 1: The plot of atomic number (y–axis) versus number of neutrons (x-axis) for stable nuclei shows a curvature towards x-axis from the line of 45o slope as the atomic number is increased. JEE-2008 and STATEMENT – 2: Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides. (A) Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1 (B) Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct Explanation for Statement–1 (C) Statement–1 is True, Statement–2 is False (D) Statement–1 is False, Statement–2 is True
37. (A)
10 200 30 40 50 60 70 80 90 100
20
40
60
80
100
120
Belt containing stable nuclei
NUMBEROFPROTONS(P)
NUMBER OF NEUTRONS (N)
N
1
P
=
38. STATEMENT – 1: For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. JEE-2008 and STATEMENT – 2: At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. (A) Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1 (B) Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct Explanation for Statement–1 (C) Statement–1 is True, Statement–2 is False (D) Statement–1 is False, Statement–2 is True 38. (D) ∆G = ∆G0 + RT lnQ At equilibrium , ∆G = 0 (∆G0 ≠ 0) ∆G0 = – RT lnK If ∆G is – ve, reaction will be spontaneous.
Paragraph for Question Nos. 39 to41 Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Applications of colligative properties are very useful in day–to–day life. One of its examples is the use of ethylene glycol and water mixture as anti–freezing liquid in the radiator of automobiles. JEE-2008 A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given : Freezing point depression constant of water (Kfwater) = 1.86 K kg mol–1 Freezing point depression constant of ethanol (Kfethanol) = 2.0 K kg mol–1 Boiling point elevation constant of water (Kbwater) = 0.52 K kg mol–1 Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol–1 Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol–1 Molecular weight of ethanol = 46 g mol–1
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non–volatile and non–dissociative.
39. The freezing point of the solution M is (A) 268.7 K (B) 268.5 K (C) 234.2 K (D) 150.9 K 39. (D)
Molality of the solution
0.1 1000 m 2.4 0.9 46 × = = × ∆Tf = Kf × m = 2 × 2.4 = 4.8
Tf = 155.7 – 4.8 = 150.9 K
40. The vapour pressure of the solution M is (A) 39.3 mm Hg (B) 36.0 mm Hg (C) 29.5 mm Hg (D) 28.8 mm Hg 40. (B) P = xA p0A = (0.9 × 40) = 36 mm of Hg
41. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is (A) 380.4 K (B) 376.2 K (C) 375.5 K (D) 354.7 K 41. (B) Molality of the solution in which mole fraction of H2O 0.9, is, 0.1 m 1000 6.17m 0.9 18 = × = ×
∆Tb = Kb × m = 0.52 × 6.17 = 3.2
∴ Tb = 373 + 3.2 = 376.2 K
42. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol–1) JEE-2008 (A) 9.65 x 104 (B) 19.3 x 104 sec (C) 28.95 x 104 sec (D) 38.6 x 104 sec 42. (B) 0.01 mole of H2 = 0.02 equivalents of H2 ≡ 0.02 F Q i t = ⇒ 10 x 10–3 = 0.02 96,500 t ×
⇒
_ 2 0.02 96,500
t
10 ×
=
⇒ t = 19.3 x 104 sec
43. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is JEE-2008 (A) CH3 (CH2)15 N+(CH3)3 Br– (B) CH3(CH2)11OSO3– Na+ (C) CH3(CH2)6 COO– Na+ (D) CH3(CH2)11 N+(CH3)3 Br– 43. (B) Detergents have more tendency to form micelles. Due to higher molar mass and greater polarity, CH3 (CH2)11 OSO3– Na+ will form micelles at lowest molar concentration.
44. Solubility product constants (Ksp) of salts of types MX, MX2 and M3X at temperature “T” are 4.0 x 10–8, 3.2 x 10–14 and 2.7 x 10–15 respectively. Solubilities (mol dm–3) of the salts at temperature “T” are in the order JEE-2008 (A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2 44. (D) _ MX M X + + Ksp = s2 = 4 x 10–8 ⇒ s = 2 x 10–4
_2
2M X M 2X + + Ksp = 4s3 = 3.2 x 10–14 ⇒ s = 2 x 10–5
_3
3M X 3M X + + Ksp = 27s4 = 2.7 x 10–15 ⇒ s= 1 x 10–4 so, MX > M3X > MX2 45. STATEMENT–1 : JEE-2008 There is a natural asymmetry between converting work to heat and converting heat to work. and STATEMENT–2 : No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. (A) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is a correct explanation for STATEMENT–1 (B) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is NOT a correct explanation for STATEMENT–1 (C) STATEMENT–1 is True, STATEMENT–2 is False (D) STATEMENT–1 is False, STATEMENT–2 is True 45. (A)
Paragraph for Question Nos. 46 to 48
In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space–filling model of this structure, called hexagonal close–packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ‘r’. JEE-2008
46. The number of atoms in this HCP unit cell is (A) 4 (B) 6 (C) 12 (D) 17
46. (B) Effective number of atoms per unit cell of HCP = 6
47. The volume of this HCP unit cell is
(A) 3 24 2r (B) 3 16 2r (C) 3 12 2r (D) 3 64 r 3 3 47. (A) The volume of HCP unit cell = 3 24 2 r
48. The empty space in this HCP unit cell is (A) 74% (B) 47.6% (C) 32% (D) 26%
48 (D) % packing fraction = 74% % of empty space = 26%.
49. Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 x 4 matrix given in the ORS. Column I Column II (A) Oribital angular momentum of the electron in a hydrogen-like atomic orbital JEE-2008 (p) Principal quantum number (B) A hydrogen-like one-electron wave function obeying Pauli principle (q) Azimuthal quantum number (C) Shape, size and orientation of hydrogen–like atomic orbitals (r) Magnetic quantum number (D) Probability density of electron at the nucleus in hydrogen-like atom (s) Electron spin quantum number 49. (A) – q (B) – p, q, r (C) – p, q, r (D) – p, q, r
50. When 20g of naphthoic acid (C11H8O2) is dissolved in 50g of benzene (Kf = 1.72 K kg mol–1), a freezing point depression of 2K is observed. The van’t Hoff factor (i) is JEE-2008 (A) 0.5 (B) 1 (C) 2 (D) 3
50. (A) ∆Tf = i Kf m
2 = i x 1.72 x
20 172
x
1000 50
∴ i = 0.5
51. The value of log10K for a reaction A B is JEE-2008 (Given: ∆rH o 298K = –54.07 kJ mol–1, ∆rS o 298K = 10 JK–1 mol–1 and R = 8.314 JK–1 mol–1; 2.303 x 8.314 x 298 = 5705) (A) 5 (B) 10 (C) 95 (D) 100 51. (B) ∆Go = ∆Ho – T∆So = –2.303 RT log10 K
or, –54.07 –
10
298 x
1000
= –
5705 1000
log10 K
∴ log10 K = 10 52. The species having bond order different from that in CO is JEE-2008 (A) NO– (B) NO+ (C) CN– (D) N2
52. (A) Except NO–, all other species have 14 electrons as there are in CO, so they have same bond order of 3. NO– has 16 e–s and has a bond order of 2
53. The percentage of p-character in the orbitals forming P–P bonds in P4 is JEE-2008 (A) 25 (B) 33 (C) 50 (D) 75
53. (D) Since the hybridisation of P in P4 is sp3, the % p-character is 75 P
P
P
P
54. STATMENT-1: Micelles are formed by surfactant molecules above the critical micellar concentration CMC. JEE-2008 because STATEMENT-2: The conductivity of a solution having surfactant molecules decreases sharply at the CMC. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement–1 (B) Statement-1 is True, Statement–2 is True; Statement-2 is NOT a correct explanation for statement–1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True
55. (B) No. of ions per unit volume decreases due to aggregation of ions at CMC, hence conductivity decreases. Paragraph for Question Nos. 56 to 58
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 x 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. JEE-2008
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 Faraday = 96500 coulombs).
56. The total number of moles of chlorine gas evolved is (A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0 56. (B) At anode: 2Cl– →Cl2↑ + 2e– At cathode: 2H2O + 2e– → H2↑ + 2OH– The solution contains 2 moles of Cl– ions, which on electrolysis produce 1 mole of Cl2 gas.
57. If the cathode is a Hg electrode, the maximum weight(g) of amalgam formed from this solution is (A) 200 (B) 225 (C) 400 (D) 446 57. (D) If Hg is used as cathode then following reactions occur At anode: 2Cl– → Cl2↑ + 2e– At Cathode: Na+ + Hg + e– → Na-Hg (Sodium amalgam) Since 2 moles of Na+ ions are there in solution, 2 moles of sodium amalgam will be formed.
∴ weight of amalgam = 223 x 2 = 446 g
58. The total charge (coulombs) required for complete electrolysis is (A) 24125 (B) 48250 (C) 96500 (D) 193000
58. (D) At anode: 2Cl– → Cl2↑ + 2e– At Cathode: 2H2O + 2e– → H2↑ + 2OH– 2 moles of e–s are involved in electrolysis ∴ total charge required = 96500 x 2 = 193000 Coulombs.
59. Match gases under specified conditions listed in Column I with their properties / laws in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 x 4 matrix given in the ORS. JEE-2008
Column I Column II (A) hydrogen gas (P = 200 atm, T = 273 K) (p) compressibility factor ≠ 1 (B) hydrogen gas (P ≈ 0 atm, T = 273 K) (q) attractive forces are dominant (C) CO2 (P = 1 atm, T = 273 K) (r) PV = nRT (D) real gas with very large molar volume (s) P(V – nb) = nRT 59. A- p, s ; B- r; C- p, q ; D- p, s
(A) H2 is the lightest gas and has very weak attractive forces, hence ‘a’ is negligible. It always shows positive deviation. (B) H2 at very low pressure i.e. p ≈ 0, shows ideal behaviour i.e. PV = nRT (C) For CO2, the following graph is valid
z
p
At low pressure, z < 1 & at high pressure, z > 1. According to conditions given at very low temperature, attractive forces are dominant.
(D) 2 a P (V b) RT v + − = at large molar volume v – b ≠ v; but 2 a P v +
≈ P
hence, the above Vanderwaal’s equation becomes P(v – b) = RT & hence z ≠1
No comments:
Post a Comment