2.4 CELDAS DE TRABAJO
3.1.2 SENSORES DE SEGURIDAD Y SUPERVISION DE SEGURIDAD
(a) salbutamol (b) salvarsan (c) diazepam (d) LSD.
ANSWERS
1. (b) : 1 mol of H20 = 18 g = 6.02 x 1023 molecules 0-06 , „„ ,n?3
Hence 0.06 e H , 0 = * 6.02x10" molecules
s 2 18
= 2 x io21 molecules.
2. (b) : Average kinetic energy depends only on temperature and does not depend upon the nature of the gas.
3. (a) : For rhombohedral system, a = b = c, a = 3 = v * 90°.
Rhombohedral system is an alternative representation of the trigonal or hexagonal crystal systems.
4. (b): 1 st line is for 3 —> 2 and 2nd line is for 4 —> 2.
5. (b) : CI" ion is smaller than B r and K+ is larger than Na4. Hence ratio of CP : K+ is smallest.
6. (b) : u = 2xl01 4 cm~\ c = v*. = v / o or, v = cu = (3 x 10IO)(2 x 1014) = 6 x 1024 s"1.
7. (b) : The difference in energy of the actual molecule (experimental value) and energy of the most stable resonating structure is called resonance energy, i.e. E0 — £3.
8. (c) : Since the element X forms XC13, X205 and Ca3X2, therefore, it must be N or P. Since it does form JVCI5, therefore, it must be N since it has no rf-orbitals to expand its covalency from 3 to 5. Therefore, A" must be N.
9. (a)
10. (c) : The sum of mole fractions of all the components = 1.
, , M7RT (M7\RT
12. (d) : AHNMT cannot be predicted unless we know i heat of dissociation of acetic acid.
13. (d) : 50 ml of 1 M KOH + 50 ml of 0.5 M H2S04
will involve complete neutralisation.
14. (c): Increasing concentration of a product favours reverse reaction.
15. (a) : Lower the equilibrium constant, greater is the stability.
16. (c) : On dilution, ionisation increases.
17. (d) : pH = 10 means [H+] = IO"10 M.
18. (b) : Here one element of KC103 is oxidised
- 2 0 +5 - }
(O —» O,) while the other is reduced (CI —» CI) and hence it is an example of intramolecular redox reaction.
19. (b) : NO, N2H4
To balance N atoms, multiply NO," by 2, we have
+5 - 2
2N03- N2H4
Total O.N. = 2 x +5 Total O.N. = 2 x -2
= +10 = - 4 To balance O.N., add \4e~ to L.H.S., we have
2N03- + 14e~ —» N2H4
or, N03- + 7e~ -> 1/2 N2H4
Therefore, number of electrons involved in the reduction of N03" ions is 7.
20. (c) : At anode (made of Pt), H:0 is oxidised more easily than S042~ ions. H+ ions produced combine with
18 CHEMISTRY TODAY | SEPTEMBER '05 44
SO„2 ions to produce H2S04. Cu2+ ions are deposited on cathode. 02 is evolved at anode.
21. (b): Equivalent conductance increases with dilution and ultimately becomes constant.
22. (d) : Ionic conductance of ion = transport number of that ion x Aeq° of strong electrolyte containing that ion = (1 - 0.40) x 160.84 = 96.504
ionic conductance _ 96.504 _ 3
Ionic mobility = — - 1 0 96,500 96,500
23. (c): Lower the activation energy, faster is the reaction.
24. (b) : As it is a reaction of 1st order,
t [A]
, 2.303, i.e.. k = log
p N0 2O5
PNjOJ
°f' 2.303
From slope, k can be calculated.
, , , 0.693 . 2.303, 25. (c) : k = — m i n = log
-t + l o g p N2O5
a
o r , '99.9%
45 2.303x45
0.693
'99.9% a -0.999a log 10 = 448 min
= 7— hours. ,1 2
26. (a) : (a) has all atoms with A — Z - N same.
27. (b) : 2H Th -> 2£f Pb + * \ a + y p 232 = 208 + 4x or, x = 6 90 = 82 + 2 x - y or, y = 4.
28. (d) : x = ^ ^ l o g — = 2 . 3 x l 0 -2m i n- 1. 30 14
Initially, activity = \N i.e.
28 = 2.3 x 10-2 x or, N = 1217 = 1200.
29. (b) : Gelatin stabilises the colloids and prevents crystallisation.
30. (a) : ZSM-5 (size selective catalyst) converts alcohol to petrol.
31. (d) : IE, of B is lower than that of Be because in B, a 2p-electron is to be removed while in Be it is the 2\-electron. Similarly, IE, of AI is lower than that of Mg because in Al, a 3/7-electron is to be removed while in Mg it is the 3,v electron. IE, of N is higher than that
of O because of extra stability of exactly half-filled 2p-orbitals in N. IE, of Mg is higher than that of Na because of higher nuclear charge and completely filled 3.?-orbital in Mg.
32. (a) : IE + EA = 275 + 86 = 361 k cal mol"1
= 361 x 4.184 = 1510.42 kJ mol"1
1510.42 O D
Electronegativity = — — = 2.797 = 2.8 33. (c) : As the reduction potential of hydrogen is lower than that of sodium, it will be discharged at the cathode and chlorine will be discharged at the anode.
34. (d) : Calamine is ZnC03 and zincite is ZnO. Both are minerals of zinc.
35. (b) : IE of H atom is 1312 kJ mol"1.
36. (b): Na,ALSi208 • *H20 + Ca2+ . , , >
- z from hard water
zeolite
CaAI2Si208 • JCH20 + 2Na"
37. (a) : H+ ions are discharged at a higher potential when Hg cathode is used than that using Pt cathode.
38. (a) : Be shows coordination number of four in BeO which is covalent and has zinc sulphide (wurtzite) structure while all others show a coordination number of six. (NaCl structure) and are ionic.
39. ( c ) : Generally Mg(OH)2 is used as an antacid. The metal is Mg which burns in C 02 atmosphere.
2Mg + C 02 -> 2MgO + C.
40. (d) : 3 02 S l'e n tfc t n c > 2 0 ,
A discharge
41. (a) : Alkaline earth metal oxides are more basic than transition metal oxides which in turn are more basic than non-metal oxides and thus the acidic character increases from left to right in a period.
42. (b) : A mixture of one part of powdered Al and three parts of Fe203 is called thermite.
43. (c) : A i d , • 6H20 d l s s o c i a t i o" >
[A1C12(H20)4]+ + [AIC14(H2 0)J-44. (c) : Water in presence of oxygen reacts with Pb (pipes) to form soluble Pb(OH)2 which gives poisonous Pb2+ ions in solution.
Pb + 02 + H20 -> Pb(OH)2
45. (b) : Element Y belongs to group 14 of the periodic table which forms two chlorides y d4 (a colourless, volatile liquid) and KC12 (a colourless solid).
46. (a) : Extent of hydrolysis increases in the order:
CC14 < MgCl2 < A1C13 < SiCl4 < PCI5.
C H E M I S T R Y T O D A Y | SEPTEMBER '05 4 5
47. (c) : Both B and AI have 2 (Is2) and 8 (2s2 2p6) electrons in their penultimate (last but one) shell respectively but the remaining elements of group 13 have 18 electrons in their penultimate shell and hence B and AI show different properties from those of remaining members of group 13.
48. (b) : Reducing character of hydrides of nitrogen family increases as the size of the central atom increases down the group. All other properties stated here decrease down the group.
49. (d) : The compound contains C, H and S as the elements which on burning give C02, H20 and S02 as oxides respectively. First oxide (C02) turns lime water milky, the second oxide (H20) turns anhydrous (white) CuS04 blue due to hydration and the third oxide (S02) dissolves in H20 giving an acid (H2S03) which lowers the pH of the solution.
C, H, S are the elements in a compound.
C + 02 — C 0 2 ; 2H2 + 02 — 2 H 2 0
(1) (II)
s + o
2so
2 (III)C 02 + Mg(OH)2 -4 MgCOj + H20
(I) lime water insoluble (milkiness)
5H20 + CuS04 CuS04 • 5H20
(II) colourless blue
S02 + HzO H2S03
(III) acid, lowers pH of the solution
50. (c) : The shape of SF3C13 molecule is octahedral.
51. ( d ) : He has lowest boiling point due to weak van der Waals forces between its atoms.
52. (c) : All noble gases except He (which has Is2) have ns2 np6 electronic configuration of the valence shell.
53. (a) : Na2Cr207 is hygroscopic and cannot be used as a primary standard in volumetric analysis.
54. (b) : IUPAC name is potassiumtrioxalato aluminate(III).
55. (c) : Cl2 + H20 HCl + HCIO;
X
AgN03 + HCl AgCl + HN03 white ppt.
2HC1 + Mg MgCl2 + H2 T Y
56. (b) : Sewage is a biodegradable pollutant.
9 io l , r u
»••«. ' O C t
7 6 4 3
2-methylspiro[4.5]deca-l,6-diene
58. (c) : Only option (c) is correct as shown below.
CI CI
I H2 I
CH3CH2 - C — CH - CH3
—-3-chloro-2-pentene (A)
CI
2-chloro-2-pentene (B)
• CH3CH2 — CH — CH2CH3 optically inactive
CI CH optically active
59. (d) : Angle increases progressively.
sp3 (109° 28'), sp2 (120°), sp( 180°)
60. (b): As the s-character increases or the ^-character decreases, the energy of the orbital decreases. Thus the order is p > sp3 > sp2 > sp > s.
61. (d) : Wurtz reaction.
62. (b) : Ozonolysis since all the three alkenes will give different products as shown below.
(•)03
CH3CH2CH — C H 2 1-buterie
CH3CH=CHCH3 (ii) Zn/H20
(>)o3
» CH3CH2CHO + HCHO
(ii) Zn/H20 > 2CH3CHO (CH3)2C = CH2 J> (CH3)2CO + HCHO
(ii) Zn/H20
CH,
63. (b) : C H3- C = CH2 + C6H6 isobutene
(CH3)3C-C6H5 ferf-butylbenzene
64. (c)
0_N a+
NaOH^ rj- ^ r "N 02
The reaction occurs by activated nucleophile substitution.
65. (b) : CC14 + 4KOH
-4KC1 > C(OH).
4 - H20
HO - CO - OH > K2C03 carbonic acid 2
18 C H E M I S T R Y TODAY | SEPTEMBER '05 46
66. (d) : Since compound Y reacts with I2 and Na2C03
to form triiodomethane, therefore, Y must be a methyl ketone. Since Y is obtained by oxidation of X with K.2Cr207, therefore, Xmust be a methylcarbinol, thus X is CH3CHOHCH3.
67. (a) : During oxidation of unsymmetrical ketones, i.e. /7-propyImethy I ketone, the keto group stays with the smaller alkyl group (Popoff's rule) i.e.
CH,CH2COOH and CH3COOH are formed.