JEE Chemistry Analytical Chemistry MCQs Set 02

Answer: (d) Cl

Explanation:
1. A clean copper wire dipped in the organic compound is heated in a flame.
2. Volatile copper halides form, which colour the flame green or blue.
3. So the test detects halogens, here chlorine among the options.

Teacher's Note:
1. B
2. e
3. i
4. l
5. s
6. t
7. e
8. i
9. n
10. '
11. s
12.
13. t
14. e
15. s
16. t
17.
18. i
19. s
20.
21. a
22.
23. q
24. u
25. i
26. c
27. k
28.
29. b
30. u
31. t
32.
33. n
34. o
35. t
36.
37. v
38. e
39. r
40. y
41.
42. s
43. p
44. e
45. c
46. i
47. f
48. i
49. c
50.
51. t
52. e
53. s
54. t
55.
56. f
57. o
58. r
59.
60. h
61. a
62. l
63. o
64. g
65. e
66. n
67. s
68. .
69.
70. I
71. t
72.
73. f
74. a
75. i
76. l
77. s
78.
79. f
80. o
81. r
82.
83. f
84. l
85. u
86. o
87. r
88. i
89. n
90. e
91. ,
92.
93. b
94. e
95. c
96. a
97. u
98. s
99. e
100.
101. c
102. o
103. p
104. p
105. e
106. r
107.
108. f
109. l
110. u
111. o
112. r
113. i
114. d
115. e
116.
117. i
118. s
119.
120. n
121. o
122. t
123.
124. v
125. o
126. l
127. a
128. t
129. i
130. l
131. e
132. .

Answer: (c) 4.23 g

Explanation:
1. \( \ce{2CuSO4.5H2O + 4KI -> Cu2I2 + I2 + 2K2SO4 + 10H2O} \).
2. Molar mass of 2 \( \ce{CuSO4.5H2O} \) = 2 × 249 = 498 g, which liberates 1 mol \( \ce{I2} \) = 254 g.
3. Iodine liberated = \( \dfrac{254}{498}\times8.3=4.23 \) g.

Teacher's Note:
1. C
2. u
3. ²
4. ⁺
5.
6. o
7. x
8. i
9. d
10. i
11. s
12. e
13. s
14.
15. i
16. o
17. d
18. i
19. d
20. e
21.
22. t
23. o
24.
25. i
26. o
27. d
28. i
29. n
30. e
31.
32. a
33. n
34. d
35.
36. i
37. t
38. s
39. e
40. l
41. f
42.
43. f
44. o
45. r
46. m
47. s
48.
49. C
50. u
51. I
52. .
53.
54. T
55. w
56. o
57.
58. m
59. o
60. l
61. e
62. s
63.
64. o
65. f
66.
67. C
68. u
69. ²
70. ⁺
71.
72. g
73. i
74. v
75. e
76.
77. o
78. n
79. e
80.
81. m
82. o
83. l
84. e
85.
86. o
87. f
88.
89. I
90. ₂
91. .

Answer: (a) \( \ce{CuSCN} \)

Explanation:
1. Copper(II) is first reduced to copper(I) and then precipitated as the sparingly soluble copper(I) thiocyanate, CuSCN.
2. CuSCN is dried and weighed, so it is the weighing form.
3. The other compounds are not used as the weighing form.

Teacher's Note:
1. T
2. h
3. e
4.
5. w
6. e
7. i
8. g
9. h
10. i
11. n
12. g
13.
14. f
15. o
16. r
17. m
18.
19. m
20. u
21. s
22. t
23.
24. b
25. e
26.
27. p
28. u
29. r
30. e
31. ,
32.
33. s
34. t
35. a
36. b
37. l
38. e
39.
40. a
41. n
42. d
43.
44. o
45. f
46.
47. k
48. n
49. o
50. w
51. n
52.
53. f
54. o
55. r
56. m
57. u
58. l
59. a
60. .
61.
62. C
63. o
64. p
65. p
66. e
67. r
68.
69. i
70. s
71.
72. d
73. e
74. t
75. e
76. r
77. m
78. i
79. n
80. e
81. d
82.
83. a
84. s
85.
86. C
87. u
88. S
89. C
90. N
91.
92. (
93. o
94. r
95.
96. a
97. s
98.
99. C
100. u
101. O
102. )
103. .

Answer: (d) 9 – 11

Explanation:
1. Each dimethylglyoxime molecule loses a proton on forming the nickel complex, \( \ce{Ni^2+ + 2 dmgH2 -> Ni(dmgH)2 + 2H+} \).
2. A basic or ammoniacal medium removes these protons and drives the reaction forward.
3. In acid the nitrogen atoms are protonated and cannot bind nickel, so the best range is pH 9 – 11.

Teacher's Note:
1. N
2. i
3. –
4. D
5. M
6. G
7.
8. i
9. s
10.
11. a
12.
13. r
14. e
15. d
16.
17. c
18. h
19. e
20. l
21. a
22. t
23. e
24.
25. f
26. o
27. r
28. m
29. e
30. d
31.
32. i
33. n
34.
35. a
36. m
37. m
38. o
39. n
40. i
41. a
42. c
43. a
44. l
45.
46. s
47. o
48. l
49. u
50. t
51. i
52. o
53. n
54. .
55.
56. S
57. t
58. r
59. o
60. n
61. g
62.
63. a
64. c
65. i
66. d
67.
68. d
69. i
70. s
71. s
72. o
73. l
74. v
75. e
76. s
77.
78. t
79. h
80. e
81.
82. p
83. r
84. e
85. c
86. i
87. p
88. i
89. t
90. a
91. t
92. e
93. .

Answer: (a) \( \ce{B(OMe)3} \)

Explanation:
1. On heating a borate with conc. sulphuric acid and methanol, volatile trimethyl borate \( \ce{B(OCH3)3} \) is formed.
2. Its vapour burns with a green-edged flame, which is the test for borate.

Teacher's Note:
1. T
2. h
3. i
4. s
5.
6. i
7. s
8.
9. t
10. h
11. e
12.
13. c
14. o
15. n
16. f
17. i
18. r
19. m
20. a
21. t
22. o
23. r
24. y
25.
26. t
27. e
28. s
29. t
30.
31. f
32. o
33. r
34.
35. b
36. o
37. r
38. a
39. t
40. e
41.
42. r
43. a
44. d
45. i
46. c
47. a
48. l
49. .
50.
51. T
52. h
53. e
54.
55. o
56. t
57. h
58. e
59. r
60.
61. c
62. o
63. m
64. p
65. o
66. u
67. n
68. d
69. s
70.
71. d
72. o
73.
74. n
75. o
76. t
77.
78. g
79. i
80. v
81. e
82.
83. a
84.
85. g
86. r
87. e
88. e
89. n
90.
91. f
92. l
93. a
94. m
95. e
96. .

Answer: (c) \( \ce{NH4+} \)

Explanation:
1. Nessler's reagent is alkaline \( \ce{K2[HgI4]} \).
2. It gives a brown or yellow precipitate of oxydimercuric ammonium iodide with ammonia or ammonium ions.

Teacher's Note:
1. N
2. e
3. s
4. s
5. l
6. e
7. r
8. '
9. s
10.
11. r
12. e
13. a
14. g
15. e
16. n
17. t
18.
19. c
20. a
21. n
22.
23. d
24. e
25. t
26. e
27. c
28. t
29.
30. a
31. m
32. m
33. o
34. n
35. i
36. a
37.
38. d
39. o
40. w
41. n
42.
43. t
44. o
45.
46. a
47. b
48. o
49. u
50. t
51.
52. o
53. n
54. e
55.
56. p
57. a
58. r
59. t
60.
61. p
62. e
63. r
64.
65. m
66. i
67. l
68. l
69. i
70. o
71. n
72. .
73.
74. K
75. e
76. e
77. p
78.
79. i
80. t
81.
82. i
83. n
84.
85. m
86. i
87. n
88. d
89.
90. w
91. i
92. t
93. h
94.
95. t
96. h
97. e
98.
99. b
100. r
101. o
102. w
103. n
104.
105. p
106. r
107. e
108. c
109. i
110. p
111. i
112. t
113. a
114. t
115. e
116. .

Answer: (b) Hot dilute HCl solution

Explanation:
1. Both \( \ce{Ag+} \) and \( \ce{Pb^2+} \) give white chlorides with HCl.
2. PbCl₂ is soluble in hot water and so dissolves in hot dilute HCl, while AgCl does not.
3. So hot dilute HCl separates the two; the other reagents give precipitates with both.

Teacher's Note:
1. P
2. b
3. C
4. l
5. ₂
6.
7. i
8. s
9.
10. s
11. o
12. l
13. u
14. b
15. l
16. e
17.
18. i
19. n
20.
21. h
22. o
23. t
24.
25. w
26. a
27. t
28. e
29. r
30.
31. b
32. u
33. t
34.
35. A
36. g
37. C
38. l
39.
40. i
41. s
42.
43. n
44. o
45. t
46. .
47.
48. T
49. h
50. i
51. s
52.
53. i
54. s
55.
56. a
57. l
58. s
59. o
60.
61. t
62. h
63. e
64.
65. b
66. a
67. s
68. i
69. s
70.
71. o
72. f
73.
74. s
75. e
76. p
77. a
78. r
79. a
80. t
81. i
82. n
83. g
84.
85. t
86. h
87. e
88. m
89.
90. i
91. n
92.
93. g
94. r
95. o
96. u
97. p
98.
99. I
100. .

Answer: (b) \( \ce{BaCl2} \)

Explanation:
1. The red gas is chromyl chloride, \( \ce{CrO2Cl2} \), which shows that the reagent is a chloride.
2. Barium gives a green (apple-green) flame: \( \ce{2BaCl2 + K2Cr2O7 + 3H2SO4} \) forms \( \ce{2CrO2Cl2} \), \( \ce{2BaSO4} \) and other products.
3. CaCl₂ gives a brick-red flame, so the reagent is BaCl₂.

Teacher's Note:
1. F
2. l
3. a
4. m
5. e
6.
7. c
8. o
9. l
10. o
11. u
12. r
13. s
14. :
15.
16. C
17. a
18.
19. b
20. r
21. i
22. c
23. k
24.
25. r
26. e
27. d
28. ,
29.
30. S
31. r
32.
33. c
34. r
35. i
36. m
37. s
38. o
39. n
40. ,
41.
42. B
43. a
44.
45. a
46. p
47. p
48. l
49. e
50.
51. g
52. r
53. e
54. e
55. n
56. .
57.
58. C
59. h
60. r
61. o
62. m
63. y
64. l
65.
66. c
67. h
68. l
69. o
70. r
71. i
72. d
73. e
74.
75. i
76. s
77.
78. t
79. h
80. e
81.
82. t
83. e
84. s
85. t
86.
87. f
88. o
89. r
90.
91. c
92. h
93. l
94. o
95. r
96. i
97. d
98. e
99. .

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Sodium fusion is used in the Lassaigne test because sodium reacts safely enough with the organic compound.
2. Potassium reacts too vigorously (even explosively), so it is not used. Both statements are true and the reason explains the assertion.

Teacher's Note:
1. S
2. o
3. d
4. i
5. u
6. m
7.
8. i
9. s
10.
11. t
12. h
13. e
14.
15. s
16. t
17. a
18. n
19. d
20. a
21. r
22. d
23.
24. m
25. e
26. t
27. a
28. l
29.
30. f
31. o
32. r
33.
34. t
35. h
36. e
37.
38. f
39. u
40. s
41. i
42. o
43. n
44. .
45.
46. L
47. a
48. s
49. s
50. a
51. i
52. g
53. n
54. e
55. '
56. s
57.
58. e
59. x
60. t
61. r
62. a
63. c
64. t
65.
66. i
67. s
68.
69. m
70. a
71. d
72. e
73.
74. b
75. y
76.
77. b
78. o
79. i
80. l
81. i
82. n
83. g
84.
85. t
86. h
87. e
88.
89. f
90. u
91. s
92. e
93. d
94.
95. m
96. a
97. s
98. s
99.
100. w
101. i
102. t
103. h
104.
105. w
106. a
107. t
108. e
109. r
110. .

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. If both N and S are present, sodium thiocyanate is formed: \( \ce{Na + C + S + N -> NaSCN} \).
2. With \( \ce{Fe^3+} \) it gives blood red \( \ce{Fe(SCN)3} \) colouration, so the red colour is due to the presence of sulphur.
3. Both statements are true and the reason explains the assertion.

Teacher's Note:
1. R
2. e
3. d
4.
5. c
6. o
7. l
8. o
9. u
10. r
11.
12. w
13. i
14. t
15. h
16.
17. f
18. e
19. r
20. r
21. i
22. c
23.
24. c
25. h
26. l
27. o
28. r
29. i
30. d
31. e
32.
33. i
34. n
35. d
36. i
37. c
38. a
39. t
40. e
41. s
42.
43. S
44. C
45. N
46. ⁻
47. ,
48.
49. w
50. h
51. i
52. c
53. h
54.
55. n
56. e
57. e
58. d
59. s
60.
61. b
62. o
63. t
64. h
65.
66. N
67.
68. a
69. n
70. d
71.
72. S
73. .
74.
75. P
76. r
77. u
78. s
79. s
80. i
81. a
82. n
83.
84. b
85. l
86. u
87. e
88.
89. i
90. n
91. d
92. i
93. c
94. a
95. t
96. e
97. s
98.
99. N
100.
101. a
102. l
103. o
104. n
105. e
106. .

Answer: (d) A is false, but R is true.

Explanation:
1. In alkaline medium \( \ce{H2S} \) ionises more, so the sulphide ion concentration is high, and Sb(III) is easily precipitated.
2. The assertion is false.
3. The concentration of \( \ce{S^2-} \) is large, not inadequate, in alkaline solution, so the reason is false.

Teacher's Note:
1. H
2. C
3. l
4.
5. s
6. u
7. p
8. p
9. r
10. e
11. s
12. s
13. e
14. s
15.
16. i
17. o
18. n
19. i
20. s
21. a
22. t
23. i
24. o
25. n
26.
27. o
28. f
29.
30. H
31. ₂
32. S
33.
34. b
35. y
36.
37. t
38. h
39. e
40.
41. c
42. o
43. m
44. m
45. o
46. n
47.
48. i
49. o
50. n
51.
52. e
53. f
54. f
55. e
56. c
57. t
58. ,
59.
60. w
61. h
62. i
63. l
64. e
65.
66. a
67. l
68. k
69. a
70. l
71. i
72.
73. i
74. n
75. c
76. r
77. e
78. a
79. s
80. e
81. s
82.
83. i
84. t
85. .
86.
87. T
88. h
89. i
90. s
91.
92. i
93. s
94.
95. u
96. s
97. e
98. d
99.
100. t
101. o
102.
103. s
104. e
105. p
106. a
107. r
108. a
109. t
110. e
111.
112. g
113. r
114. o
115. u
116. p
117. s
118.
119. i
120. n
121.
122. q
123. u
124. a
125. l
126. i
127. t
128. a
129. t
130. i
131. v
132. e
133.
134. a
135. n
136. a
137. l
138. y
139. s
140. i
141. s
142. .

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Borax bead gives coloured metaborates with many transition metals, but aluminium gives a colourless bead, so the test is not suitable. The assertion is true.
2. Al₂O₃ is indeed insoluble in water, so the reason is true.
3. Insolubility in water has nothing to do with the colour of the bead, so it is not the explanation.

Teacher's Note:
1. T
2. h
3. e
4.
5. b
6. o
7. r
8. a
9. x
10.
11. b
12. e
13. a
14. d
15.
16. t
17. e
18. s
19. t
20.
21. d
22. e
23. p
24. e
25. n
26. d
27. s
28.
29. o
30. n
31.
32. c
33. o
34. l
35. o
36. u
37. r
38. e
39. d
40.
41. m
42. e
43. t
44. a
45. b
46. o
47. r
48. a
49. t
50. e
51.
52. f
53. o
54. r
55. m
56. a
57. t
58. i
59. o
60. n
61.
62. b
63. y
64.
65. d
66. -
67. b
68. l
69. o
70. c
71. k
72.
73. m
74. e
75. t
76. a
77. l
78.
79. i
80. o
81. n
82. s
83. .
84.
85. A
86. l
87. ³
88. ⁺
89.
90. h
91. a
92. s
93.
94. n
95. o
96.
97. d
98.
99. e
100. l
101. e
102. c
103. t
104. r
105. o
106. n
107. s
108. .

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Starch forms an intense blue adsorption complex with iodine, so it indicates the end point. The assertion is true.
2. Starch is a polysaccharide, so the reason is true.
3. Being a polysaccharide is not why it indicates iodine, so it is not the explanation.

Teacher's Note:
1. T
2. h
3. e
4.
5. b
6. l
7. u
8. e
9.
10. c
11. o
12. l
13. o
14. u
15. r
16.
17. d
18. i
19. s
20. a
21. p
22. p
23. e
24. a
25. r
26. s
27.
28. a
29. t
30.
31. t
32. h
33. e
34.
35. e
36. n
37. d
38.
39. p
40. o
41. i
42. n
43. t
44.
45. w
46. h
47. e
48. n
49.
50. a
51. l
52. l
53.
54. t
55. h
56. e
57.
58. i
59. o
60. d
61. i
62. n
63. e
64.
65. i
66. s
67.
68. r
69. e
70. d
71. u
72. c
73. e
74. d
75.
76. b
77. y
78.
79. t
80. h
81. i
82. o
83. s
84. u
85. l
86. p
87. h
88. a
89. t
90. e
91. .
92.
93. A
94. d
95. d
96.
97. s
98. t
99. a
100. r
101. c
102. h
103.
104. n
105. e
106. a
107. r
108.
109. t
110. h
111. e
112.
113. e
114. n
115. d
116.
117. p
118. o
119. i
120. n
121. t
122. .

Answer: (c) A is true, but R is false.

Explanation:
1. Potassium dichromate is pure, stable and not hygroscopic, so its standard solution can be prepared directly. The assertion is true.
2. These properties, not high solubility, make it a primary standard, so the reason is false.

Teacher's Note:
1. A
2.
3. p
4. r
5. i
6. m
7. a
8. r
9. y
10.
11. s
12. t
13. a
14. n
15. d
16. a
17. r
18. d
19.
20. m
21. u
22. s
23. t
24.
25. b
26. e
27.
28. s
29. t
30. a
31. b
32. l
33. e
34. ,
35.
36. a
37. v
38. a
39. i
40. l
41. a
42. b
43. l
44. e
45.
46. i
47. n
48.
49. a
50.
51. p
52. u
53. r
54. e
55.
56. f
57. o
58. r
59. m
60.
61. a
62. n
63. d
64.
65. n
66. o
67. n
68. -
69. h
70. y
71. g
72. r
73. o
74. s
75. c
76. o
77. p
78. i
79. c
80. .
81.
82. S
83. o
84. l
85. u
86. b
87. i
88. l
89. i
90. t
91. y
92.
93. i
94. s
95.
96. n
97. o
98. t
99.
100. t
101. h
102. e
103.
104. c
105. r
106. i
107. t
108. e
109. r
110. i
111. o
112. n
113. .

Answer: (d) A is false, but R is true.

Explanation:
1. Antimony sulphide dissolves in yellow ammonium sulphide forming ammonium thioantimonate: \( \ce{Sb2S3 + 3(NH4)2S + 2S -> 2(NH4)3SbS4} \).
2. So the assertion is false and the reason is also false.

Teacher's Note:
1. S
2. u
3. l
4. p
5. h
6. i
7. d
8. e
9. s
10.
11. o
12. f
13.
14. a
15. r
16. s
17. e
18. n
19. i
20. c
21. ,
22.
23. a
24. n
25. t
26. i
27. m
28. o
29. n
30. y
31.
32. a
33. n
34. d
35.
36. t
37. i
38. n
39.
40. d
41. i
42. s
43. s
44. o
45. l
46. v
47. e
48.
49. i
50. n
51.
52. y
53. e
54. l
55. l
56. o
57. w
58.
59. a
60. m
61. m
62. o
63. n
64. i
65. u
66. m
67.
68. s
69. u
70. l
71. p
72. h
73. i
74. d
75. e
76.
77. a
78. s
79.
80. t
81. h
82. i
83. o
84. -
85. s
86. a
87. l
88. t
89. s
90. .
91.
92. T
93. h
94. i
95. s
96.
97. s
98. e
99. p
100. a
101. r
102. a
103. t
104. e
105. s
106.
107. g
108. r
109. o
110. u
111. p
112.
113. I
114. I
115. B
116.
117. f
118. r
119. o
120. m
121.
122. I
123. I
124. A
125. .

Answer: (c) A is true, but R is false.

Explanation:
1. The \( \ce{As2S3} \) sol is negatively charged by adsorbed \( \ce{S^2-} \) ions, and \( \ce{Fe^3+} \) carries a high positive charge that neutralises it and coagulates the sol. The assertion is true.
2. The coagulation is by charge neutralisation, not by formation of \( \ce{Fe2S3} \), so the reason is false.

Teacher's Note:
1. H
2. a
3. r
4. d
5. y
6. –
7. S
8. c
9. h
10. u
11. l
12. z
13. e
14.
15. r
16. u
17. l
18. e
19. :
20.
21. t
22. h
23. e
24.
25. i
26. o
27. n
28.
29. o
30. f
31.
32. o
33. p
34. p
35. o
36. s
37. i
38. t
39. e
40.
41. c
42. h
43. a
44. r
45. g
46. e
47.
48. a
49. n
50. d
51.
52. h
53. i
54. g
55. h
56. e
57. r
58.
59. v
60. a
61. l
62. e
63. n
64. c
65. y
66.
67. h
68. a
69. s
70.
71. a
72.
73. g
74. r
75. e
76. a
77. t
78. e
79. r
80.
81. c
82. o
83. a
84. g
85. u
86. l
87. a
88. t
89. i
90. n
91. g
92.
93. p
94. o
95. w
96. e
97. r
98. .
99.
100. F
101. e
102. ³
103. ⁺
104.
105. i
106. s
107.
108. v
109. e
110. r
111. y
112.
113. e
114. f
115. f
116. e
117. c
118. t
119. i
120. v
121. e
122.
123. f
124. o
125. r
126.
127. t
128. h
129. e
130.
131. n
132. e
133. g
134. a
135. t
136. i
137. v
138. e
139.
140. A
141. s
142. ₂
143. S
144. ₃
145.
146. s
147. o
148. l
149. .

Answer: (c) A is true, but R is false.

Explanation:
1. Alcohol vapour reduces orange dichromate to green chromium(III): \( \ce{K2Cr2O7 + 4H2SO4 -> K2SO4 + Cr2(SO4)3 + 4H2O + 3[O]} \) and \( \ce{C2H5OH + [O] -> CH3COOH} \).
2. So the assertion is true. The colour change is due to oxidation of alcohol, not complexation, so the reason is false.

Teacher's Note:
1. O
2. r
3. a
4. n
5. g
6. e
7.
8. t
9. o
10.
11. g
12. r
13. e
14. e
15. n
16.
17. i
18. s
19.
20. t
21. h
22. e
23.
24. s
25. i
26. g
27. n
28. a
29. l
30.
31. i
32. n
33.
34. t
35. h
36. e
37.
38. b
39. r
40. e
41. a
42. t
43. h
44.
45. a
46. n
47. a
48. l
49. y
50. s
51. e
52. r
53. .
54.
55. I
56. t
57.
58. i
59. s
60.
61. a
62.
63. r
64. e
65. d
66. o
67. x
68.
69. c
70. h
71. a
72. n
73. g
74. e
75.
76. o
77. f
78.
79. c
80. h
81. r
82. o
83. m
84. i
85. u
86. m
87.
88. f
89. r
90. o
91. m
92.
93. +
94. 6
95.
96. t
97. o
98.
99. +
100. 3
101. .

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Cyanide reduces Cu(II) to Cu(I) and forms the very stable complex \( \ce{[Cu(CN)4]^3-} \) which does not give enough \( \ce{Cu+} \) to precipitate as sulphide.
2. Cadmium forms the less stable \( \ce{[Cd(CN)4]^2-} \), which is decomposed by \( \ce{H2S} \) giving yellow CdS.
3. Both statements are true and the masking of copper by reduction and complexation explains the separation.

Teacher's Note:
1. T
2. h
3. i
4. s
5.
6. i
7. s
8.
9. a
10.
11. s
12. t
13. a
14. n
15. d
16. a
17. r
18. d
19.
20. m
21. a
22. s
23. k
24. i
25. n
26. g
27.
28. m
29. e
30. t
31. h
32. o
33. d
34.
35. f
36. o
37. r
38.
39. s
40. e
41. p
42. a
43. r
44. a
45. t
46. i
47. n
48. g
49.
50. C
51. d
52.
53. f
54. r
55. o
56. m
57.
58. C
59. u
60. .
61.
62. T
63. h
64. e
65.
66. s
67. t
68. a
69. b
70. i
71. l
72. i
73. t
74. y
75.
76. o
77. f
78.
79. t
80. h
81. e
82.
83. c
84. o
85. p
86. p
87. e
88. r
89.
90. c
91. y
92. a
93. n
94. i
95. d
96. e
97.
98. c
99. o
100. m
101. p
102. l
103. e
104. x
105.
106. i
107. s
108.
109. t
110. h
111. e
112.
113. k
114. e
115. y
116. .