Question Bank (Solutions): Organic Chemistry : Some Basic Principles & Techniques | Chemistry Class 11 - NEET PDF Download

 Page 1


—
 
5 4 3 2 1 
1. (b) : HC C — CH CH — CH 3 
Pent-2-en-4-yne 
(10 ?-bonds and 3?-bonds) 
14. (a) : 
2. (b) : 
3. (d) : There are four double bonds. Hence, no. of 
?-electrons = 2 × 4 = 8. 
15. (a) : 
CH
3   
 
4 
CH
3
 
 
2 1 
 
4-Ethyl-3-propylhex-1-ene 
4. (d) : C
7 
H  –
6
C – C
5
 H CH–
3
CH –C      CH 
sp
3 
3
 
sp
3   
sp
2 
sp
2 
sp
3 
sp sp 
CH
3
 
? C 2 - sp, C 3 - sp
3
, C 5 - sp
2 
and C 6 - sp
3
 
sp
3 
sp
2 
sp
2 
sp
3 
sp sp 
16. (a) : 
5. (d) : CH
3 
CH CH CH
2 
C CH 17. (b) : 
6 5 4 3 2 1 
The state of hybridisation of carbon in 1, 3 and 5 position 
are sp, sp
3 
and sp
2
. 
6. (d) : 
7. (d) : 
8. (c) : 
9. (a) : Tetrachloroethene being an alkene has sp
2
-
hybridised C-atoms and hence  the  angle  Cl  – C – Cl  is 
120° while in tetrachloromethane, carbon is sp
3 
hybridised, therefore the angle Cl – C – Cl is 109°28???
10.  (a) : 
(C 6H 7O 2N) 
Hence, it is homocyclic (as the ring system is made of 
one type of atoms, i.e., carbon) but not aromatic. 
11. (d) : 
If a molecule contains both carbon-carbon double or 
triple bonds, the two are treated as per in seeking the 
lowest number combination. However, if the sum of 
numbers turns out to be the same starting from either 
of the carbon chain, then lowest number is given to the 
C C double bond. 
18. (c) : 
Since the sum of numbers starting from either side of the 
carbon chain turns out to be the same, so lowest number 
is given to the C C end. 
19. (c) : 
It is 2,3-dimethylpentanoyl chloride. 
20. (a) : 
4-Ethyl-3-methylheptane 
CH 3 
12. (c) : CH 3 — CH — CH 2 — 
(iso-butyl group) 
13. (d) : 
IUPAC name of the compound is 
3- ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid. 
21.   (a) : 
22.  (b) : 
The double bond gets priority over triple bond. Therefore, 
correct IUPAC name is 1-hexene-5-yne. 
23. (a) : 
C 
E 
Hints & Explanations 
Page 2


—
 
5 4 3 2 1 
1. (b) : HC C — CH CH — CH 3 
Pent-2-en-4-yne 
(10 ?-bonds and 3?-bonds) 
14. (a) : 
2. (b) : 
3. (d) : There are four double bonds. Hence, no. of 
?-electrons = 2 × 4 = 8. 
15. (a) : 
CH
3   
 
4 
CH
3
 
 
2 1 
 
4-Ethyl-3-propylhex-1-ene 
4. (d) : C
7 
H  –
6
C – C
5
 H CH–
3
CH –C      CH 
sp
3 
3
 
sp
3   
sp
2 
sp
2 
sp
3 
sp sp 
CH
3
 
? C 2 - sp, C 3 - sp
3
, C 5 - sp
2 
and C 6 - sp
3
 
sp
3 
sp
2 
sp
2 
sp
3 
sp sp 
16. (a) : 
5. (d) : CH
3 
CH CH CH
2 
C CH 17. (b) : 
6 5 4 3 2 1 
The state of hybridisation of carbon in 1, 3 and 5 position 
are sp, sp
3 
and sp
2
. 
6. (d) : 
7. (d) : 
8. (c) : 
9. (a) : Tetrachloroethene being an alkene has sp
2
-
hybridised C-atoms and hence  the  angle  Cl  – C – Cl  is 
120° while in tetrachloromethane, carbon is sp
3 
hybridised, therefore the angle Cl – C – Cl is 109°28???
10.  (a) : 
(C 6H 7O 2N) 
Hence, it is homocyclic (as the ring system is made of 
one type of atoms, i.e., carbon) but not aromatic. 
11. (d) : 
If a molecule contains both carbon-carbon double or 
triple bonds, the two are treated as per in seeking the 
lowest number combination. However, if the sum of 
numbers turns out to be the same starting from either 
of the carbon chain, then lowest number is given to the 
C C double bond. 
18. (c) : 
Since the sum of numbers starting from either side of the 
carbon chain turns out to be the same, so lowest number 
is given to the C C end. 
19. (c) : 
It is 2,3-dimethylpentanoyl chloride. 
20. (a) : 
4-Ethyl-3-methylheptane 
CH 3 
12. (c) : CH 3 — CH — CH 2 — 
(iso-butyl group) 
13. (d) : 
IUPAC name of the compound is 
3- ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid. 
21.   (a) : 
22.  (b) : 
The double bond gets priority over triple bond. Therefore, 
correct IUPAC name is 1-hexene-5-yne. 
23. (a) : 
C 
E 
Hints & Explanations 
 
The group should get priority over methyl group. 
? Correct IUPAC name is 
30.   (c) : Enolic   form   of   ethyl   acetoacetate   has 18 
?-bonds and 2 ?-bonds. 
31. (b) 
32. (b) : Only four structural isomers are possible for 
monochlorinated diphenylmethane. 
24.   (a) : 
25.   (a) : 
As —COOH group is highest priority group, it is 
numbered one. So, the IUPAC name is 3-amino-5- 
heptenoic acid. 
26.  (b) : 
27.  (b) : 
28. (a) : ?-Hydrogen at bridge carbon never participate 
in tautomerism. Thus, only (III) exhibits tautomerism. 
33. (b) : It is a special type of functional isomerism, in 
which both the isomers are represented by one and the 
same substance and are always present in equilibrium. It is 
exhibited by nitroalkane (RCH 2NO 2) and isonitroalkane. 
34. (a) : There are 7 isomers in C 4H 10O. Out of these, 4 
are alcohols and 3 are ethers. 
CH 3 CH 2 – O – CH 2 CH 3, 
CH 3OCH 2 CH 2 CH 3, 
 
CH 3 CH 2 CH 2 CH 2OH, 
29. (b) : In keto-enol tautomerism, 
(I) 
here, ?-H participates. 
(II) 
35. (d) : Isomers must have same molecular formula 
but different structural formula. 
36. (b) : 5-chain isomers are obtained from alkane 
C
6
H
14
.
 
(i) CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 
(ii) 
(iii) (iv) 
(v) 
+ 
37. (d) : tert-Butyl carbocation, (CH3 )3 C
+
 is more 
here, ?-H participates. 
stable than sec-butyl carbocation (CH 3)2CH due to 
hyperco
+
njugation. 
+
 
(III) 
(CH 3) 3C has nine C — H bonds while (CH 3) 2CH has six 
C — H bonds. Thus, there is more hyperconjugative 
structures in tert-butyl carbocation. 
38. (c) : Among the given carbocations, 
here, ?-H participates (p-tautomerism). CH  — C
+
H— CH — CH — CH is most stable 
3 2 2 3 
v 
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—
 
5 4 3 2 1 
1. (b) : HC C — CH CH — CH 3 
Pent-2-en-4-yne 
(10 ?-bonds and 3?-bonds) 
14. (a) : 
2. (b) : 
3. (d) : There are four double bonds. Hence, no. of 
?-electrons = 2 × 4 = 8. 
15. (a) : 
CH
3   
 
4 
CH
3
 
 
2 1 
 
4-Ethyl-3-propylhex-1-ene 
4. (d) : C
7 
H  –
6
C – C
5
 H CH–
3
CH –C      CH 
sp
3 
3
 
sp
3   
sp
2 
sp
2 
sp
3 
sp sp 
CH
3
 
? C 2 - sp, C 3 - sp
3
, C 5 - sp
2 
and C 6 - sp
3
 
sp
3 
sp
2 
sp
2 
sp
3 
sp sp 
16. (a) : 
5. (d) : CH
3 
CH CH CH
2 
C CH 17. (b) : 
6 5 4 3 2 1 
The state of hybridisation of carbon in 1, 3 and 5 position 
are sp, sp
3 
and sp
2
. 
6. (d) : 
7. (d) : 
8. (c) : 
9. (a) : Tetrachloroethene being an alkene has sp
2
-
hybridised C-atoms and hence  the  angle  Cl  – C – Cl  is 
120° while in tetrachloromethane, carbon is sp
3 
hybridised, therefore the angle Cl – C – Cl is 109°28???
10.  (a) : 
(C 6H 7O 2N) 
Hence, it is homocyclic (as the ring system is made of 
one type of atoms, i.e., carbon) but not aromatic. 
11. (d) : 
If a molecule contains both carbon-carbon double or 
triple bonds, the two are treated as per in seeking the 
lowest number combination. However, if the sum of 
numbers turns out to be the same starting from either 
of the carbon chain, then lowest number is given to the 
C C double bond. 
18. (c) : 
Since the sum of numbers starting from either side of the 
carbon chain turns out to be the same, so lowest number 
is given to the C C end. 
19. (c) : 
It is 2,3-dimethylpentanoyl chloride. 
20. (a) : 
4-Ethyl-3-methylheptane 
CH 3 
12. (c) : CH 3 — CH — CH 2 — 
(iso-butyl group) 
13. (d) : 
IUPAC name of the compound is 
3- ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid. 
21.   (a) : 
22.  (b) : 
The double bond gets priority over triple bond. Therefore, 
correct IUPAC name is 1-hexene-5-yne. 
23. (a) : 
C 
E 
Hints & Explanations 
 
The group should get priority over methyl group. 
? Correct IUPAC name is 
30.   (c) : Enolic   form   of   ethyl   acetoacetate   has 18 
?-bonds and 2 ?-bonds. 
31. (b) 
32. (b) : Only four structural isomers are possible for 
monochlorinated diphenylmethane. 
24.   (a) : 
25.   (a) : 
As —COOH group is highest priority group, it is 
numbered one. So, the IUPAC name is 3-amino-5- 
heptenoic acid. 
26.  (b) : 
27.  (b) : 
28. (a) : ?-Hydrogen at bridge carbon never participate 
in tautomerism. Thus, only (III) exhibits tautomerism. 
33. (b) : It is a special type of functional isomerism, in 
which both the isomers are represented by one and the 
same substance and are always present in equilibrium. It is 
exhibited by nitroalkane (RCH 2NO 2) and isonitroalkane. 
34. (a) : There are 7 isomers in C 4H 10O. Out of these, 4 
are alcohols and 3 are ethers. 
CH 3 CH 2 – O – CH 2 CH 3, 
CH 3OCH 2 CH 2 CH 3, 
 
CH 3 CH 2 CH 2 CH 2OH, 
29. (b) : In keto-enol tautomerism, 
(I) 
here, ?-H participates. 
(II) 
35. (d) : Isomers must have same molecular formula 
but different structural formula. 
36. (b) : 5-chain isomers are obtained from alkane 
C
6
H
14
.
 
(i) CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 
(ii) 
(iii) (iv) 
(v) 
+ 
37. (d) : tert-Butyl carbocation, (CH3 )3 C
+
 is more 
here, ?-H participates. 
stable than sec-butyl carbocation (CH 3)2CH due to 
hyperco
+
njugation. 
+
 
(III) 
(CH 3) 3C has nine C — H bonds while (CH 3) 2CH has six 
C — H bonds. Thus, there is more hyperconjugative 
structures in tert-butyl carbocation. 
38. (c) : Among the given carbocations, 
here, ?-H participates (p-tautomerism). CH  — C
+
H— CH — CH — CH is most stable 
3 2 2 3 
v 
Organic Chemistry - Some Basic Principles and Techniques 
carbocation. As is consists of maximum number of 
?-hydrogens and stablised by hyperconjugation. 
39. (a, b) : –I effect increases on increasing the 
electronegativity of atom. 
? – NH 2 < – OR < – F ( –I effect) 
Also, – NR 2 < – OR < – F (–I effect) 
40. (c) : –NO 2 group is meta-directing, thus will 
stabilize a electrophile at m-position. 
41. (c) 
42. (d) : Nucleophiles are electron rich species hence, 
they are Lewis bases. 
43. (c) : 
44. (a) : Nucleophile will attack a stable carbocation 
(S N1 reaction). 
45. (a) : Hyperconjugation can occur only in compound 
III as it has ?-hydrogen atom. 
46. (d) : is most stable due to 
hyperconjugation. 
47. (c) 
48. (a) : Greater the number of electron donating alkyl 
groups (+I effect), greater is the stability of carbocations. 
+I effect is in the order : 
More the number of hyperconjugative structures of 
carbocations, more is the stability. 
Hence, the order of stability of carbocations is 
5 < 4 < 3 < 1 < 2. 
49. (a) 
50. (b) : More the number of hyperconjugative 
structures, the greater is the stability. 
51. (c) : Nucleophilic substitution reaction involves the 
displacement of a nucleophile by another. 
 
52. (b) : +R effect of —OH group is greater than that 
o f —OCH 3 group. 
53. (c) 
54. (c) : Higher the no. of electron releasing groups 
lower will be stability of carbanion, and vice-versa. So, 
the order of stability of carbanions is 
55. (a) : In case of different nucleophiles, but present in 
the same group in the periodic table, then larger is the 
atomic mass, higher is the nucleophilicity. Hence, the 
increasing order of nucleophilicity of the halide ions is 
F
– 
< Cl
– 
< Br
– 
< I
–
. 
56. (d) : 
3° carbon is more stable due to the stabilization of the 
charge by three methyl groups (or +I 
– 
effect). It can also 
be explained on the basis of hyperconjugation. Greater 
the number of hyperconjugative ?-H atoms, more will 
be the hyperconjugative structures and more will be the 
stability. 
57. (c) : 3º > 2º > 1º more the delocalisation of positive 
charge, more is its stability. 
58. (d) : All the properties mentioned in the question 
suggest that it is a benzene molecule. Since in benzene all 
carbons are sp
2
-hybridised, therefore, C – C – C angle is 
120º. 
59. (b) : Paper chromatography is a type of partition 
chromatography. 
60. (c) : T he o- and p-nitrophenols are separated by 
steam distillation since o-isomer is steam volatile due to 
intramolecular H-bonding while p-isomer is not steam 
volatile due to association of molecules by intermolecular 
H-bonding. 
Page 4


—
 
5 4 3 2 1 
1. (b) : HC C — CH CH — CH 3 
Pent-2-en-4-yne 
(10 ?-bonds and 3?-bonds) 
14. (a) : 
2. (b) : 
3. (d) : There are four double bonds. Hence, no. of 
?-electrons = 2 × 4 = 8. 
15. (a) : 
CH
3   
 
4 
CH
3
 
 
2 1 
 
4-Ethyl-3-propylhex-1-ene 
4. (d) : C
7 
H  –
6
C – C
5
 H CH–
3
CH –C      CH 
sp
3 
3
 
sp
3   
sp
2 
sp
2 
sp
3 
sp sp 
CH
3
 
? C 2 - sp, C 3 - sp
3
, C 5 - sp
2 
and C 6 - sp
3
 
sp
3 
sp
2 
sp
2 
sp
3 
sp sp 
16. (a) : 
5. (d) : CH
3 
CH CH CH
2 
C CH 17. (b) : 
6 5 4 3 2 1 
The state of hybridisation of carbon in 1, 3 and 5 position 
are sp, sp
3 
and sp
2
. 
6. (d) : 
7. (d) : 
8. (c) : 
9. (a) : Tetrachloroethene being an alkene has sp
2
-
hybridised C-atoms and hence  the  angle  Cl  – C – Cl  is 
120° while in tetrachloromethane, carbon is sp
3 
hybridised, therefore the angle Cl – C – Cl is 109°28???
10.  (a) : 
(C 6H 7O 2N) 
Hence, it is homocyclic (as the ring system is made of 
one type of atoms, i.e., carbon) but not aromatic. 
11. (d) : 
If a molecule contains both carbon-carbon double or 
triple bonds, the two are treated as per in seeking the 
lowest number combination. However, if the sum of 
numbers turns out to be the same starting from either 
of the carbon chain, then lowest number is given to the 
C C double bond. 
18. (c) : 
Since the sum of numbers starting from either side of the 
carbon chain turns out to be the same, so lowest number 
is given to the C C end. 
19. (c) : 
It is 2,3-dimethylpentanoyl chloride. 
20. (a) : 
4-Ethyl-3-methylheptane 
CH 3 
12. (c) : CH 3 — CH — CH 2 — 
(iso-butyl group) 
13. (d) : 
IUPAC name of the compound is 
3- ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid. 
21.   (a) : 
22.  (b) : 
The double bond gets priority over triple bond. Therefore, 
correct IUPAC name is 1-hexene-5-yne. 
23. (a) : 
C 
E 
Hints & Explanations 
 
The group should get priority over methyl group. 
? Correct IUPAC name is 
30.   (c) : Enolic   form   of   ethyl   acetoacetate   has 18 
?-bonds and 2 ?-bonds. 
31. (b) 
32. (b) : Only four structural isomers are possible for 
monochlorinated diphenylmethane. 
24.   (a) : 
25.   (a) : 
As —COOH group is highest priority group, it is 
numbered one. So, the IUPAC name is 3-amino-5- 
heptenoic acid. 
26.  (b) : 
27.  (b) : 
28. (a) : ?-Hydrogen at bridge carbon never participate 
in tautomerism. Thus, only (III) exhibits tautomerism. 
33. (b) : It is a special type of functional isomerism, in 
which both the isomers are represented by one and the 
same substance and are always present in equilibrium. It is 
exhibited by nitroalkane (RCH 2NO 2) and isonitroalkane. 
34. (a) : There are 7 isomers in C 4H 10O. Out of these, 4 
are alcohols and 3 are ethers. 
CH 3 CH 2 – O – CH 2 CH 3, 
CH 3OCH 2 CH 2 CH 3, 
 
CH 3 CH 2 CH 2 CH 2OH, 
29. (b) : In keto-enol tautomerism, 
(I) 
here, ?-H participates. 
(II) 
35. (d) : Isomers must have same molecular formula 
but different structural formula. 
36. (b) : 5-chain isomers are obtained from alkane 
C
6
H
14
.
 
(i) CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 
(ii) 
(iii) (iv) 
(v) 
+ 
37. (d) : tert-Butyl carbocation, (CH3 )3 C
+
 is more 
here, ?-H participates. 
stable than sec-butyl carbocation (CH 3)2CH due to 
hyperco
+
njugation. 
+
 
(III) 
(CH 3) 3C has nine C — H bonds while (CH 3) 2CH has six 
C — H bonds. Thus, there is more hyperconjugative 
structures in tert-butyl carbocation. 
38. (c) : Among the given carbocations, 
here, ?-H participates (p-tautomerism). CH  — C
+
H— CH — CH — CH is most stable 
3 2 2 3 
v 
Organic Chemistry - Some Basic Principles and Techniques 
carbocation. As is consists of maximum number of 
?-hydrogens and stablised by hyperconjugation. 
39. (a, b) : –I effect increases on increasing the 
electronegativity of atom. 
? – NH 2 < – OR < – F ( –I effect) 
Also, – NR 2 < – OR < – F (–I effect) 
40. (c) : –NO 2 group is meta-directing, thus will 
stabilize a electrophile at m-position. 
41. (c) 
42. (d) : Nucleophiles are electron rich species hence, 
they are Lewis bases. 
43. (c) : 
44. (a) : Nucleophile will attack a stable carbocation 
(S N1 reaction). 
45. (a) : Hyperconjugation can occur only in compound 
III as it has ?-hydrogen atom. 
46. (d) : is most stable due to 
hyperconjugation. 
47. (c) 
48. (a) : Greater the number of electron donating alkyl 
groups (+I effect), greater is the stability of carbocations. 
+I effect is in the order : 
More the number of hyperconjugative structures of 
carbocations, more is the stability. 
Hence, the order of stability of carbocations is 
5 < 4 < 3 < 1 < 2. 
49. (a) 
50. (b) : More the number of hyperconjugative 
structures, the greater is the stability. 
51. (c) : Nucleophilic substitution reaction involves the 
displacement of a nucleophile by another. 
 
52. (b) : +R effect of —OH group is greater than that 
o f —OCH 3 group. 
53. (c) 
54. (c) : Higher the no. of electron releasing groups 
lower will be stability of carbanion, and vice-versa. So, 
the order of stability of carbanions is 
55. (a) : In case of different nucleophiles, but present in 
the same group in the periodic table, then larger is the 
atomic mass, higher is the nucleophilicity. Hence, the 
increasing order of nucleophilicity of the halide ions is 
F
– 
< Cl
– 
< Br
– 
< I
–
. 
56. (d) : 
3° carbon is more stable due to the stabilization of the 
charge by three methyl groups (or +I 
– 
effect). It can also 
be explained on the basis of hyperconjugation. Greater 
the number of hyperconjugative ?-H atoms, more will 
be the hyperconjugative structures and more will be the 
stability. 
57. (c) : 3º > 2º > 1º more the delocalisation of positive 
charge, more is its stability. 
58. (d) : All the properties mentioned in the question 
suggest that it is a benzene molecule. Since in benzene all 
carbons are sp
2
-hybridised, therefore, C – C – C angle is 
120º. 
59. (b) : Paper chromatography is a type of partition 
chromatography. 
60. (c) : T he o- and p-nitrophenols are separated by 
steam distillation since o-isomer is steam volatile due to 
intramolecular H-bonding while p-isomer is not steam 
volatile due to association of molecules by intermolecular 
H-bonding. 
2 
2 
 
61. (b) : Sublimation method is used for those organ P
1
V
1 
? 
P
2
V
2
 
substances which pass directly from solid to vapour state 
on heating and vice-versa on cooling e.g. benzoic acid, 
naphthalene, camphor,  anthracene,  etc.  Naphthalene  
is volatile and benzoic acid is non-volatile due to the 
T
1
 
V ? 
P
1
V
1
T
2
T
1
P
2   
 
T
2
 
? 
700 ? 40 ? 273 
300 ? 760 
? 33.52 mL 
formation of the dimer. 
22400 mL of N 2 at STP weighs = 28 g 
62. (b) : In steam distillation of toluene, the pressure 
of toluene in vapour is less than pressure of barometer, 
because it is carried out when a solid or liquid is insoluble 
in water and is volatile with steam but the impurities are 
? 33.52 mL of N 2 
at STP weighs = 
28 ? 33.52
 
22400 
= 0.0419 g 
non-volatile. 
63. (d) : In the IR spectroscopy, each functional group 
appears at a certain peak (in cm
–1
). So, cyclohexanone 
can be identified by carbonyl peak. 
64. (b) : 3Na 4[Fe(CN) 6] + 2Fe 2(SO 4) 3 
Sodium ferrocyanide 
Fe 4[Fe(CN) 6] 3 + 6Na 2SO 4 
Ferric ferrocyanide 
(Prussian blue) 
65. (a) : In case of Lassaigne’s test of halogens, it is 
% of N ?
??Mass of nitrogen at STP 
? 100
 
Mass of organic compound taken 
? 
0.0419 
? 100 ? 16.76% 
0.25 
70. (a) : H 2SO 4 + 2NH 3 ? (NH 4) 2SO 4
10 mL of 1 M H 2SO 4 = 10 mmol 
[ M × V (mL) = mmol] 
Acid used for the absorption of ammonia 
= 10 mL of 2 N (or 1 M) H 2SO 4
% of N = 
1.4 ? N ? V 
? 
1.4 ? 2 ? 10 
? 37.33%
 
necessary to remove sodium cyanide and sodium 
sulphide from the sodium extract if nitrogen and sulphur 
W 0.75 
are present. This is done by boiling the sodium extract 
with conc. HNO 3. 
NaCN + HNO 3 ? NaNO 3 + HCN ? ?
Na 2S + 2HNO 3 ? 2NaNO 3 + H 2S ? ?
66. (b) : Sodium cyanide (Na + C + N ? NaCN). 
67. (d) 
68. (b) : 2CuSO 4 + K 4[Fe(CN) 6] ? Cu 2[Fe(CN) 6] 
chocolate ppt. 
71. (b) : Given : V 1 = 55 mL, V 2 = ? 
P 1 = 715 – 15 = 700 mm, P 2 = 760 mm 
T 1 = 300 K, T 2 = 273 K 
General gas equation, 
P
1
V
1 
? 
P
2
V
2
 
T
1 
T
2
 
Volume of nitrogen at STP, 
V ? 
P 1V 1T 2 
? 
700 ? 55 ? 273 
? 46.099 mL
 
P 2 T 1 760 ? 300 
69. (a) : Mass of organic compound = 0.25 g 
Experimental values, At STP 
+ 2K 2 SO 4 
%  of  nitrogen  = 
V
2
 
8W 
compound. 
, where W = the mass of organic 
V 1 = 40 mL, V 2 = ? 
T 1 = 300 K, T 2 = 273 K 
P 1 = 725 – 25 = 700 mm, P 2 = 760 mm 
% of N =
 46.099 
? 16.46 
8 ? 0.35 
72. (a) 
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