Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
AIEEE 2004 Chemistry Solved Paper
Show Para
Hide Para
Share question:
© examsnet.com
Question : 1
Total: 75
6.02
×
10
20
molecules of urea are present in
100
ml
of its solution. The concentration of urea solution is
(Avogadro constant,
N
A
=
6.02
×
10
23
mol
−
1
)
0.02
M
0.01
M
0.001
M
0.01
M
Validate
Solution:
Moles
(
n
)
=
molecules
N
A
⇒
6.02
×
10
20
6.02
×
10
23
⇒
0.001
Concentration
(
M
)
=
moles
volume
Given volume of urea is
100
ml
=
0.1
litre
[Note: While calculating Concentration volume should always be in litre.]
⇒
0.001
0.1
⇒
0.01
M
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
Next Question
Examsnet