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
JEE Advanced 2020 Concept Recapitulation Test 4 Paper 2
Show Para
Hide Para
Share question:
© examsnet.com
Question : 22
Total: 54
A solution is prepared by dissolving
1.5
g
of a monoacidic base into
1.5
k
g
of water at
300
K
, which showed a depression in freezing point by
0.165
∘
C
. When
0.496
g
of the same base titrated, after dissolution, requires
40
m
l
of semimolar
H
2
S
O
4
solution. If
K
f
of water is
1.86
K
k
g
m
o
l
−
1
, then select the correct statement(s) out of the following (assuming molarity = molality):
The
p
H
of the solution of weak base is
11.903
lonisation constant of the base is
8
×
10
−
4
The osmotic pressure of the aq. Solution of base is
2.167
atm
The base is
10
%
ionized in aq. Solution
Validate
Solution:
Calculation of mol. wt. of BOH:
eq. of the base
=
eq. of
H
2
S
O
4
used
0.496
m
8
×
1
=
40
×
10
−
3
×
1
2
×
2
⇒
m
B
=
12.4
g
m
o
l
−
1
Now:
∆
T
f
=
i
x
K
t
×
m
0.165
=
i
×
1.86
×
1.5
12.4
×
1000
150
i
=
0.165
×
12.4
1.86
=
1
−
1
=
1
+
α
∴
α
=
0.1
So,
(
O
H
−
)
=
C
α
=
0.08
×
0.1
=
8
×
10
−
9
∴
p
H
=
11.903
So,
K
b
=
C
α
2
=
0.8
×
(
0.1
)
2
=
8
×
10
−
4
Also,
π
=
i
C
R
T
=
1.1
×
0.08
×
0.0821
×
300
=
2.167
a
t
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
Prev Question
Next Question