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JEE Main Chemistry Class 11 Some Basic Concepts of Chemistry Part 2 Questions
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Question : 7
Total: 64
Molarity (M) of an aqueous solution containing
x
g
of anhyd.
CuSO
4
in
500
mL
solution at
32
∘
C
is
2
×
10
−
1
M
. Its molality will be ______
×
10
−
3
m
.
(nearest integer).
[Given density of the solution
=
1.25
g
∕
mL
.]
[9 Apr 2024 Shift 1]
Your Answer:
Validate
Solution:
M
s
o
l
n
=
v
s
o
l
n
×
d
s
o
l
n
=
500
×
1.25
=
625
g
Mass of solute
(
x
)
=
0.2
×
0.5
×
159.5
=
15.95
n
solute
=
0.1
Mass of solvent
=
Mass of solution
−
Mass of solute
=
625
−
15.95
=
609.05
m
=
0.1
609.05
1000
m
=
0.164
=
164
×
10
−
3
© examsnet.com
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