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JEE Main Chemistry Class 11 Solutions Part 2 Questions
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© examsnet.com
Question : 18
Total: 80
The mass of sodium acetate
(
CH
3
COONa
)
required to prepare
250
mL
of
0.35
M
aqueous solution is ______g. (Molar mass of
CH
3
COONa
is
82.02
g
mol
−
1
)
[30-Jan-2024 Shift 1]
Your Answer:
Validate
Solution:
Moles
=
Molarity
×
Volume in litres
=
0.35
×
0.25
Mass
=
moles
×
molar mass
=
0.35
×
0.25
×
82.02
=
7.18
g
Ans. 7
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