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JEE Main 9 Jan 2020 Shift 1 Solved Paper
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© examsnet.com
Question : 47
Total: 75
The molarity of
H
N
O
3
in a sample which has density 1.4 g/mL and mass percentage of 63% is _____ .
(Molecular Weight of
H
N
O
3
=
63
)
Your Answer:
Validate
Solution:
100 gm soln → 63 gm
H
N
O
3
100
1.4
m
L
→ 1 mole
H
N
O
3
Molarity
=
1
100
1.4
×
1
1000
=
14
M
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