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JEE Main 3 Apr 2025 Shift 2 Paper
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© examsnet.com
Question : 69
Total: 75
10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask.
[3 Apr 2025 Shift 2]
10 M HCl solution
20 M HCl solution
Neutral solution
0.2 M NaCl solution
Validate
Solution:
10 mL of 2 M NaOH reacts with 20 mL of 1 M HCl to give completely neutralised 30 mL of
NaCl
(
20
m
mole) solution.
10 mL of this solution has
‌
20
30
×
10
=
‌
20
3
m
‌
mol
of NaCl
In Final 100 mL solution :
‌
[
HCl
]
=
‌
2
100
×
1000
=
20
M
‌
[
NaCl
]
=
‌
20
3
×
100
=
‌
1
15
M
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