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NCERT Class XII Chemistry
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Question : 30 of 53
Marks: +1, -0
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:  
M=
w2×1000
M2×V

⇒0.15=
w2×1000
122×250

or, w2=
1.5×122
4
=4.57g
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