SELECT * FROM question_mgmt as q WHERE id=1715 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
We know Molarity = moles of the solute / volume of solution
Putting the given values in above equation,we get
0.15 = (mole of benzoic acid / 250) x 1000
Or
moles of benzoic acid = (0.15 x 250) / 1000
= 0.0375 mol of benzoic acid
Also molecular mass of benzoic acid (C6H5COOH)
= 7 × 12 + 6 × 1 + 2 × 16
= 122 g/mol
Therefore amount of benzoic acid = 0.0375 x 122 = 4.575 g
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