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) CBSE Free NCERT Solution of 12th chemistry Solutions calculate the amount of benzoic acid c6h5cooh re

Question:

Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Answer:

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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