Calculate The Amount Of Benzoic Acid (C6H5COOH) Required For Preparing 250 Ml Of 0.15 M Solution In Methanol.
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 ml of 0.15 M
solution in methanol.
asked Jun 13 12:00:00 AM
Bhushan Mudliyar
1 Answers
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Answer-
We know molarity = Moles of the Solute/ volume of solution
Moles of Benzoic Acid = 0.15*250/1000
= 0.0375 mol of benzoic acid
Also molecular mass of banzoic acid (C6H5COOH)
= 7 * 12 + 6 * 1 + 2 * 16
= 122g/mol
Therefore amount of banzoic acid = 0.0375 * 122
= 4.575g
asked Oct 15 12:00:00 AM
Sangeeta
Q: 22 A: 1924