Calculate The Amount Of Benzoic Acid (C6H5COOH) Required For Preparing 250 Ml Of 0.15 M Solution In Methanol.

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Bhushan Mudliyar 180 Days Ago
29

Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 ml of 0.15 M

solution in methanol.

asked Jun 13 12:42:27 PM

Bhushan Mudliyar

Q: 533 A: 0

1 Answers

  • 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

    56 Days Ago

    asked Oct 15 11:25:10 AM

    Sangeeta

    Q: 22 A: 1924

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