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

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

Solution

In this problem molarity = 0.15M is given 

Let take volume of solution = 1 liter  = 1000 mL

Us the above formula we get number of moles of solute = 0.15 moles

 Molar mass of benzoic acid (C₆H₅COOH) = 12 × 6 + 5 × 1 + 12 + 16 + 16 +1

       = 72 + 5 + 12 + 32 + 1

                                                                = 122 g mol-1

Mass of 0.15 mole of benzoic acid = number of moles x molar mass

     =0.15 × 122 g

     = 18.3 g

Thus, 1000 mL of the solution has mass of benzoic acid = 18.3 g

So        250 mL of the solution has mass of benzoic acid = 18.3 × 250/1000

  = 4.58 g.