Question

How many milliliters of 0.332 M H3PO4 will contain the following? (a) 0.35 mol H3PO4 mL H3PO4 the tolerance is +/-9% (b) 46.3 g H3PO4 mL H3PO4 the tolerance is +/-9% (c) 5.61 x 1022 molecules of H3PO4 mL H3PO4 the tolerance is +/-9%

Answer 1

a) no of mol of H3PO4 = 0.35 mol

          n = M*V

      volume(v) = n/M = 0.35/0.332 = 1.05 L

                   = 1050 ml

b) no of mol of H3PO4 = w/mwt = 46.3/98 = 0.472 mol

     volume(v) = n/M = 0.472/0.332 = 1.422 L

                = 1422 ml

c) no of mol of H3PO4 = (5.61*10^22)/(6.023*10^23) = 0.093 mol

    ( 1 mol = 6.023*10^23 molecules)

      volume(v) = n/M = 0.093/0.332 = 0.28 L

                          = 280 ml