Question

Express the concentration of a 0.0400M aqueous solution of fluride,F-, in mass percentage and in parts per million.Assume the density of the solution is 1.00f/mL.

Answer 1

Molar mass of fluoride is 19g/mol

Given concentration =0.0400 M

                             = 0.0400 mol /L

We know that number of moles,n = mass/molarmass

So mass of fluoride,m = 0.0400 mol x 19 g/mol

                                = 0.76 g

So 0.76 g of F^- Present in 1L=1000 mL of solution

                                          = 1000 mL x 1.00 g/mL

                                          = 1000 g

So mass percentage = mass of F^- / mass of solution

                                = 0.76 g / 1000 g

                                = 0.00076 %

0.76 g present in 1000 mL = 0.76 g/1L

                                       = 760 mg/1L

                                      = 760 ppm                            since 1ppm = mg/L