Question

the density of a 6.27 M aqueous acetic acid solution is 1.045 g/ml. determine the molarity of this solution and mole fraction of acetic acid

Answer 1

the weight of the acetic acid

   = 6.27 X 60

    =376.2 g

The weight of the solution

   =1000 X 1.045

    =1045

         Weight of the water =(1045-376.2)

                                           =668.8g

density of water is 0.9982 g/ml

so, vol of water is 668.8/0.9982 =670 ml

so molarity of the solution is (376.2/60) x (1000/670) =9.35 M

moles of acetic acid 376.2/60 = 6.27 mol

moles of water 668.8/18 = 37.15 mol

therefore mole fraction of acetic acid is

moles of acetic acid /( moles of acetic acid + moles of water)

6.27/(6.27+37.15) =0.144