Question

Ethanol is a commonly used for human in alcoholic human beverages in dilute concentrations ranging from 3-40%. An aqueous solution of ethanol (C2H6O) is 24.0% by mass and has a density of 0.84g/mL. Calculate its molarity, molality, and mole fraction of ethanol

Answer 1

The aqueous solution contains 24% ethanol by mass.

Hence, in 100 g solution, 24 g ethanol and 76 g water are present.

Now, density of ethanol = 0.84 g/ mL

Hence, 24 g ethanol = 24/0.84 mL = 28.57 mL

Now, 76 g water = 76 mL

Total volume of the solution = (76 + 28.57) mL

                                               = 104.57 mL

Now, in 104.57 mL solution 28.57 mL ethanol is present.

Hence, in 1000 mL solution = (28.57 x 1000)/104.57 mL ethanol

                                              = 273.21 mL ethanol

                                             = (273.21 x 0.84 ) g

                                              = 229.49 g

                                              = 229.49/46 mol (Molar weight of ethanol = 46 g/mol)

= 4.989 mol

Hence, the molarity of the solution is 4.989.

Now,    76 g water contains 24 g ethanol

Hence, 1000 g water contains = (24 x 1000)/76 g ethanol

                                                = 315.78 g ethanol

                                                = 6.86 mol

Hence, molality of the solution is 6.86.

Now, Aqueous solution contains 24% ethanol by mass.

Hence, in 100 g:

Ethanol = 24 g

             = 24/46 mol (Molar weight of ethanol = 46 g/mol)

            = 0.52 mol

Water = 100-24 = 76 g

          = 76/18 mol (Molar weight of water = 18 g/mol)

         = 4.22 mol

Hence, mole fraction of ethanol = 0.52/(0.52+4.22)

                                                   = 0.109