Question

A solution is prepared by dissolving 25.00 g of acetic acid in 750.0 g of water. The density of the resulting solution is 1.105 g/ml.

How would I calculate the molality, of the solution, the mole fraction of acetic acid in the solution, and what is the concentration of acetic acid in ppm?

Answer 1

molality of a solution is defined as the number of moles of solute present in 1 Kg of solvent .

Thus molality (m) = number of moles / mass of solvent in Kg

moles of acetic acid = mass/ molar mass

= 25g/60 g/mol

and mass of water = 750g = 0.75 L

thus molaity (m ) = (25/60)/0.75

= 0.56

mole fraction of acetic acid in solution = moles of acetic acid / total moles of solution.

moles of acetic acid = 25/60 =0.4167

moles of water = 750/18= 41.67

Thus mole fraction = 0.4167 /(0.4167+41.67)

= 0.0099

Concentration in ppm

ppm is parts per million

that is how muc solute is preent in one million parts of solvent by weight.

750 g of water has 20 g of acetic acid

10⁶ g water has = 20x 10⁶/750

=2.6 x10⁴ ppm