physical chemistry

Question

A solution was prepared by dissolving 25 cm3 of ethanol (C2H5OH, mw = 46.0, d = 0.786 g/cm3) in water to give a solution of final volume 200 cm3 (Solution A).

The density of the Solution A is 0.952 g/cm3.

i) Calculate the molarity, molality, wt/wt percent and the concentration in ppm of ethanol in that solution.

Answer 1

Solution A

25 mls C₂H₅OH x 0.786 g / ml = 19.65 g C₂H₅OH

moles C₂H₅OH = 19.65 g x 1 mol / 46.0 g = 0.427 moles C₂H₅OH

Volume = 200 ml = 0.200 L

(i) Molarity of C₂H₅OH = moles / L = 0.427 mols / 0.200 L = 2.14 M

(ii) Molality of C₂H₅OH = mols C₂H₅OH / kg water

mass of solution A = 200 ml x 0.952 g / ml = 190 g

mass of water = 190 g - mass of C₂H₅OH = 190 g - 19.65 = 171 g H2O = 0.171 kg H₂O

Molality of C₂H₅OH = 0.427 mols / 0.171 kg = 2.50 m

(iii) wt/wt % of C₂H₅OH = mass C₂H₅OH / mass solution (x100%) = 19.65 g / 190 g (x100%) = 10.3%

(iv) ppm C₂H₅OH = 19.65 g / 200 ml = 0.0983 g / ml x 1000 ml / L x 1000 mg / g = 98300 mg /L = 98300 ppm