Question

Chapter 5 (Simple Mixtures Place Print Name: Ebony Thornton 1. Given that p H ,0) 0.02308 atm and p(H,0) 0.02239 atm in a solution in which 0.122 kg of a non-volatile solute (M-241 g mol ) is dissolved in 0.920 kg water at 293 K, calculate the activity and activity coefficient of water in the solution.

Answer 1

Cible : p*(4,0) = 0.02308 atm + CH,0) = 0.02239 etmi wit of non volatile solut = 0.122 kg = 122g. - = 2419 mot soluti) wat. of H2O = 0.920 kg - 9209 - M- 18.019/mud (nee) To calculate Activity of the - Lucas ho = 0.02239 atm P&O 0.023 of atm o who = 0.970 who= 0.970 Activity coefficient in To calculate activity - Coefficient we have to calculate male fraction of to XA NHO Mazot salute S 720/8.02 g/mol 9207 + 1229 18.02g/ml 241g/mol

2 0.99 51.054 51.054 + 0.506 Activity coefficient = WHO Xnzo 0.970 = 0.979 0.99 Actunity coefficient = 0.919 or 0.98