Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.19 m Na2SO4(aq). Constants may be found here.

Answer 1

Freezing point depresion = i*kf*m, m= molality, kf= Freezing point depression constant = 1.86 deg.c/m

i = Van't Hoff factor for Na2SO4 ( it dissociates into 2Na+ ions and 1 SO4-2 ions, i= 2+1=3

hence freeziing point depression = 3*1.86*2.19=12.2

Freezing point =Freezing point of water- Freezing point depression= 0-12.2 =-12.2 deg.c

Boiling point elevation = i*kb*m,kb= boiling point elevation constant =0.512deg,c/m

Hence Boiling point elevation = 3*2.19*0.512 =3.4

Boiling point = boiling point of water+ boiling point elevation = 100+3.4= 103.4 deg.c