Question

4. At 25 °C a 0.1 molal solution of CH3COOH
is 1.35 % dissociated in an aqueous
solution. Calculate freezing point and
osmotic pressure of the solution assuming
molality and molarity to be identical.

​

Answer 1

Answer:

Molality =0.1 m

Degree of dissociation of CH3COOH=1.35%

CH3COOH⇋ CH3COO−+H+ ;       α%=1.35%

100                       0                           0

100−1.35          1.35                  1.35

∴i=100100−1.35+2×1.35 =100101.35=1.0135

∴ΔTf=1.0135×1.86×0.1=0.1885

∴Tfsolution=−0.1885oC

π=iCRT=1.0135×0.1×0.0821×298=2.48 atm

If no dissociation is assumed i=1

∴ Both △Tf & π are little bit higher if i=1.0135 is considered