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 ABNORMAL MOLAL MASS:
 IN CASE OF ASSOCIATION
 IN CASE OF DISSOCIATION
 Van’t Hoff factor ‘i’
Experimentally determined value would be less than theoretically expected & the
molar mass calculated much higher than expected.
Association of solute in solution:
(A)n⇆nA
The concentration of A(unassociated)molecule=C(1-α)
The concentration of A (associated)molecule =C α/n
∴Total concentration of solute particle =C (1-α)+ C α/n
=C(1- α+ α/n)
the vant hoff factor i is defind as,
i=
i= C(1- α+ α/n)
C
=1+ α(1/n -1)
i.e.
Experimentally determined value would be higher than theoretically
expected & the molar mass calculated much lower than expected.
Dissociation of solute in solution:
AxBy ⇆ xA + yB
Where z+ & z- are charges on cation and anion respectively.
Concentration of undissociated AxBy = (1-α)C
Concentration of dissociated ions = x Cα+y Cα
∴ total concentration of solute particles=C[(1-α)+xα+yα]
Vant hoff factor ‘i’
i= C[(1-α)+xα+yα]
C
i =1+α(x+y-1)
If x+y=1
i=1+α(n-1)
z+ z-
∴ α = i-1
n-1
molar mass.pptx

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molar mass.pptx

  • 1.
  • 2.  ABNORMAL MOLAL MASS:  IN CASE OF ASSOCIATION  IN CASE OF DISSOCIATION  Van’t Hoff factor ‘i’
  • 3. Experimentally determined value would be less than theoretically expected & the molar mass calculated much higher than expected. Association of solute in solution: (A)n⇆nA The concentration of A(unassociated)molecule=C(1-α) The concentration of A (associated)molecule =C α/n ∴Total concentration of solute particle =C (1-α)+ C α/n =C(1- α+ α/n) the vant hoff factor i is defind as, i= i= C(1- α+ α/n) C =1+ α(1/n -1) i.e.
  • 4. Experimentally determined value would be higher than theoretically expected & the molar mass calculated much lower than expected. Dissociation of solute in solution: AxBy ⇆ xA + yB Where z+ & z- are charges on cation and anion respectively. Concentration of undissociated AxBy = (1-α)C Concentration of dissociated ions = x Cα+y Cα ∴ total concentration of solute particles=C[(1-α)+xα+yα] Vant hoff factor ‘i’ i= C[(1-α)+xα+yα] C i =1+α(x+y-1) If x+y=1 i=1+α(n-1) z+ z- ∴ α = i-1 n-1