19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1C.Calculate the van't Hoff tion factor and dissocition constant of fluoroacetic acid.

Asked by prashant.jain | 2nd Sep, 2011, 10:03: PM

Expert Answer:

Number of moles of CH2FCOOH = 19.5 /78 =0.25

Molality (m) = (0.25 x 1000)/500 = 0.50 mol/Kg

 

Now,                Tf =Kf x m   =1.86 x 0.50

                         =0.93 K

Van't Hoff factor = (observed freezing point depression /calculated frezing point depression)

   = 1/0.93  =1.0753

CH2FCOOH dissociates as:- CH2FCOOH  CH2FCOO-  + H+

For dissociation, = (i-1) /( n-1)

Here i=1.0753 and n=2

= 1.0753-1 = 0.0753

Therefore, Ka = [CH2FCOO-] [H+] / [CH2FCOOH]

                   = C2 /(1-)

                  = 0.50 x (0.0753)2 / 0.9247 =3.07 x 10-3

Answered by  | 7th Sep, 2011, 03:46: PM

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