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Question 33

19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

Answer

Molecular mass of CH2FCOOH

14 + 19 + 12 + 16 + 16 + 1 = 78 g/mol

Now, Moles of CH2FCOOH = 19.5 / 78

= 0.25

Taking the volume of the solution as 500 mL, we have the concentration:

C = (0.25 / 500) X 1000

Therefore Molality =  0.50m

So now putting the value in the formula :

ΔTf = Kf x m

=1.86 x 0.50 = 0.93K

Van’t hoff factor = observed freezing point depression / calculated freezing point depression

= 1 / 0.93 = 1.0753

 

Let α be the degree of dissociation of CH2FCOOH

Now total number of moles  = m(1-a) + ma +ma = m(1+a)

Or

i = α(1+α) / α = 1 +α  = 1.0753

Therefore α = 1.0753- 1

= 0.0753

Now the Value of Ka is given as:

K= [CH2FCOO-][H+/  CH2FCOOH

(Cα x Cα) / (C (1-α))

= Cα2 / (1-α)

Ka = 0.5 X (0.0753)2  (1-0.0753)

= 0.5 X 0.00567 / 0.09247

= 0.00307 (approx.)

= 3 X 10-3

 

 

 

 

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