19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Solution:

Given values in problem 

Use the formula of molar mass, to get observed mass we get 

After calculation we get observed mass

M₂ = 72.54 g mol ^-1

Real molar mass of CH₂FCOOH  = 14 + 2+17 ++12 +16+16+1 = 78 g mol ^{– 1}

Hence, van’t Hoff factor, I  =  M2 (calculated)/ M2 (Observed)

                             = 78/72.54   = 1.0753

Let α is the degree of dissociation of CH₂FCOOH

 

Number of mole of CH₂FCOOH 

Number of moles =19.5/78 = 0.25 moles

Volume of CH₂FCOOH in liter  = 500 / 1000 = 0.5 liter

 Now use the formula 

Plug the value we get 

(our molarity is C )

C = 0.25/05  = 0.5 M

 

 

Solutions

K_a  = 0.00307  = 3.07 × 10 ^-3

I  =  0.0753