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 vant Hoff factor and dissociation constant of fluoroacetic acid.

Given- w1 = 500g


W2 = 19.5g


Kf = 1.86 K kg mol-1


Molar mass of CH2FCOOH = 12 + 2 + 19 + 12 + 16 + 16 + 1


= 78 g mol-1


The depression in freezing point is calculated by,


→ ff (where, m is the molality)





∴ f (calculated) = 0.93


To find out the vant Hoff’s factor, we use the formula,


→



∴


CH2FCOOH → CH2FCOO- + H +


To find out the degree of dissociation α, we use


→


∴


With degree of dissociation, known to us, we can easily calculate the dissociation constant, Ka using the formula,




∴


Thus, the vant Hoff’s factor is 1.07 an the dissociation constant is 2.634x10-3


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