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

K_f = 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, K_a using the formula,

∴

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