2.33    19.5 \; g of CH_{2}FCOOH is dissolved in 500\; g  of water. The depression in the freezing point of water observed is 1.0^{\circ}C. Calculate the van’t  Hoff factor and dissociation constant of fluoroacetic acid.

Answers (1)

Firstly we need to calculate molality in order to get vant's hoff factor.

So moles of CH2FCOOH :

                                                \frac{19.5}{78} = 0.25

We need to assume volume of solution to be nearly equal to 500 mL.         (as 500 g water is present)

Now, we know that :            \Delta T_f = i\ K_f\ m

or                                           i = \frac{1}{0.93} = 1.0753

Now for dissociation constant :- 

                                                    a = i - 1 = 1.0753 - 1 = 0.0753

 and,                                            K_a = \frac{Ca^2}{1-a}

Put values of C and a in the above equation, we get :

                                                 K_a = 3\times10^{-3}      

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