Question

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°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer 1

_____________________________________________________________
W₁ = 500 g 
W₂ = 19.5 g 
K(f) = 1.86 K kg mol⁻¹
ΔT(f) = 1K

As we know ,

$M_2 =\frac{K_f . W_2 . 1000}{/\ T_f . W_1}$

$\frac{1.86 X 19.5 X 1000}{500 X 1 }$

72.54 g mol⁻¹

I.e.       Observed mass of CH₂FCOOH is  72.4 g mol⁻¹

Calculated molar mass of CH₂FCOOH is = 14 + 19 + 12 + 16 + 16 + 1 
                                                                  = 78 g mol⁻¹

Van't Hoff Factor 

$i = \frac{M_2_(cal)}{M_2_(obs)}$

78 / 72 . 54 = 1.0753

____________________________________________________________
Let α Be the degree of disassociation 

CH₂FCOOH            ⇄     CH₂FCOO⁻    +       H⁺

Initial concentration             0                          0

At Equilibrium                      Cα                       Cα

Total = C(1+α)

∴ $i = \frac{C(1+a)}{C}$

i = 1 + α 

α = i - 1 

   = 1.0753 - 1 
    
   = 0.0753

Value of K(α) is given is 

$K_a= \frac{[CH_2FCOOH^-][H^+]}{CH_2FCOOH}$

$K_a = \frac{Ca . Ca }{C(1-a)}$

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

By Volume of the solution  as 500 mL 

The concentration will be :

$C = \frac{ \frac{19.5}{78} }{500} X1000M$
   
          = 0.5 M 

Therefore, 

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

$K_a = \frac{0.5 X (0.0753)^2}{1-0.0753}$

K(α)       = 0.00307 (aprox.)
  
              = 3. 07 X 10⁻³
____________________________________________________________

Answer 2

Explanation:

see the attachment...answer is there..