Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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19.5 g of CH2FCOOH 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. - Chemistry

19.5 g of CH2FCOOH 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.

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Solution

It is given that:

w1 = 500 g

w2 = 19.5 g

Kf = 1.86 Kkg mol-1 

ΔTf = 1K

We know that:

`M_2 = (K_fxxw_2xx1000)/(triangleT_fxxw_1)`

`= (1.86 "K kg mol"^(-1) xx 19.5 g xx 1000 "g kg" ^(-1))/(500g xx 1K)`

= 72.54 gmol-1

Therefore, observed molar mass of  CH2FCOOH , (M2)obs = 72.54 g mol

The calculated molar mass of CH2FCOOH is 

(M2)cal = 14 + 19 + 12 + 16 + 16 + 1

= 78 g mol-1

Therefore, van’t Hoff factor,` i = ((M_2)_(cal))/(M_2)_(obs)`

`= (78 gmol^(-1))/(72.54 g mol^(-1))`

= 1.0752

Let α be the degree of dissociation of CH2FCOOH 

                                  CH2FCOOH ↔ CH2FCOO- +H+

Initial Conc              Cmol L-1         0                0

At equilibrium         C(1-α)            Cα              Cα         Total = C(1+α)

`:.i = (C(1+alpha))/C`

`=> i = 1+alpha`

`=>alpha = i -1`

= 1.0753 -1

 = 0.0753

Now, the value of Ka is given as:

`K_alpha = ([CH_2FCOO^(-)][H^(+)]) /[CH_2FCOOH] `

` = (Calpha.Calpha)/(C(1-alpha))`

`= (Calpha^2)/(1-alpha)`

Taking the volume of the solution as 500 mL, we have the concentration:

`C = (19.5/78)/500 xx 1000M`

= 0.5M

Therefore, `K_alpha = (Calpha^2)/(1-alpha)`

`= (0.5xx (0.0753)^2)/(1-0.0753)`

`= (0.5xx0.00567)/0.9247`

= 0.00307 (approximately)

= 3.07 x  10-3

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 12 Chemistry Textbook
Chapter 2 Solutions
Q 33 | Page 61
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