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19.5 g of $CH_2FCOOH$ is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00$^{\large\circ}$ C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

$\begin{array}{1 1}1.0753,3.07\times 10^{-3}\\2.0753,2.07\times 10^{-3}\\0.0753,0.07\times 10^{-3}\\4.0753,4.07\times 10^{-3}\end{array} $

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$W_2=19.5g$
$W_1=500g$
$\Delta T_f$ (observed) =$1.0^{\large\circ}C$
$K_f$ for water =$1.86Kkg mol^{-1}$
$M_2$(observed)=$\large\frac{1000 K_f W_2}{W_1 \Delta T_f}$
$\Rightarrow \large\frac{1000\times 1.86\times 19.5}{500\times 1.0}$
$\Rightarrow 72.54 gmol^{-1}$
$M_2$(calculated)=$CH_2FCOOH$
$\Rightarrow 12+2+19+12+32+1$
$\Rightarrow 78gmol^{-1}$
Van't Hoff factor (i)=$\large\frac{M_2(calculated)}{M_1(observed)}$
$\Rightarrow \large\frac{78}{72.54}$$=1.0753$
Calculation of dissociation constant : Suppose degree of dissociation at the given concentration is $x$
$CH_2FCOOH\leftrightharpoons CH_2FCOO^-+H^+$
$CH_2FCOOH \Rightarrow n(1-x)$
$CH_2FCOO^-\Rightarrow nx$
$H^+\Rightarrow nx$
Total number of moles =$n(1-x+x+x)=n(1+x)$
$x=i-1$
$\;\;\;=1.0753-1=0.0753$
$K_a=\large\frac{[CH_2F COO^-][H^+]}{[CH_2FCOOH]}$
$\Rightarrow \large\frac{nx\times nx}{n(1-x)}$
$\Rightarrow \large\frac{nx^2}{1-x}$
$n=\large\frac{19.5}{78}\times \frac{1}{500}\times $$1000=0.5M$
$K_a=\large\frac{(0.5)(0.0753)^2}{1-0.0753}$
$\Rightarrow 3.07\times 10^{-3}$
answered Aug 5, 2014 by sreemathi.v
 

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