Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
$\Delta T_f$ (observed) =$1.0^{\large\circ}C$
$K_f$ for water =$1.86Kkg mol^{-1}$
$M_2$(observed)=$\large\frac{1000 K_f W_2}{W_1 \Delta T_f}$
$\Rightarrow \large\frac{1000\times 1.86\times 19.5}{500\times 1.0}$
$\Rightarrow 72.54 gmol^{-1}$
$\Rightarrow 12+2+19+12+32+1$
$\Rightarrow 78gmol^{-1}$
Van't Hoff factor (i)=$\large\frac{M_2(calculated)}{M_1(observed)}$
$\Rightarrow \large\frac{78}{72.54}$$=1.0753$
Calculation of dissociation constant : Suppose degree of dissociation at the given concentration is $x$
$CH_2FCOOH\leftrightharpoons CH_2FCOO^-+H^+$
$CH_2FCOOH \Rightarrow n(1-x)$
$CH_2FCOO^-\Rightarrow nx$
$H^+\Rightarrow nx$
Total number of moles =$n(1-x+x+x)=n(1+x)$
$K_a=\large\frac{[CH_2F COO^-][H^+]}{[CH_2FCOOH]}$
$\Rightarrow \large\frac{nx\times nx}{n(1-x)}$
$\Rightarrow \large\frac{nx^2}{1-x}$
$n=\large\frac{19.5}{78}\times \frac{1}{500}\times $$1000=0.5M$
$\Rightarrow 3.07\times 10^{-3}$
answered Aug 5, 2014 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App