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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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Calculate the depression in the freezing point of water when 10 g of $CH_3CH_2CHClCOOH$ is added to 250 g of water. $K_a = 1.4 \times 10^{–3}, K_f = 1.86 K kg mol^{–1}$.

$\begin{array}{1 1}0.65^{\large\circ}\\0.75^{\large\circ}\\0.55^{\large\circ}\\0.45^{\large\circ}\end{array} $

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Molar mass of $CH_3CH_2CHClCOOH=12+3+12+2+12+1+35.5+12+32+1$
$\Rightarrow 122.5g mol^{-1}$
Number of moles of $CH_3CH_2CHClCOOH =\large\frac{10}{122.5}$
$\Rightarrow 8.16\times 10^{-2}$ mole
$\therefore$ Molality of the solution (m)
$m=\large\frac{8.16\times 10^{-2}mol}{250}$$\times 1000g kg ^{-1}$
$\;\;\;\;=0.3264$
If $x$ is the degree of dissociation of $CH_3CH_2CHClCOOH$,then
$CH_3CH_2CHClCOOH\leftrightharpoons CH_3CH_2CHClCOO^-+H^+$
To calculate Van't Hoff factor
Total moles of particles =$n[1-x+x-x]$
$\Rightarrow n[1+x]$
$i=\large\frac{n(1-x)}{n}$$=1+x$
We know that
$x=\sqrt{\large\frac{K_a}{x}}$
$\;\;=\sqrt{\large\frac{1.4\times 10^{-3}}{0.3264}}$$=0.065$
$i=1+0.065=1.065$
$\Delta T_f=i K_f m$
$\Rightarrow 1.065\times 1.86 \times 0.3264$
$\Rightarrow 0.65^{\large\circ}$
answered Aug 5, 2014 by sreemathi.v
 
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