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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. $K_a = 1.4 \times 10^{–3}, K_f = 1.86 K kg mol^{–1}$.

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

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$
$i=1+0.065=1.065$
$\Delta T_f=i K_f m$
$\Rightarrow 1.065\times 1.86 \times 0.3264$
$\Rightarrow 0.65^{\large\circ}$