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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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0.5g KCl was dissolved in 100g water and the solution originally at $20^{\large\circ}C$,froze at $-0.24^{\large\circ}C$.Calculate the percentage ionization of salt.$(K_f$ per 1000g of water =1.86K)

$\begin{array}{1 1}92\%\\82\%\\72\%\\62\%\end{array} $

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$\Delta T_f=0-(-0.24^{\large\circ}C)=0.24^{\large\circ}C$
$M_2=\large\frac{1000K_f W_2}{\Delta T_f W_1}$
$\;\;\;\;\;\;=\large\frac{1000\times 1.86\times 0.5}{0.24\times 100}$
$\;\;\;\;\;\;=38.75gmol^{-1}$
Theoretical molar mass of KCl =39+35.5
$\Rightarrow 74.5gmol^{-1}$
$i=\large\frac{\text{Calculated mol mass}}{\text{Theoretical mol mass}}$
$\;\;\;=\large\frac{74.5}{38.75}$$=1.92$
$KCl \leftrightharpoons K^++Cl^-$
$1\;mole\quad 0\quad 0$
$1-\alpha\quad\; \alpha\quad\;\alpha$
Total number of moles after dissociation =$1+\alpha$
$i=\large\frac{1+\alpha}{1}$
$i-1=\alpha=1.92-1$
$\Rightarrow 0.92$
Percentage dissociation =92%
answered Aug 8, 2014 by sreemathi.v
 

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