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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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Depression in freezing points of $0.1\;M$ aqueous solution of $HCl,CuSO_4$ and $K_2SO_4$ are

$\begin{array}{1 1}(a)\;\text{in the ratio of 1 : 2 : 3}\\(b)\;\text{in the ratio of 2 : 4 : 4}\\(c)\;\text{in the ratio of 2 : 2 : 3}\\(d)\;\text{equal}\end{array}$

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Answer: $(C)\;\text{2 : 2 : 3}$
Depression of Freezing point, $\Delta\; T_f = i\;K_f\; m$, where $i$ is the van't Hoff factor, $K_f$ is the Freezing Point Depression Constant and $m$ is the molality.
$\Rightarrow \Delta T_f\propto i$, the van't Hoff Factor.
$i=2$ for $HCl$, as it will give 2 ions $H^{+}$ and $Cl^{-}$.
$i=2$ for $CuSO_4$ as it will also give 2 ions, $Cu^{2+}$ and $SO_4^{2-}$.
$i=3$ for $K_2SO_4$, as it will give 3 ions, two $K^{+}$ ions and one $SO_4^{2+}$ ion.
answered Feb 17, 2014 by sreemathi.v
edited Jul 15, 2014 by balaji.thirumalai
 

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