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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The freezing point (in$^{\large\circ}C)$ of a solution containing 0.1g of $K_3[Fe(CN)_6]$ (mol .wt .329) in 100g of water $(K_t=1.86Kkgmol^{-1})$

$\begin{array}{1 1}(a)\;-5.7\times 10^{-2}\\(b)\;-5.7\times 10^{-3}\\(c)\;-1.2\times 10^{-2}\\(d)\;-2.3\times 10^{-2}\end{array}$

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Answer: -0.0226 $^{\circ}$C
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.
$\Rightarrow$ Molarity $= \large\frac{W}{M} $$ \times\large\frac {1000}{V\;mL}$$ = \large\frac{0.1}{329}$$\times \large\frac{100}{1000}$$ =0.003039$
$\Rightarrow \Delta T_f = 4 \times 1.86 \times 0.003039 = 0.02261$ which is equal to $T_0 - T_f$
Since, $T_0 = 273^{\circ}K = 0^{\circ}C \rightarrow T_f = -0.0226^{\circ}C$
answered Feb 13, 2014 by sreemathi.v
edited Jul 17, 2014 by balaji.thirumalai
 

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