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Assuming $ 100\%$ ionization , the increasing order of the freezing point of the solution will be

$\begin{array}{1 1} 0.10\;mol\;kg^{-1}\;Ba_3(PO_4)_2 < 0.10\;mol\;kg^{-1} Na_2SO_4 < 0.10 \;mol\;kg^{-1} KCl\\0.10\;mol\;kg^{-1}\;KCl < 0.10\;mol\;kg^{-1} Na_2SO_4 <0.10 \;mol \;kg^{-1} Ba_3 (PO_4)_2\\ 0.10\;mol\;kg^{-1}\;Na_2(SO_4) < 0.10\;mol\;kg^{-1} Ba_3(PO_4)_2 < 0.10 mol\;kg^{-1} KCl \\ 0.10\;mol\;kg^{-1}\;KCl < 0.10\;mol\;kg^{-1} Ba_3(PO_4)_2 < 0.10 \;mol\;kg^{-1} Na_2SO_4 \end{array} $</span>

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Answer : $ 0.10\;mol\;kg^{-1}\;Ba_3(PO_4)_2 < 0.10\;mol\;kg^{-1} Na_2SO_4 < 0.10 \;mol\;kg^{-1} KCl$
Assuming complete dissociation $(- \Delta T_t) _{Ba_3(PO_4)_2} > (- \Delta T_t)_{Na_2SO_4} > (- \Delta T_t ) _{KCL}$
Hence $(T_t)_{Ba_3(PO_4)_2} < (T_t )Na_2SO_4 < (T_t)_{KCl}$
answered Aug 5, 2014 by meena.p
 

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